Algebraic Structures
Algebra begins when a bare set is equipped with a rule for combining its elements. This chapter builds that idea up axiom by axiom: binary operations and closure first, then associativity, neutral and inverse elements, until the simple structures (semigroups, monoids, groups) and the composite ones (rings, fields, modules, vector spaces, algebras) emerge as successive strengthenings of the same scheme. The second half of the chapter studies groups in detail — the structure that physics uses to express symmetry — including their basic examples, the maps that preserve them, the symmetric group, the decompositions of a group into conjugacy classes and cosets, the quotient group, and the direct product.
Binary operations and algebraic structures
Binary operations
Consider two sets \(A\) and \(B\). We may ask whether, starting from \(A\) and \(B\), it is possible to define a set containing all the possible combinations of two elements of \(A\) and of \(B\) respectively. For example, let \(A=\set{a,b}\) and \(B=\set{\alpha,\beta}\). Is the collection of the pairings \(\left(a,\alpha\right)\), \(\left(a,\beta\right)\), \(\left(b,\alpha\right)\) and \(\left(b,\beta\right)\) then a set? The answer is yes, owing to the arbitrariness in the definition of a set (Definition 3.28). According to the historical account, René Descartes was the first to think of ordering pairs of numbers in this way.
The Cartesian product of two sets \(A\) and \(B\), denoted \(A\times B\), is the set containing as elements all the possible pairs of combinations of elements of \(A\) and \(B\). The elements of \(A\times B\) are called \(2\)-tuples. The generalization is immediate: the elements of the Cartesian product \(A_1\times\ldots\times A_n\) are called \(n\)-tuples. Rests on Definition 3.28.
As an example, note that it is also possible to construct the Cartesian product of a set with itself,
where it must be noted that \(\left(a,b\right)\neq\left(b,a\right)\), since they are distinct combinations.
Consider the sets \(A\) and \(X\). A binary operation \(\ast\) is a map of the form
Consider a set \(A\) and an arbitrary set \(X\), together with a binary operation \(\ast:A\times A\longrightarrow X\). The set \(A\) together with the operation \(\ast\) is called an algebraic structure, and is denoted \((A,\ast)\). Rests on Definition 4.2.
For example, the set \(\set{0,1}\subset\Z\) together with ordinary addition,
forms an algebraic structure whose binary operation is made explicit by the table
The closure axiom
Consider an algebraic structure \((A,\ast)\). If the binary operation has not previously been fixed explicitly, in general \(a\ast b\) need not be an element of \(A\). The current theories of physics, however, rest on a mathematical formalism in which sets lacking such an internal mechanism are not of interest, and we therefore present the formalism only for algebraic structures that possess one.
A binary operation \(\ast:A\times A\longrightarrow X\) is an internal binary operation (IBO) if and only if \(X=A\). An algebraic structure \((A,\ast)\) whose operation \(\ast\) is an IBO is said to be closed, to be a magma, or to have closure with respect to \(\ast\). Closure is expressed formally by the axiom
When no ambiguity is possible or relevant, we shall refer to a closed algebraic structure simply as an algebraic structure, and we shall denote an algebraic structure \((A,\ast)\) simply by \(A\), the name of its underlying set.
The associativity axiom
Just as happens with the real numbers, we can formalize the fact that the product of one number with the product of two others equals the product of the first two with the last.
An algebraic structure \((A,\ast)\) satisfies the property of associativity if and only if
Rests on Definition 4.3.
The associativity axiom is genuinely important because of the following property. Consider the equation \(a=b\ast c\). With associativity available we may operate on the whole equation with a further quantity \(d\); that is, \(a\ast d=(b\ast c)\ast d\), or equivalently \(a\ast d=b\ast(c\ast d)\). If we did not have that property we would be left with something like \(a\ast d=b\ast c\ast d\), an expression devoid of meaning, since we have not defined \(\ast\) as a “ternary operation”.
The commutativity axiom
For the real numbers, the order of the factors does not alter the product. We can formalize this for algebraic structures through the following axiom.
An IBO \(\ast\) is commutative — equivalently, an algebraic structure \((A,\ast)\) is abelian with respect to \(\ast\) — if and only if the following axiom holds:
Rests on Definition 4.4.
The neutral-element axiom and uniqueness
In order to construct more sophisticated structures, we may demand the existence of an element \(e_A\) whose operation with any element is irrelevant.
An element \(e_A\) of an algebraic structure \((A,\ast)\) is a neutral element if and only if it satisfies the axiom
Rests on Definition 4.3.
We must observe that the axiom does not assert commutativity among all the elements of the set: commutation is demanded only for the neutral element. Many of the structures we shall define make use of this axiom, which will let us uncover properties usable throughout mathematical physics. On occasion, when we wish to specify the symbol employed for the neutral element, we shall denote the algebraic structure by \((A,\ast,e_A)\).
The neutral element of an algebraic structure is unique. Rests on Definition 4.9.
Derives Proposition 4.10. Consider two neutral elements \(e_A\) and \(f_A\). By the axiom we have \(e_A=e_A\ast f_A\), and furthermore
hence \(e_A=f_A\).
∎The inverse-element axiom
Let \((A,\ast,e_A)\) be an algebraic structure equipped with the neutral-element axiom. A further axiom we may impose is that to each element \(a\) there be assigned a dual element \(a^{-1}\) such that, when the two are operated, the neutral element results.
An algebraic structure \((A,\ast,e_A)\) satisfies the inverse-element axiom if and only if
and we call \(a^{-1}\) an inverse element of \(a\). Rests on Definition 4.9.
Substructures
Consider a closed algebraic structure \((A,\ast)\), and let \(S\subseteq A\) be a non-empty subset. The IBO \(\ast\) induces a new binary operation \(\ast_S: S\times S\longrightarrow A\). Then \((S,\ast_S)\) is an algebraic structure, but not necessarily a closed one. If, however, we restrict the codomain, the operation \(\ast'_S: S\times S\longrightarrow S\) is by definition an IBO, and \((S,\ast'_S)\) is a closed algebraic structure. When no ambiguity is possible we shall simply write \(\ast'_S=\ast\).
\((S,\ast)\) is a substructure of \((A,\ast)\) if and only if
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\(S\neq\varnothing\);
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\(\forall\ a,b\in S,\ a\ast b\in S\).
We further introduce the notation
Rests on Definition 4.4.
Maps between algebraic structures
Product of maps
Consider the algebraic structures \((A,\ast_A)\) and \((B,\ast_B)\). As we know (Definition 3.52), \(\mathcal{F}(A,B)\) is the set of all maps from \(A\) to \(B\). Let us define the IBO
Given the maps \(f\) and \(g\), we have thereby defined a new map \(f\ast g\in\mathcal{F}(A,B)\), which we shall call the product of maps.
Neutral map
Let \((A,\ast_A)\) be an algebraic structure and \((B,\ast_B,e_B)\) an algebraic structure satisfying the neutral-element axiom. Consider a map of the form
that is, a map assigning to every element of \(A\) the neutral element of \(B\). We shall call this map the neutral map, neutral function, or neutral operator. Now consider a map \(f:A\longrightarrow B\). We see that
so, since \(a\) is arbitrary, we simply have \(f\ast\Theta=f\), and analogously \(\Theta\ast f=f\).
Opposite map
The previous two headings supplied \(\mathcal{F}(A,B)\) with an operation and with a neutral element for it. One ingredient is missing before the set of maps is a group in its own right, and it is the pointwise inverse.
Let \((A,\ast_A)\) be an algebraic structure and let \((B,\ast_B,e_B)\) satisfy both the neutral-element axiom and the inverse-element axiom Equation (4.5). Given \(f:A\longrightarrow B\), define
which we call the opposite map of \(f\): it inverts the value of \(f\) at each point separately, and does nothing else.
The superscript is \(\mathrm{op}\), and not \(-1\), on purpose. The symbol \(f^{-1}\) is already spoken for twice over in Logic, Sets, and Maps: it is the inverse map of a bijection (Definition 3.51), which exists only when \(f\) is bijective and satisfies \(f^{-1}\circ f=\id_A\); and it is the preimage (Definition 3.53), which returns a subset. The opposite map Equation (4.6) is neither. It is defined for every \(f\), bijective or not, it involves no composition, and the inversion happens in the codomain \(B\) rather than between \(A\) and \(B\). The three notions coincide in no interesting case, and conflating them is the commonest slip in this part of the subject. Rests on Equation (4.6), Definition 3.51 and Definition 3.53.
For \(f^{\mathrm{op}}\) to be a map at all, the value \(\left(f(a)\right)^{-1}\) must be determined by \(f(a)\) — the existence and uniqueness demanded by Definition 3.43. Existence is the inverse-element axiom. Uniqueness is not automatic: Equation (4.5) asserts that an inverse exists, not that it is unique, and the proof of uniqueness (Proposition 4.23) needs associativity. We therefore assume throughout this heading that \((B,\ast_B)\) is associative — in practice that \(B\) is a group, Definition 4.21 — and note that without it Equation (4.6) selects one inverse among possibly several and is a definition only after such a selection has been made.
Granting that, the opposite map does for \(\mathcal{F}(A,B)\) exactly what \(a^{-1}\) does for \(B\):
and since \(a\in A\) was arbitrary, \(f\ast f^{\mathrm{op}}=\Theta\). The same computation with the factors exchanged gives \(f^{\mathrm{op}}\ast f=\Theta\).
Let \(A\) be a non-empty set and \((B,\ast_B,e_B)\) a group. Then \((\mathcal{F}(A,B),\ast)\) is a group, with neutral element the neutral map \(\Theta\) and with the inverse of \(f\) given by \(f^{\mathrm{op}}\). If \(B\) is abelian, so is \(\mathcal{F}(A,B)\). Rests on Definition 4.21, Equation (4.6) and Proposition 4.23.
Derives Proposition 4.14. Each group axiom is checked pointwise, which is available because two maps are equal exactly when they agree at every point.
Closure. \(f\ast g\) was defined as a map \(A\longrightarrow B\), so it lies in \(\mathcal{F}(A,B)\).
Associativity. For every \(a\in A\),
the middle step being associativity in \(B\).
Neutral element. \(f\ast\Theta=\Theta\ast f=f\) was shown under the previous heading.
Inverses. \(f\ast f^{\mathrm{op}}=f^{\mathrm{op}}\ast f=\Theta\) was shown above, and \(f^{\mathrm{op}}\) is well defined because inverses in a group are unique by Proposition 4.23.
Commutativity. If \(B\) is abelian then \((f\ast g)(a)=f(a)\ast_B g(a)=g(a)\ast_B f(a)=(g\ast f)(a)\) for every \(a\), so \(f\ast g=g\ast f\).
∎This small proposition is used more often in physics than its size suggests. It is the reason that the \(B\)-valued functions on any set inherit \(B\)'s algebraic structure without further argument: real-valued functions on a region form an abelian group under pointwise addition, and with the scalar multiplication of Linear Algebra and Representation Theory a vector space — which is what makes function spaces the natural home of fields (Hilbert Spaces). When \(B\) is a Lie group rather than \(\R\), the same pointwise construction produces the group of gauge transformations, the maps from spacetime into the gauge group, whose structure is the subject of Lie Groups, Lie Algebras, and Fibre Bundles and whose physical content occupies Generalized Classical Field Theory. Nothing in the construction requires \(A\) to carry any structure at all; only the codomain does the work.
Kernel and image
Let \((A,\ast_A)\) be an algebraic structure and \((B,\ast_B,e_B)\) an algebraic structure satisfying the neutral-element axiom, and consider a map \(f:A\longrightarrow B\). The kernel of \(f\) is the set
Rests on Definition 4.9.
Let \((A,\ast_A)\) and \((B,\ast_B)\) be algebraic structures and \(f:A\longrightarrow B\) a map. The image of \(f\) is the set
As a set this is Definition 3.44, the image of a map; it is restated here only to fix the notation \(\im{f}\) used throughout this chapter. Nothing about the structures enters the definition, and nothing is claimed by it — that the image of a homomorphism is itself a structure of the same kind is a separate statement and needs its own proof. Rests on Definitions 3.43 and 3.44.
Consider two algebraic structures satisfying the neutral-element axiom, \((A,\ast_A,e_A)\) and \((B,\ast_B,e_B)\), and let \(f:A\longrightarrow B\) be a homomorphism. Then
Derives Proposition 4.17.
(\(\Rightarrow\)) We must prove that if \(\ker{f}=\set{e_A}\) then \(f\) is injective. Indeed, since \(e_A\in\ker{f}\),
On the other hand, let \(a_1,a_2\in A\) and suppose that \(f(a_1)=f(a_2)\). Operating on both sides with \(\ast_B\, f(a^{-1}_{2})\),
so \(a_1\ast_Aa^{-1}_2\in\ker{f}\); hence \(a_1\ast_Aa^{-1}_2=e_A\), and therefore \(a_1=a_2\). This proves injectivity.
(\(\Leftarrow\)) Conversely, suppose \(f\) is injective. Because \(f\) is a homomorphism and \(e_A=e_A\ast_Ae_A\),
and operating on the right with \(\ast_B\,\left(f(e_A)\right)^{-1}\) leaves \(e_B=f(e_A)\) — the computation stated on its own account in Proposition 4.46. Hence \(e_A\in\ker{f}\), that is, \(\set{e_A}\subseteq\ker{f}\). Conversely, let \(a\in\ker{f}\). Then \(f(a)=e_B=f(e_A)\), and injectivity gives \(a=e_A\); thus \(\ker{f}\subseteq\set{e_A}\). The two inclusions together give \(\ker{f}=\set{e_A}\).
∎Simple algebraic structures
The conditions defining an algebraic structure form a purely abstract set of axioms. The axioms we now impose are chosen because the resulting structures are relevant to physics: they possess properties that we can link with nature.
Semigroups, monoids, and groups
An algebraic structure \((A,\ast)\) is a semigroup if and only if
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\(\forall\ a,b\in A,\quad a\ast b\in A\);
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\(\forall\ a,b,c\in A,\quad a\ast(b\ast c)=(a\ast b)\ast c\);
that is, \((A,\ast)\) is a magma whose elements moreover satisfy associativity under \(\ast\). Rests on Definitions 4.4 and 4.6.
For example, the set of natural numbers together with addition as internal binary operation forms a semigroup with infinitely many elements, the closure and associativity properties being clearly satisfied; this semigroup is moreover abelian.
The abelian semigroup formed by the set \(\set{0,1}\subset\Z\) together with the usual multiplication has the internal operation
Here closure is clear, and associativity follows[The reader is invited to carry out the proof of associativity explicitly. To do so one must verify the eight possibilities \((0\cdot0)\cdot0\stackrel{?}{=}0\cdot(0\cdot0)\), \((0\cdot0)\cdot1\stackrel{?}{=}0\cdot(0\cdot1)\), etc.] from the associativity of \(\Z\). Rests on Definition 4.18.
An algebraic structure \((A,\ast)\) is a monoid if and only if
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\(\forall\ a,b\in A,\quad a\ast b\in A\);
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\(\forall\ a,b,c\in A,\quad a\ast(b\ast c)=(a\ast b)\ast c\);
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\(\forall\ a\in A,\ \exists\ e_A \in A\mid a\ast e_A=a\quad\wedge\quad e_A\ast a=a\);
that is, \((A,\ast)\) is a semigroup possessing in addition a neutral element (which, as we have seen, is unique). Rests on Definitions 4.9 and 4.18.
Returning to the earlier example, \((\N,+)\) is not a monoid, since that set contains no element serving as a neutral element for addition. However, if we adjoin the number zero to the naturals, that very element fulfils the property with respect to addition, so \((\N\cup\set{0},+,0)\) is a monoid. Another example, easy to check, is the monoid \((\N,\cdot,1)\). Finally, the example of Equation (4.11) is furthermore an abelian monoid.
An algebraic structure \((A,\ast)\) is a group if and only if
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\(\forall\ a,b\in A,\quad a\ast b\in A\);
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\(\forall\ a,b,c\in A,\quad a\ast(b\ast c)=(a\ast b)\ast c\);
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\(\forall\ a\in A,\ \exists\ e_A \in A\mid a\ast e_A=a\quad\wedge\quad e_A\ast a=a\);
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\(\forall\ a\in A,\ \exists\ a^{-1} \in A\mid a\ast a^{-1}=e_A\quad\wedge\quad a^{-1}\ast a=e_A\);
that is, \((A,\ast)\) is a monoid in which every element of the set moreover possesses an inverse element (and since associativity holds by axiom, the inverse of each element is unique). Rests on Definitions 4.11 and 4.20.
As an example, upon adjoining the negative numbers to \((\N\cup\set{0},+,0)\) we obtain the group \((\Z,+,0)\): every element \(k\in\Z\) has its inverse element \(-k\), giving for each of them \(k+(-k)=-k+k=0\).
The set \(\set{1,-1}\subset\Z\) together with ordinary multiplication has the internal operation
Here closure and associativity are clear. The (as we know, unique) neutral element is the number \(1\), which is thus an inverse element for itself, since \(1\cdot1=1\). An inverse element for \(-1\) is \(-1\) itself, since \((-1)\cdot(-1)=1\). Rests on Definition 4.21.
Basic properties of groups
Since in a group we have associativity at our disposal, the following important properties valid in any group can be proved. Strictly speaking they hold in any structure with neutral element, inverses, and associative elements; but as we shall not encounter structures that fail to close, we state them here as properties of a group.
Given an element of a group \(A\), its inverse is unique. Rests on Definition 4.21.
Derives Proposition 4.23. Let \(a\in A\) and let \(a^{-1},\ \tilde{a}^{-1}\) be two inverse elements of \(a\). By the axiom, \(a\ast a^{-1}=e_A\) and \(a\ast \tilde{a}^{-1}=e_A\), so that, operating on the left with \(a^{-1}\ast\),
and thus \(a^{-1}=\tilde{a}^{-1}\).
∎The inverse of the neutral element is the neutral element itself. Rests on Definition 4.21 and Proposition 4.23.
Derives Proposition 4.24. By the neutral-element axiom we have \(e_A\ast e_A=e_A\). Operating on both sides with \(\ast\, e_A^{-1}\) we obtain \(e_A\ast(e_A\ast e_A^{-1})=e_A^{-1}\), that is, \(e_A=e_A^{-1}\). Since the inverse is unique, we conclude that the inverse of the neutral element is itself.
∎In a group,
and analogously
Rests on Definition 4.21.
Derives Proposition 4.25. Starting from \(a\ast b=e_A\) and operating on the right with \(\ast\,b^{-1}\),
and the converse and the second equivalence follow analogously.
∎The inverse of the inverse of an element is the element itself:
Rests on Proposition 4.25.
Derives Proposition 4.26. In the context of the preceding proposition, from Equation (4.14) we have
and substituting into Equation (4.13) we obtain \(\left(a^{-1}\right)^{-1}=a\).
∎Let \(a,b\in G\). The inverse of the product of \(a\) and \(b\) is
Rests on Definition 4.21 and Proposition 4.23.
Derives Proposition 4.27. We have
and analogously on the other side. Hence, since the inverse is unique, \(b^{-1}\ast a^{-1}\) is the inverse of \(a\ast b\).
∎Composite algebraic structures
Axioms for composite algebraic structures
Distributivity axioms
Let \((A,+)\) and \((V,\bar +)\) be two algebraic structures, and consider further a map \(\circ\) of the form
We say that \(\circ\) is distributive with respect to \(\bar +\) if and only if
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\(\forall\ a\in A,\ \forall\ u,v\in V,\ a\circ (u\,\bar +\,v)=a\circ u\, \bar +\, a\circ v\);
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\(\forall\ a,b\in A,\ \forall\ v \in V,\ (a+b)\circ v= a\circ v\, \bar +\, b\circ v\).
Rests on Definition 4.3.
That the structure \(A\) acts from the left is only a convention, or the traditional usage; nothing prevents the creation of structures with distributivity from the left with respect to \(\circ\) together with distributivity from the right with respect to \(\circ\) (or even another binary map). In this book we shall use only left distributivity, and we shall mention it explicitly when absolutely necessary. Note also that if the structures \((A,+)\) and \((V,\bar +)\) are identical — that is, \(A=V\) and there is an isomorphism between \((A,+)\) and \((A,\bar +)\) — then \(\circ\) becomes an internal binary operation; in this context, if the magma \((A,\circ)\) is abelian, the two axioms above are completely equivalent.
Action axioms
Let \((A,\ast,e)\) be an algebraic structure with neutral element and \(X\) an arbitrary set. Consider the map \(\bullet\) of the form
We say that \(\bullet\) is an action of the structure \(A\) on the set \(X\) if and only if \(\bullet\) satisfies
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\(\forall\ x\in X,\ e\bullet x=x\);
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\(\forall\ x\in X,\ \forall\ a,b\in A,\ (a\ast b)\bullet x=a\bullet (b\bullet x)\).
Rests on Definition 4.9.
Composite algebraic structures
Consider an abelian group \((A,+,0_A)\) and a semigroup \((A,\circ)\). We say that \(\left[(A,+,0_A),\circ,(A,\circ)\right]\stackrel{\text{not.}}{=}(A,+,\circ)\) is a ring if and only if
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\(\forall\ a,b,c\in A,\quad a\circ (b+c)=a\circ b+a\circ c\);
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\(\forall\ a,b,c\in A,\quad (a+b)\circ c= a\circ c+b\circ c\);
that is, if and only if the property of distributivity of \(\circ\) with respect to \(+\) holds. Rests on Definitions 4.8, 4.18, 4.21 and 4.28.
For example, in \(\Z\) multiplication distributes with respect to addition, so \((\Z,+,\cdot)\) is a ring. Further:
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Upon imposing commutativity of the IBO \(\circ\) — that is, if \((A,\circ)\) becomes an abelian semigroup — we say that \((A,+,\circ)\) is a commutative ring. \((\Z,+,\cdot)\) is a commutative ring.
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Imposing the existence of a neutral element \(1_A\) for the semigroup — that is, if \((A,\circ,1_A)\) becomes a monoid — we say that \((A,+,\circ)\) is a ring with unity. Since in \(\Z\) the number 1 is a neutral element for multiplication, \((\Z,+,\cdot)\) is a (commutative) ring with unity. Another example is the real numbers: \((\R,+,0)\) is an abelian group, and \((\R,\cdot,1)\) is an abelian monoid. As is well known, multiplication distributes with respect to addition, and consequently \((\R,+,\cdot)\) is a (commutative) ring with unity.
Consider a ring with unity \(\left[(A,+,0_A),\circ,(A,\circ,1_A)\right]=(A,+,\circ)\) and an abelian group \((V,\bar{+},0_V)\). Let there be, in addition, the binary map
We say that \([(A,+,\circ),\cdot,(V,\bar{+})]\) is a module \(V\) over \(A\) if and only if the binary map \(\cdot\) distributes with respect to \(\bar +\) and, in addition, \(\cdot\) is an action of the monoid \((A,\circ,1_A)\) on the group \((V,\bar{+},0_V)\) regarded as a set. That is, if and only if
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\(\forall\ a\in A,\ \forall\ u,v\in V,\quad a(u\,\bar{+}\,v)=au\,\bar{+}\,av\);
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\(\forall\ a,b\in A,\ \forall\ u\in V,\quad (a+b)u=au\,\bar{+}\,bu\);
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\(\forall\ a,b\in A,\ \forall\ u\in V,\quad a(bu) = (a\circ b)u\);
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\(\forall\ u\in V,\quad 1_Au=u\).
Consider a ring with unity \((A,+,\circ)\), that is, such that \((A,+,0_A)\) is an abelian group and \((A,\circ,1_A)\) is a monoid. We say that \(\left[(A,+,0_A),\circ,(A\setminus\set{0_A},\circ,1_A)\right] \stackrel{\text{not.}}{=}(A,+,\circ)\) is a field if for every element of the monoid \((A,\circ,1_A)\) there exists an inverse element, except for the neutral element \(0_A\) of \((A,+)\); that is, if in addition
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\(\forall\ a\in A\setminus\{0_A\},\ \exists\ a^{-1}\in A\mid a\circ a^{-1}=1_A \ \wedge\ a^{-1}\circ a=1_A\),
which means that the monoid \((A,\circ)\) almost becomes a group — as it would, completely, if an inverse existed for \(0_A\). Rests on Definitions 4.11, 4.20 and 4.30.
-
If the monoid \((A,\circ)\) is moreover abelian, we say that \((A,+,\circ)\) is a commutative field.
Consider a module \([(A,+,\circ),\cdot,(V,\bar{+})]\). We say that \([(A,+,\circ),\cdot,(V,\bar{+})]\) is a vector space \(V\) over \(A\) if and only if the ring with unity \((A,+,\circ)\) is a commutative field; that is, if and only if the following further axioms hold:
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\(\forall\ a\in A\setminus\{0_A\},\ \exists\ a^{-1}\mid a\circ a^{-1}=1_A\ \wedge \ a^{-1}\circ a=1_A\);
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\(\forall\ a,b\in A,\ a\circ b=b\circ a\).
As a characterization, a vector space is thus a commutative field (arising from a ring with unity in which inverses now exist except for \(0_A\)) together with a distributive action of the field's monoid-with-inverses on an abelian group.
Vector spaces are very important in physics. In classical physics they manifest an abstract arena serving to model nature at each point of the Universe; in the context of quantum physics, they serve to model the possible states of a system. Even if a system is confined to some smaller set of states, we can model that smaller system as a vector-space substructure.
Consider a vector space \([(A,+,\circ),\cdot,(V,\bar{+})]\), together with the IBO
We say that \([(A,+,\circ),\cdot,(V,\bar{+},\times)]\) is an algebra \(V\) over \(A\) if and only if the following axioms hold:
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\(\forall\ u,v,w\in V,\ u\times(v\,\bar{+}\,w)= u\times v\,\bar{+}\,u\times w\);
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\(\forall\ u,v,w\in V,\ (u\,\bar{+}\,v)\times w= u\times w\,\bar{+}\,v\times w\);
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\(\forall\ a\in A,\ \forall\ u,v\in V,\ a(u\times v)=(au)\times v=u\times(av)\).
Thus the algebra \([(A,+,\circ),\cdot,(V,\bar{+},\times)]\) is such that
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\((A,+,\circ)\) is a ring with unity;
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\((V,\bar{+})\) is an abelian group;
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\((V,\times)\) is a magma;
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\(A\) acts on \(V\) through the operation \(\cdot\)\,;
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various distributivity properties hold among the operations, as specified above.
Let us examine the properties that may be added, giving rise to remarkable kinds of algebras.
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If \((V,\bar{+},\times)\) is a ring — that is, if \((V,\times)\) is moreover a semigroup — we say that \([(A,+,\circ),\cdot,(V,\bar{+},\times)]\) is an associative algebra. There holds
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\(\forall\ u,v,w\in V,\ u\times (v\times w)=(u\times v)\times w\).
-
-
If \((V,\bar{+},\times)\) is a ring with unity — that is, if \((V,\times)\) is moreover a monoid — we say that \([(A,+,\circ),\cdot,(V,\bar{+},\times)]\) is a unital associative algebra. There holds
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\(\forall\ u\in V,\ \exists\ \identity_V\in V\mid \identity_V\times u=u\ \wedge\ u\times\identity_V=u\).
-
-
If the axioms
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\(\forall\ u,v\in V,\ u\times v=- v\times u\),
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\(\forall\ u,v,w\in V,\ u\times(v\times w)\,\bar{+}\, v\times(w\times u)\,\bar{+}\,w\times(u\times v)=0_V\)
hold, we say that \([(A,+,\circ),\cdot,(V,\bar{+},\times)]\) is a Lie algebra.
-
Algebras are a most interesting algebraic structure for physics, since — like groups — they provide an elegant mathematical machinery for the symmetries of a physical system, classical or quantum.
Group theory: further definitions
Group theory is an immense branch of mathematics that studies these structures in detail. Because of the axioms imposed, groups are potentially applicable to situations in geometry and in nature that involve symmetry. There is a large literature on groups containing a finite number of elements.
Such groups display a complexity in their structure that is tied directly to their cardinality and to the prime decomposition of the cardinal. There is a certain beauty in this: arithmetic appears unavoidably within the theory of algebraic structures, even though the two theories study different things. To organize these finite groups there are classification theorems for abelian groups, and for the non-abelian ones there are the Sylow theorems. A problem to solve may be, for example: find all abelian groups of cardinal 8.
Groups
This restates Definition 4.21 with each axiom displayed and labelled, so that the rest of the chapter can cite them one at a time; nothing is added to the content of that definition. Consider a set \(G\), in general of finite or infinite cardinality, together with a binary operation \(\ast\). We say that \((G,\ast)\) is a group if and only if it satisfies the following axioms.
-
Closure:
\begin{equation}\tag{4.17} \forall\ a,b\in G,\ a\ast b\in G\ep \end{equation}This property is exactly equivalent to having taken the binary operation \(\ast\) to be an IBO, or the structure \((G,\ast)\) to be a magma.
-
Associativity:
\begin{equation}\tag{4.18} \forall\ a,b,c\in G,\ a\ast(b\ast c)=(a\ast b)\ast c\ep \end{equation} -
Existence of a neutral element:
\begin{equation}\tag{4.19} \forall\ a\in G,\ \exists\ e \in G\mid a\ast e=a\quad\wedge\quad e\ast a=a\ep \end{equation} -
Existence of inverse elements:
\begin{equation}\tag{4.20} \forall\ a\in G,\ \exists\ a^{-1} \in G\mid a\ast a^{-1}=e \quad\wedge\quad a^{-1}\ast a=e\ep \end{equation}
Rests on Definition 4.21.
Let us examine some examples whose underlying sets have infinite cardinality. The structure \((\R^{\ast},\cdot)\) is a group whose neutral element, in multiplicative notation, is \(1\in\R\). On the other hand we can verify that \((\Q^{\ast},\div)\), where
is not a group, since the associativity axiom fails: indeed,
To denote a group we shall use the notations \((G,\ast)\) and \(G=\{e,a,b,c,\ldots\}\), where \(e\) is the neutral element, though we may also use the index notation \(G=\{g_i\}_{i\in I\subset\N}\). When no ambiguity is possible we shall denote the operation \(\cdot\) by simple juxtaposition.
Consider a group \((G,\ast)\) and a subset \(H\subseteq G\). We say that \((H,\ast)\) is a subgroup of \((G,\ast)\) if and only if
-
\(\forall\ a,b\in H,\ a\ast b\in H\);
-
\(\forall\ a\in H,\ \exists\ e_H \in H\mid a\ast e_H=a\ \wedge\ e_H\ast a=a\) — which is completely equivalent to saying that the neutral element of \((G,\ast)\) lies in \(H\), since a neutral element of \(G\), if it exists, is unique;
-
\(\forall\ a\in H,\ \exists\ a^{-1} \in H\mid a\ast a^{-1}=e\ \wedge\ a^{-1}\ast a=e\) — which is equivalent to saying that the inverse elements of the elements of \(H\) belong to \(H\).
Rests on Definition 4.21.
Since all the elements of the group \((G,\ast)\) satisfy associativity, we may dispense with that condition in asserting that a subgroup is indeed a group. In particular a subgroup is a substructure, for \(H\) is closed by definition. We therefore introduce the notation
Given a group \(G=\set{e,a,\ldots}\) it is clear that \(\set{e}\leq G\) and that \(G\leq G\); these subgroups are known as the trivial subgroups.
Cayley tables
The Cayley table of a finite group is the resource we shall use to symbolize how the elements of the group operate. Let \((\set{e,a,b,c,\ldots},\cdot)\) be a group. Its Cayley table is given by
Rests on Definition 4.21.
The Cayley table of a group stores the whole “machinery” of the group: it contains all the information needed to describe the group's internal product explicitly. Two important observations about Cayley tables are the following.
-
The elements in any one row or column of a Cayley table cannot repeat. Indeed, suppose that within one column we had \(ab=x\) and also \(ac=x\). This says \(ab=ac\), and since we are in a group, \(a\) has an inverse. Multiplying on the left by the inverse of \(a\) we obtain \(b=c\), which is a contradiction, since the elements of the group as a set must all be distinct.
-
For an abelian group the Cayley table is symmetric with respect to the diagonal \(e\), \(a^2\), \(b^2\), \(c^2,\ldots\) — as can be seen trivially from the general Cayley table.
Basic groups
The first finite groups
The trivial group
Consider the algebraic structure \((\set{e},\ast)\), where \(e\) satisfies the neutral-element axiom. Trivially this set satisfies the four group axioms; we call it the trivial group. Since every group contains a neutral element (by definition), given a group \((G,\cdot,e)\) we can always affirm that the set formed by the neutral element of \(G\) is a subgroup of \(G\), that is, \(\set{e}\leq G\). The Cayley table of the trivial group is
For example, consider \(0\in\Z\) together with the IBO \(+\). We have
that is, the structure \((\{0\},+)\) is a trivial group. This does not happen for an arbitrary IBO defined on \(\Z\): for example, \((\{0\},\cdot)\) is not a group, since \(0\) has no multiplicative inverse, so the inverse axiom fails.
The group of two elements
Consider the algebraic structure \((\set{e,a},\ast)\), where \(e\) satisfies the neutral-element axiom. Let us see how the elements of \(\set{e,a}\) must operate for the group axioms to be satisfied. Clearly we have
But what must the result of \(a\ast a\) be? That is, supposing that \(\set{e,a,a\ast a}\) is a group, let us deduce the conditions \(a\) must satisfy. The options are:
-
the option \(a\ast a=a\) is not valid, since we know the elements of a row or column of a Cayley table cannot repeat;
-
on the other hand, to respect closure, \(a\ast a=e\) is the only other option for \(\set{e,a}\) to be a group of cardinal 2.
Thus the Cayley table of a group of two elements is
We have therefore proved that this is the only manner, or configuration, in which a set of two elements can form a group. For example, consider the elements \(1,-1\in\Z\) together with the IBO \(\cdot\)\,. We have
that is, the structure \((\{1,-1\},\cdot)\) is a group of two elements. By contrast, \((\{1,-1\},+)\) is not a group, closure being clearly violated since in this structure \(1+1=2\). Indeed, one may regard the question “how much is one plus one?” as a trick question, to be answered with the question “in which algebraic structure?”.
The group of three elements
In this case we have
Since, however, the elements of a row or column cannot repeat, we conclude that \(a\ast b=e\), and in the same way \(b\ast a=e\). Filling in the remaining entries by the same logic we find \(a\ast a=b\) and \(b\ast b=a\), so the general Cayley table for a group of three elements is
which is clearly unique. As an example, the set of the three cube roots of unity together with complex multiplication forms a group, with the Cayley table of a group of three elements:
The group $\Z$
In the discussion of the definition of a group we mentioned that \((\Z,+,0)\) is a group. This group is very fruitful for explaining the ideas of subgroup and of quotient spaces, and we therefore study it in detail.
The subgroup $n\Z$
Let \(n\in\Z\), and define the set \(n\Z=\{nk\mid k\in\Z\}\); that is, \(n\Z\) consists of all the multiples of \(n\) within \(\Z\). For example,
\(n\Z\) is a subgroup of \(\Z\). Rests on Definition 4.36.
Derives Proposition 4.38. We check the requirements:
-
\(n\Z\neq\varnothing\), since \(0\in n\Z\) (take \(k=0\)).
-
Trivially the neutral element \(0\) lies in \(n\Z\), since the number zero is a multiple of every integer.
-
Finally, for every element of \(n\Z\) there exists an inverse: if \(a\in n\Z\) is a multiple of \(n\), so is \(-a\). Indeed, we may write \(a=nl\) with \(l\in\Z\); then, since distributivity holds in \(\Z\),
\begin{align*} a+(-a) & = nl+(-nl)\\ & = nl+n(-l)\\ & = n(l-l)\\ & = 0\ec \end{align*}so the inverse axiom is satisfied.
Congruence modulo $n$
As human beings we do not count the hours from the beginning of the Universe. As a reference for indicating clock time we speak of \(k\) hours past midnight: a day is completed when 24 hours have elapsed since midnight, and hour number 25 we begin to call hour 1 again. We shall build definitions that say that 25 is mathematically equivalent to 1 (though not equal), every 24 hours.
Let \(k,l\in\Z\). They are congruent modulo \(n\) when
The relation Equation (4.21) is an equivalence relation. Rests on Definitions 3.58 and 4.21.
Derives Proposition 4.40. We verify the three properties of Definition 3.58:
-
\(k\sim_n k\), since \(k-k=0\in n\Z\); reflexivity therefore holds.
-
If \(k\sim_n l\), we have
\begin{align*} k-l\in n\Z \iff & \ k-l=np,\ p\in\Z,\\ \iff & \ l-k=n(-p),\ p\in\Z,\\ \iff & \ l-k=nq,\ q\in\Z\ec \end{align*}because \(\Z\) is a group, so there exists \(-p=q\in\Z\). Thus
\[ k\sim_n l\implies l\sim_n k\ec \]and symmetry holds.
-
If \(k\sim_n l\) and \(l\sim_n p\), we have
\begin{align*} k-l&=nq,\ q\in\Z,\\ l-p&=nr,\ r\in\Z\ep \end{align*}Adding the equations member by member gives \(k-p=n(q+r)\), and by closure \(q+r=s\in\Z\); hence \(k-p\in n\Z\). Thus
\[ k\sim_n l\ \wedge\ l\sim_n p\implies k\sim_n p\ec \]and transitivity holds.
We have proved that the relation defined is an equivalence relation. We shall say that two elements equivalent under this relation are congruent modulo \(n\). In the earlier example, 25 is congruent to 1 modulo 24.
The quotient space $\Z/n\Z$
As we have seen, congruence modulo \(n\) is an equivalence relation. Let \(k\in\Z\). We denote by \(\overline{k}\) the class of \(k\), given by
As we know, the quotient space is the set of all the classes. Let us analyse the possible classes making up our quotient space:
-
\(\overline{0}\) consists of the multiples of \(n\);
-
\(\overline{1}\) consists of the multiples of \(n\), plus 1;
-
\(\overline{n}\) consists of the multiples of \(n\), plus \(n\) — that is, simply the multiples of \(n\). Thus
\begin{equation}\tag{4.22} \overline{0}=\overline{n}\ep \end{equation} -
\(\overline{n+1}\) consists of the multiples of \(n\), plus \(n+1\) — that is, simply the multiples of \(n\), plus 1.
Hence the elements of our quotient space are the classes (none repeated)
The set $\Z/n\Z$ as an abelian group
Define the internal binary operation
\(\left(\frac{\Z}{n\Z},\overline{+},\overline{0}\,\right)\) is an abelian group. Rests on Definition 4.21, Definition 4.8 and Proposition 4.40.
Derives Proposition 4.41. To avoid overloading the notation we write \(\overline{+}\) simply as \(+\). We have:
-
There is closure by construction: the class \(\overline{k+l}=\overline{k}+\overline{l}\) always lies in the quotient space, since the class of \(k+l\) can always be written as the class of \(n\) plus something, and finally as the class of that something. For example, for \(n=5\),
\[ \overline{3}+\overline{8}=\overline{3+8}=\overline{11}=\overline{5+6} =\overline{6}=\overline{5+1}=\overline{1}\in \frac{\Z}{5\Z}\ep \] -
There is associativity, since
\begin{align*} (\overline{k}+\overline{l})+\overline{m}&=\overline{k+l}+\overline{m}\\ &=\overline{(k+l)+m}\\ &=\overline{k+(l+m)}\\ &=\overline{k}+\overline{l+m}\\ &=\overline{k}+(\overline{l}+\overline{m})\ep \end{align*} -
The neutral element \(\overline{0}\) exists: indeed, \(\overline{k}+\overline{0}=\overline{k+0}=\overline{k}\).
-
An inverse element \(\overline{-k}\) exists for each element \(\overline{k}\) of the set: indeed, \(\overline{k}+\overline{-k}=\overline{k-k}=\overline{0}\). For example, for \(n=5\) we have
\[ \frac{\Z}{5\Z}=\left\{\overline{0},\overline{1},\overline{2}, \overline{3},\overline{4}\right\}\ec \]and the inverse element of \(\overline{1}\) is \(\overline{4}\), that is, \(\overline{-1}=\overline{4}\) — not by imposition, but by the definition of the class: the multiples of 5, minus 1, are the same numbers (the same class) as the multiples of 5, plus 4.
-
The IBO \(\overline{+}\) is commutative, owing to the commutativity of \(\Z\). Indeed,
\begin{align*} \overline{k}+\overline{l} & = \overline{k+l}\\ & = \overline{l+k}\\ & = \overline{l}+\overline{k}\ep \end{align*}
In conclusion, \(\left(\frac{\Z}{n\Z},\overline{+},\overline{0}\,\right)\) is an abelian group, distinct from the group \(\left(\Z,+,0\right)\).
Functions on groups
The group of maps into a group
The construction this section needs has already been carried out in the generality of arbitrary algebraic structures: the pointwise product of Section 4.1.2, the neutral map \(\Theta\), and the opposite map Equation (4.6) make \(\mathcal{F}(A,B)\) a group whenever \(B\) is one, by Proposition 4.14. Taking \(A\) to be a group \(G\) and \(B\) a group \(H\) costs nothing extra and adds nothing new: \(G\)'s own operation plays no part, because the pointwise construction never operates on the argument.
What is genuinely new here is that the maps of interest between groups are not all of \(\mathcal{F}(G,H)\) but the homomorphisms, and those are a different matter.
Let \(G\) and \(H\) be groups and let \(\operatorname{Hom}(G,H)\) denote the set of group homomorphisms from \(G\) to \(H\). If \(H\) is abelian, then \(\operatorname{Hom}(G,H)\) is a subgroup of \((\mathcal{F}(G,H),\ast)\), and it is abelian. If \(H\) is not abelian, then \(\operatorname{Hom}(H,H)\) is not even closed under \(\ast\). Rests on Definition 4.43, Proposition 4.14, Proposition 4.27 and Definition 4.36.
Derives Proposition 4.42. Let \(f,g\in\operatorname{Hom}(G,H)\) and \(a,b\in G\). Expanding both sides,
The two agree for all \(a,b\) precisely when \(f(b)\ast_H g(a) =g(a)\ast_H f(b)\), that is, when the images of \(f\) and \(g\) commute with one another.
If \(H\) is abelian this holds automatically, so \(f\ast g\) is again a homomorphism. The neutral map \(\Theta\) is a homomorphism, since \(\Theta(a\ast_G b)=e_H=e_H\ast_H e_H=\Theta(a)\ast_H\Theta(b)\); and \(f^{\mathrm{op}}\) is one because \(f^{\mathrm{op}}(a\ast_G b)=\left(f(a)\ast_H f(b)\right)^{-1} =f(b)^{-1}\ast_H f(a)^{-1}=f^{\mathrm{op}}(a)\ast_H f^{\mathrm{op}}(b)\), the last step using commutativity again. So \(\operatorname{Hom}(G,H)\) contains the neutral element and is closed under the operation and under opposites, which is what it is to be a subgroup; it is abelian because \(\mathcal{F}(G,H)\) is, by Proposition 4.14.
If \(H\) is not abelian, pick \(a,b\in H\) with \(a\ast_H b\neq b\ast_H a\) and take \(f=g=\id_H\), which is certainly a homomorphism. The displayed condition then reads \(b\ast_H a=a\ast_H b\), which fails, so \(\id_H\ast\,\id_H\) is not a homomorphism and \(\operatorname{Hom}(H,H)\) is not closed.
∎The proposition is the reason the theory of group homomorphisms does not reduce to the theory of maps. For abelian targets one may add homomorphisms and stay inside the class — which is what makes \(\operatorname{Hom}(G,\R)\) and its relatives vector spaces, and ultimately what makes cohomology groups groups. For non-abelian targets the pointwise product leaves the class immediately, and the structures that survive are built from composition instead. The remainder of this section develops the notions that do work in both cases.
Homomorphisms
Consider the groups \((G_1,\ast_1)\) and \((G_2,\ast_2)\), together with a map \(f:G_1\longrightarrow G_2\). We say that \(f\) is a group homomorphism from \(G_1\) to \(G_2\) if and only if \(f(a_1\ast_1b_1)=f(a_1)\ast_2f(b_1)\) — that is, if between the two distinct groups \(f\) preserves the multiplication law. Rests on Definitions 3.43 and 4.21.
Consider the groups \((G_1,\ast_1)\) and \((G_2,\ast_2)\) and let \(f:G_1\longrightarrow G_2\) be a group homomorphism. We say that \(f\) is an isomorphism from \(G_1\) to \(G_2\) if and only if the homomorphism is a bijection; we then say that \(G_1\) and \(G_2\) are isomorphic. Rests on Definitions 3.47 and 4.43.
Isomorphism between groups is of great importance, because it is the concept expressing that two groups operate in exactly the same way even though their elements are distinct. What totally determines a group is the operation among its elements — its Cayley table — and what an isomorphism does is precisely to preserve the operation. For example, consider the set \(\set{1,\ii,-1,-\ii}\), where \(\ii=\sqrt{-1}\in\C\). It is easy to verify that this set, together with the usual multiplication, forms an abelian group whose Cayley table is
On the other hand, it is easy to compute the Cayley table of the (abelian) group \(\frac{\Z}{4\Z}\):
Comparing the two tables, we see that these groups are essentially the same thing: they are exactly the same up to the symbols we have used to designate the elements of the sets. This may seem magical, but so it is: taking two sets with no relation whatsoever and adjoining the group axioms, an intrinsic operational structure appears when the sets are viewed as algebraic structures. The situation has a precise mathematical underpinning: these groups are isomorphic. This is the true meaning of isomorphisms. The explicit form of the isomorphism between \(\set{1,\ii,-1,-\ii}\) and \(\frac{\Z}{4\Z}\) is given by \(f(x)=\overline{-(2\ii/\pi)\ln x}\) with \(x\in\set{1,\ii,-1,-\ii}\), resulting in
One verifies, moreover, the preservation of the product law:
Let \(f:G_1\longrightarrow G_2\) be a group isomorphism. We say that \(f\) is an automorphism if and only if \(G_1=G_2\) — that is, the homomorphism \(f\) is moreover a bijection sending elements of a group to the group itself. Rests on Definition 4.44.
Since every group is isomorphic to itself, we can say that for every group there exists a trivial automorphism sending each element to itself; that is, the bijection
is a (trivial, but canonical) example of an automorphism.
Properties of homomorphisms
Under a group homomorphism, the image of the neutral element is the neutral element of the codomain:
Rests on Definition 4.43.
Derives Proposition 4.46. For the neutral element of the domain we have \(f(e_1)=f(e_1\ast_1e_1)\), and since \(f\) is a homomorphism, operating with \(\ast_2\,(f(e_1))^{-1}\),
Analogously to the preceding case, the image of an inverse element of the domain is the inverse of the image in the codomain:
Rests on Definition 4.43 and Proposition 4.46.
Derives Proposition 4.47. Indeed, operating on the left with \((f(a_1))^{-1}\ast_2\),
The symmetric group
The symmetric group
Consider a set \(X\), and let \(S_X\) be the set of all the possible bijections from \(X\) to \(X\), that is,
Rests on Definition 3.47.
Let us recall the following properties (Section 3.3):
-
the composition of two bijective maps is a bijective map;
-
the composition of maps is associative — in particular, so is the composition of bijective maps;
-
there exists the identity map, whose composition with any map returns that same map — in particular, this works for any bijective map;
-
every bijective map possesses an inverse map.
From this we may conclude that \((S_X,\circ)\) is a group under composition of maps, which we shall call the symmetric group.
The permutation group
We define the permutation group as a symmetric group \(S_G\) in which \(G\) is a finite group. Thus
Consider \(G=\{g_i\}^n_{i=1}\). As discussed in the definition of the symmetric group, the elements of \(S_G\) are bijections. Since \(G\) here has a finite number of elements, a possible bijection is, for example, for \(n=5\),
that is, in some manner \(f\) permutes the elements. We can symbolize this permutation as
As an alternative notation for permutations we introduce
where the index \(a_i\) denotes the new place of the element. For example, one may have \(ag_2=g_1\) (that is, \(a_2=1\)).
\(\abs{S_{G}}=\abs{G}!\)\,. Rests on Definition 4.49.
Derives Proposition 4.50.
Write \(n=\abs{G}\) and \(G=\set{g_1,\ldots,g_n}\). For \(0\leq k\leq n\) let \(A_k\) denote the set of injective maps from \(\set{g_1,\ldots,g_k}\) into \(G\), so that \(A_0\) has exactly one element, the empty map. We claim that
the right-hand side being read as \(1\) when \(k=0\).
The claim holds for \(k=0\). Assume it for some \(k<n\) and consider the restriction map \(A_{k+1}\longrightarrow A_k\) that forgets the value at \(g_{k+1}\). An injective \(u\in A_k\) takes \(k\) pairwise distinct values, so the injective extensions of \(u\) to \(\set{g_1,\ldots,g_{k+1}}\) are exactly the maps sending \(g_{k+1}\) to one of the elements of \(G\setminus\set{u(g_1),\ldots,u(g_k)}\), a set of \(n-k\) elements. Every \(u\in A_k\) therefore has exactly \(n-k\) preimages under the restriction map, and every element of \(A_{k+1}\) has exactly one image, whence
which is Equation (4.29) for \(k+1\). By induction it holds for every \(k\leq n\); in particular
It remains to identify \(A_n\) with \(S_G\). An element of \(A_n\) is an injective map \(f:G\longrightarrow G\); its image has exactly \(n\) elements, and the only subset of the \(n\)-element set \(G\) with \(n\) elements is \(G\) itself, so \(f\) is also surjective and hence a bijection. Conversely every bijection of \(G\) is injective. Thus \(A_n=S_G\) and \(\abs{S_G}=n!=\abs{G}!\)\,.
∎The permutation group of order $3!$
The smallest symmetric group that is not abelian is the one built on a three-element set. Take \(X=\set{1,2,3}\) and write \(S_3=S_X\); by Proposition 4.50 it has \(3!=6\) elements. It is small enough to be written out completely, and large enough that everything the rest of this chapter defines — subgroups, conjugacy, normality, quotients — can be read off its table.
Let \(i_1,\ldots,i_r\) be distinct elements of a finite set \(X\). The cycle \((i_1\,i_2\,\ldots\,i_r)\) is the bijection of \(X\) sending \(i_1\longmapsto i_2\), \(i_2\longmapsto i_3\), …, \(i_r\longmapsto i_1\) and leaving every other element of \(X\) fixed; a cycle of length two is a transposition. A permutation is also written in two-line form, the second line listing the images of the entries of the first. Products are compositions of maps, so \(fg\) means \(f\circ g\): the right-hand factor acts first. Rests on Definitions 3.47 and 4.48.
In this notation the six elements of \(S_3\) are
Three of them are transpositions, two are three-cycles, and \(r^{3}=e\), so \(r^{-1}=r^{2}\) and \(t_i^{-1}=t_i\) for each \(i\). Composing the maps of Equation (4.30) pair by pair gives the Cayley table (Definition 4.37) shown in Table 4.1.
| $\circ$ | $e$ | $r$ | $r^{2}$ | $t_{1}$ | $t_{2}$ | $t_{3}$ |
|---|---|---|---|---|---|---|
| $e$ | $e$ | $r$ | $r^{2}$ | $t_{1}$ | $t_{2}$ | $t_{3}$ |
| $r$ | $r$ | $r^{2}$ | $e$ | $t_{3}$ | $t_{1}$ | $t_{2}$ |
| $r^{2}$ | $r^{2}$ | $e$ | $r$ | $t_{2}$ | $t_{3}$ | $t_{1}$ |
| $t_{1}$ | $t_{1}$ | $t_{2}$ | $t_{3}$ | $e$ | $r$ | $r^{2}$ |
| $t_{2}$ | $t_{2}$ | $t_{3}$ | $t_{1}$ | $r^{2}$ | $e$ | $r$ |
| $t_{3}$ | $t_{3}$ | $t_{1}$ | $t_{2}$ | $r$ | $r^{2}$ | $e$ |
Every group with fewer than six elements is abelian, and \(S_3\) has six elements and is not abelian. Rests on Definition 4.21, Definition 4.8 and Proposition 4.50.
Derives Proposition 4.52. That \(S_3\) has six elements is Proposition 4.50, and Table 4.1 exhibits \(r\,t_{1}=t_{3}\neq t_{2}=t_{1}r\), so \(S_3\) is not abelian.
For the first assertion, let \(G\) be a group that is not abelian and choose \(x,y\in G\) with \(xy\neq yx\). We show that \(G\) has at least six elements. First, the five elements
are pairwise distinct. Indeed \(x\neq e\) and \(y\neq e\), since the neutral element commutes with everything, and \(x\neq y\), since every element commutes with itself. If \(xy=e\) then \(y=x^{-1}\) by Proposition 4.25, and then \(yx=e=xy\), excluded; the same argument rules out \(yx=e\). Cancelling \(x\) on the left in \(xy=x\), or \(y\) on the right in \(xy=y\), would give \(y=e\) or \(x=e\), and likewise for \(yx\); finally \(xy\neq yx\) by choice.
It remains to produce a sixth element. Consider \(x^{2}\). It cannot equal \(x\) (that gives \(x=e\)), nor \(xy\) (that gives \(x=y\)), nor \(yx\) (cancelling \(x\) on the right gives \(x=y\)), nor \(y\) — for \(x^{2}=y\) would make \(xy=x^{3}=yx\). So either \(x^{2}\) lies outside the five elements listed, and \(G\) has at least six, or \(x^{2}=e\). Applying the same argument to \(y\), we may assume \(x^{2}=y^{2}=e\), so that \(x^{-1}=x\) and \(y^{-1}=y\). Then \(xyx\) is the sixth element: \(xyx=e\) gives \(yx=x^{-1}=x\) and so \(y=e\); \(xyx=x\) gives \(yx=e\) and so \(y=x^{-1}=x\); \(xyx=y\) gives \(xy=yx^{-1}=yx\); \(xyx=xy\) gives \(x=e\); and \(xyx=yx\) gives \(xy=y\) and so \(x=e\) — every case being excluded. Hence \(\abs{G}\geq6\).
∎\(S_3\) has exactly six subgroups: the two trivial ones \(\set{e}\) and \(S_3\); the three two-element subgroups \(\set{e,t_i}\), \(i=1,2,3\); and the three-element subgroup
Derives Proposition 4.53. Each of the six listed sets is a subgroup: \(\set{e}\) and \(S_3\) trivially, \(\set{e,t_i}\) because \(t_i\,t_i=e\), and \(A_3\) because the upper-left \(3\times3\) block of Table 4.1 takes values in \(A_3\) and contains \(e\), with \(r^{-1}=r^{2}\).
Conversely let \(H\leq S_3\) be a subgroup and suppose \(H\neq\set{e}\). If \(H\) contains \(r\) or \(r^{2}\) then, being closed and containing inverses, it contains \(\set{e,r,r^{2}}=A_3\). If in addition \(H\) contains some \(t_i\), then it contains the three products \(r\,t_i\), \(r^{2}t_i\) and \(t_i\), which by Table 4.1 are the three distinct transpositions; hence \(H=S_3\). So a subgroup containing a three-cycle is \(A_3\) or \(S_3\).
Suppose instead that \(H\) contains no three-cycle. Then \(H\) consists of \(e\) and transpositions. It cannot contain two distinct transpositions: the product of two distinct transpositions is, by Table 4.1, a three-cycle, which would then lie in \(H\). Hence \(H=\set{e,t_i}\) for a single \(i\). The list is therefore complete.
∎Anticipating Definition 4.60, \(A_3\) is normal in \(S_3\): conjugation by an element of \(A_3\) maps \(A_3\) into itself because \(A_3\) is a subgroup, and Table 4.1 gives \(t_{1}r\,t_{1}^{-1}=t_{1}r\,t_{1}=r^{2}\), with the same outcome for \(t_{2}\) and \(t_{3}\). The two-element subgroups are not normal: the same table gives \(r\,t_{1}r^{-1}=r\,t_{1}r^{2}=t_{2}\), which is outside \(\set{e,t_{1}}\). The three of them are therefore conjugate to one another in the sense of Definition 4.59. Moreover \(S_3\) is the smallest group possessing a subgroup that is not normal: in an abelian group \(aHa^{-1}=\set{aha^{-1}\mid h\in H}=\set{h\mid h\in H}=H\) for every subgroup \(H\) and every \(a\), so all subgroups are normal, and by Proposition 4.52 every group of fewer than six elements is abelian.
The elements of \(S_3\) are not merely relabellings: they are the symmetries of a figure.
Let \(\Delta\subset\R^{2}\) be an equilateral triangle with vertices labelled \(1,2,3\), and let \(T\) be the set of isometries of the plane that map \(\Delta\) onto itself, a group under composition. The map sending each element of \(T\) to the permutation it induces on the vertex labels is an isomorphism \(T\longrightarrow S_3\). Rests on Definition 4.44 and Equation (4.30).
Derives Proposition 4.55. An isometry mapping \(\Delta\) onto itself maps the vertex set onto itself, since a vertex is characterized metrically as a point of \(\Delta\) at maximal distance from the centroid; call \(\pi(u)\in S_3\) the resulting permutation of the labels. Composition of isometries induces composition of the permutations, so \(\pi\) is a homomorphism (Definition 4.43).
\(\pi\) is injective. If \(\pi(u)=e\) then \(u\) fixes the three vertices, and these are not collinear; since a point of the plane is determined by its distances to three non-collinear points, and \(u\) preserves all distances, \(u\) fixes every point of the plane. So \(\ker\pi=\set{\id}\), and Proposition 4.17 gives injectivity.
\(\pi\) is surjective. The rotation through \(2\pi/3\) about the centroid of \(\Delta\) carries \(\Delta\) onto itself and induces the three-cycle \(r\) of Equation (4.30); its square induces \(r^{2}\). The reflection in the median through the vertex \(i\) carries \(\Delta\) onto itself and induces the transposition \(t_i\). Together with the identity these six isometries realize all six elements of \(S_3\), so \(\pi\) is onto and hence an isomorphism.
∎The six isometries just exhibited are all of \(T\), since \(\pi\) is a bijection. This is the first instance in this chapter of the pattern that makes group theory the language of symmetry in physics: an abstract group defined by a multiplication table turns out to be the symmetry group of a concrete object, and the two descriptions carry exactly the same information.
Every finite group of order \(n\) is isomorphic to a subgroup of order \(n\) of the permutation group \(S_n\). Rests on Definitions 4.36, 4.44 and 4.49.
Derives Theorem 4.56.
Let \((G,\cdot,e)\) be a group with \(\abs{G}=n\). For each \(a\in G\) define the left translation
Each \(p_a\) is a bijection. It is a map into \(G\) by closure. It is injective: if \(ag=ag'\), then operating on the left with \(a^{-1}\) and using associativity gives \(g=g'\). It is surjective: given \(h\in G\), the element \(a^{-1}h\) lies in \(G\) and \(p_a(a^{-1}h)=a(a^{-1}h)=h\). By Proposition 3.48 it is therefore a bijection, so \(p_a\in S_G\); its inverse map is \(p_{a^{-1}}\).
The assignment \(a\longmapsto p_a\) is a homomorphism. For \(a,b,g\in G\),
and since \(g\) was arbitrary, \(p_{ab}=p_a\circ p_b\). Writing \(P:G\longrightarrow S_G\) for the map \(a\longmapsto p_a\), this says precisely that \(P\) is a group homomorphism into \((S_G,\circ)\) (Definition 4.43).
\(P\) is injective. If \(p_a\) is the identity permutation then in particular \(ae=e\), so \(a=e\). Hence \(\ker P=\set{e}\), and Proposition 4.17 gives injectivity.
The image is a subgroup of order \(n\). The set \(P(G)=\set{p_a\mid a\in G}\) contains \(p_e=\id_G\), is closed under composition because \(p_a\circ p_b=p_{ab}\), and contains the inverse \(p_a^{-1}=p_{a^{-1}}\) of each of its elements; it is therefore a subgroup of \(S_G\) (Definition 4.36). Since \(P\) is injective, \(P(G)\) has exactly \(n\) elements, and \(P:G\longrightarrow P(G)\) is a bijective homomorphism, that is, an isomorphism (Definition 4.44).
Finally, any bijection \(\varphi:G\longrightarrow\set{1,\ldots,n}\) induces an isomorphism \(S_G\longrightarrow S_n\) by \(f\longmapsto\varphi\circ f\circ\varphi^{-1}\) — a bijection with inverse \(h\longmapsto\varphi^{-1}\circ h\circ\varphi\), and a homomorphism because the two inner factors \(\varphi^{-1}\varphi\) cancel in a product. It carries \(P(G)\) to a subgroup of \(S_n\) of order \(n\) isomorphic to \(G\).
∎Decompositions of a group
Conjugacy classes
Consider a group \(G\). Define the relation
The relation Equation (4.33) is an equivalence relation. Rests on Definition 3.58, Definition 4.21 and Proposition 4.27.
Derives Proposition 4.57. We verify the three properties:
-
Reflexivity holds, since
\[ a=eae=eae^{-1}\ec \]that is, thanks to the existence of the neutral element, \(a\sim a\).
-
If \(a\sim b\), it is because there exists \(p\in G\) such that \(b=pap^{-1}\). Multiplying on the right by \(p\) and then on the left by \(p^{-1}\),
\begin{align*} b & = pap^{-1}\\ bp & = pa\\ p^{-1}bp & = a\ep \end{align*}Hence, by the existence of inverse elements in \(G\), there exists \(q=p^{-1}\) such that \(a=qbq^{-1}\), so \(b\sim a\); symmetry therefore holds.
-
Supposing \(a\sim b\) and \(b\sim c\), there exist \(p,q\in G\) such that
\[ b=pap^{-1},\quad c=qbq^{-1}\ec \]and substituting the first into the second,
\begin{align*} c & = q(pap^{-1})q^{-1}\\ & = (qp)a(p^{-1}q^{-1})\\ & = (qp)a(qp)^{-1} && \text{by Equation (4.16)}\ep \end{align*}Hence by closure there exists \(r=qp\in G\) such that \(c=rar^{-1}\), so \(a\sim c\), and transitivity holds.
We call conjugacy classes the equivalence classes given by the relation Equation (4.33). We say that two elements of a conjugacy class are conjugate; that is, \(a\) and \(b\) are conjugate if and only if \(a\sim b\). We have
Rests on Proposition 4.57 and Definition 3.59.
Consider the group \(G\) and \(H,K\leq G\). We say that \(K\) and \(H\) are conjugate subgroups if and only if
that is,
which we shall symbolize simply as \(K=pHp^{-1}\), \(p\in G\). Rests on Definitions 4.36 and 4.58.
Let \(G\) be a group and \(H\leq G\). We say that \(H\) is a normal (or invariant) subgroup in \(G\) if and only if it is conjugate to itself by all the elements of the group, that is, if \(H=pHp^{-1}\) for all \(p\in G\). We introduce the notation
Given a group \(G\), it is easy to see that the trivial subgroups are also normal subgroups in \(G\); that is, \(\set{e}\unlhd G\) and \(G\unlhd G\). It is worth mentioning that in the group-theory literature there are many ways of defining what a normal subgroup is; one may define it as one wishes, provided the other definitions are mentioned as possible characterizations.
Cosets
Consider the group \((G,\cdot,e)\) and \(H\leq G\). Define the relation
The relation Equation (4.34) is an equivalence relation. Rests on Definitions 3.58 and 4.36.
Derives Proposition 4.61. We verify the three properties:
-
Reflexivity holds because \(H\) is a subgroup of \(G\), so the neutral element exists in \(H\) and
\[ a=ae\ec \]whence \(a\sim_H a\).
-
Suppose \(a\sim_H b\). Then there exists \(h\in H\) such that \(b=ah\). Since \(H\) is a subgroup of \(G\) (and in essence a group), \(h^{-1}\) exists. Operating on the right with \(\cdot\,h^{-1}\),
\begin{align*} b & = ah\\ bh^{-1} & = a\ep \end{align*}Then there exists \(k=h^{-1}\in H\) such that \(a=bk\); hence \(b\sim_H a\) and symmetry holds.
-
If \(a\sim_H b\) and \(b\sim_H c\), it is because there exist \(h,k\in H\) such that
\[ b=ah,\quad c=bk\ep \]Substituting the first into the second,
\begin{align*} c & = (ah)k\\ & = a(hk)\\ & = al\ec \end{align*}and thus, since \(H\) is a subgroup, by closure there exists \(l=hk\in H\); we conclude that \(a\sim_H c\), so transitivity holds.
We call left lateral class or left coset each equivalence class given by the relation Equation (4.34). We have
and, in view of the fact that \(h\) ranges over \(H\), we shall symbolize this class as
Rests on Proposition 4.61 and Definition 3.59.
Analogously we can define \(Ha\) as a right lateral class or right coset of \(G\). The study of left and right cosets is entirely parallel, so we shall refer only to left cosets, which we call simply lateral classes or cosets.
Finally, we must mention that it is trivial to verify that
Compatibility between left and right cosets
Consider the group \(G\) and \(H\leq G\). A natural question we may ask is: under which conditions is a left coset equal to a right coset? That is, under which conditions does \(aH=Ha\) hold? We may suspect a possible condition (perhaps necessary and sufficient) by looking at the definition of a normal subgroup. Does
hold?
Indeed it does. We must prove the two implications.
-
[\(\Rightarrow\))] Supposing \(H\unlhd G\), we have \(H=aHa^{-1}\) for all \(a\). Then the elements of \(H\) can be written as
\[ h'=aha^{-1}\ec \]whence
\begin{align*} Ha & = \set{h'a\mid h'\in H}\\ & = \set{(aha^{-1})a\mid h\in H}\\ & = \set{ah\mid h\in H}\\ & = aH\ep \end{align*} -
[\(\Leftarrow\))] Recall that
\[ aH=\set{ah\mid h\in H},\quad Ha=\set{h'a\mid h'\in H}\ec \]and supposing \(aH=Ha\) — that is, that these sets have the same elements — we can always write an element of \(aH\) as
\[ ah=h'a\ec \]whence
\begin{align*} aHa^{-1} & = \set{aha^{-1}\mid h\in H}\\ & = \set{(h'a)a^{-1}\mid h'\in H}\\ & = \set{h'\mid h'\in H}\\ & = H,\ \forall\,a\ec \end{align*}and thus \(H\unlhd G\).
In conclusion, we have found a characterization of the normality of a subgroup:
From Equation (4.36) a further characterization of the normality of a subgroup can be seen:
From Equation (4.36) (writing the equality of the two cosets elementwise).
The quotient group
Product of cosets
Consider a group \(G=\set{e,a,b,\ldots}\) and a subgroup \(H\leq G\). When we introduced the coset equivalence relation we did not analyse the quotient space. That quotient space is the set of all the cosets,
We define the IBO product of cosets as
This product provides yet another characterization of the normality of a subgroup:
The prescription \(aH\ast bH=abH\) needs a word of care before it can be called an operation at all. A coset is a set, and \(a\) is merely one of its elements; the recipe reads that element off, multiplies, and forms a new coset. Nothing so far guarantees that a different choice of representatives returns the same answer — and in fact that is exactly what normality is.
Let \(G\) be a group and \(H\leq G\). The prescription \(aH\ast bH=abH\) is independent of the representatives \(a\) and \(b\) chosen, and so defines an IBO on \(G/H\), if and only if \(H\unlhd G\). This is Equation (4.38). Rests on Definition 4.62, Definition 4.60 and Equation (4.35).
Derives Proposition 4.63. Throughout we use the fact that the cosets are the equivalence classes of Equation (4.34) (Proposition 4.61), so that
(\(\Leftarrow\)) Let \(H\unlhd G\) and let \(a'\in aH\), \(b'\in bH\), say \(a'=ah_1\) and \(b'=bh_2\) with \(h_1,h_2\in H\). Normality applied with \(p=b^{-1}\) gives \(H=b^{-1}Hb\), so \(b^{-1}h_1b\in H\); call this element \(h_3\), so that \(h_1b=bh_3\). Then
and \(h_3h_2\in H\) by closure, so \(a'b'\in abH\). By Equation (4.39), \(a'b'H=abH\): the answer does not depend on the representatives, and \(\ast\) is a well-defined IBO on \(G/H\).
(\(\Rightarrow\)) Suppose the prescription is independent of representatives, and let \(a\in G\) and \(h\in H\) be arbitrary. Since \(ah\in aH\) and \(a^{-1}\in a^{-1}H\), evaluating \(aH\ast a^{-1}H\) on the representatives \(ah\) and \(a^{-1}\) must give the same coset as evaluating it on \(a\) and \(a^{-1}\), that is,
the last step by Equation (4.35). By Equation (4.39) this says \(aha^{-1}\in H\). As \(h\) ranged over \(H\), we have \(aHa^{-1}\subseteq H\) for every \(a\in G\); applying the same inclusion with \(a^{-1}\) in place of \(a\) gives \(a^{-1}Ha\subseteq H\), which upon multiplying by \(a\) on the left and \(a^{-1}\) on the right reads \(H\subseteq aHa^{-1}\). The two inclusions give \(H=aHa^{-1}\) for every \(a\in G\), which is Definition 4.60.
∎The quotient group
The quotient space \(G/H\) by itself has no further relevance for our study; but if we add the condition that \(H\) be normal in \(G\), then \(G/H\), together with the product of cosets, is a group.
Let \(H\unlhd G\). Then \((G/H,\ast)\) is a group, known as the quotient group of \(G\) over \(H\). Rests on Definition 4.60, Equation (4.38) and Equation (4.35).
Derives Theorem 4.64. We verify the four axioms:
-
Taking \(H\) normal in \(G\), by Equation (4.38) we see the following: by the closure of \(G\), \(ab\) also belongs to \(G\); hence
\[ aH\ast bH=abH\in G/H\ec \]so the closure property holds for \((G/H,\ast)\).
-
The associativity of \((G/H,\ast)\) is a consequence of the associativity of \(G\). Indeed,
\begin{align*} (aH\ast bH)\ast cH & = abH\ast cH\\ & = (ab)cH\\ & = a(bc)H\\ & = aH\ast bcH\\ & = aH\ast(bH\ast cH)\ep \end{align*} -
The existence of the neutral element is due to the existence of the neutral element \(e\) in \(G\). From Equation (4.35) we see that \(eH=H\). Hence, for an arbitrary element of \(G/H\), say \(aH\), there exists \(H\in G/H\) such that
\[ H\ast aH=eH\ast aH=eaH=aH, \quad aH\ast H=aH\ast eH=aeH=aH\ep \] -
It remains to prove the existence of inverse elements. Let \(a\in G\). Since \(G\) is a group, \(a^{-1}\) exists. Now, \(a^{-1}H\) is the inverse of \(aH\), since
\[ aH\ast a^{-1}H=aa^{-1}H=H, \quad a^{-1}H\ast aH=a^{-1}aH=H\ep \]
Direct products
Direct product
Let \((G_1,\ast_1,e_1)\) and \((G_2,\ast_2,e_2)\) be groups. We define the direct product of the groups as the Cartesian product of the sets \(G_1\) and \(G_2\) together with the binary operation
\((G_1\times G_2,\ast,(e_1,e_2))\) is a group. Rests on Definitions 4.21 and 4.65.
Derives Proposition 4.66. We verify the four axioms:
-
The binary operation \(\ast\) is an IBO, since \(a_1\ast_1b_1\in G_1\) and \(a_2\ast_2 b_2\in G_2\) by closure. Hence always \((a_1\ast_1b_1,a_2\ast_2 b_2)\in G_1\times G_2\), so there is closure with respect to \(\ast\).
-
We have
\begin{align*} ((a_1,a_2)\ast(b_1,b_2))\ast(c_1,c_2) & = (a_1\ast_1b_1,a_2\ast_2b_2)\ast(c_1,c_2)\\ & = ((a_1\ast_1b_1)\ast_1c_1,(a_2\ast_2b_2)\ast_2c_2)\\ & = (a_1\ast_1(b_1\ast_1c_1),a_2\ast_2(b_2\ast_2c_2))\\ & = (a_1,a_2)\ast(b_1\ast_1c_1,b_2\ast_2c_2)\\ & = (a_1,a_2)\ast((b_1,b_2)\ast(c_1,c_2))\ec \end{align*}so associativity holds.
-
We can also see that the neutral element \((e_1,e_2)\in G_1\times G_2\) exists, since
\begin{align*} (a_1,a_2)\ast(e_1,e_2)&=(a_1\ast_1e_1,a_2\ast_2e_2)=(a_1,a_2)\ec\\ (e_1,e_2)\ast(a_1,a_2)&=(e_1\ast_1a_1,e_2\ast_2a_2)=(a_1,a_2)\ep \end{align*} -
Finally, given \((a_1,a_2)\in G_1\times G_2\), its inverse \((a_1,a_2)^{-1}=(a^{-1}_1,a^{-1}_2)\) exists. Indeed,
\[ (a_1,a_2)\ast(a_1,a_2)^{-1}=(a_1,a_2)\ast(a^{-1}_1,a^{-1}_2) =(a_1\ast_1a^{-1}_1,a_2\ast_2a^{-1}_2)=(e_1,e_2)\ec \]and analogously \((a_1,a_2)^{-1}\ast(a_1,a_2)=(e_1,e_2)\).
Consider a group \(G\) and \(H,K\leq G\). We define the product of subgroups \(HK\) as the set
Rests on Definition 4.36.
The direct product of two abelian groups is an abelian group. Rests on Definitions 4.8 and 4.65.
Derives Proposition 4.68.
Let \((G_1,\ast_1)\) and \((G_2,\ast_2)\) be abelian. That \(G_1\times G_2\) is a group is Proposition 4.66; commutativity is inherited componentwise, since for all \(a_1,b_1\in G_1\) and \(a_2,b_2\in G_2\)
the middle step applying Equation (4.3) in each factor separately.
∎Write \(\Z_n\) for the additive group \(\Z/n\Z\) of Proposition 4.41 and denote the class \(\overline{k}\) simply by \(k\). Both factors are abelian, so by Proposition 4.68 the eight-element group \(\Z_2\times\Z_4\) is abelian, and its Cayley table (Table 4.2) is symmetric about its diagonal. The operation is addition in each slot separately,
so the neutral element is \((0,0)\) and the inverse of \((i,j)\) is \((-i,-j)\) read in the respective factors.
Note that adding any element of this group to itself four times returns \((0,0)\), since \(4i\) is a multiple of \(2\) and \(4j\) a multiple of \(4\). No single element therefore runs through all eight, so \(\Z_2\times\Z_4\) is not isomorphic to \(\Z_8\) — two abelian groups of the same order need not be isomorphic, which is why the classification of finite abelian groups mentioned in Section 4.4 has content. Rests on Proposition 4.68, Proposition 4.41 and Definition 4.37.
| $\ast$ | $(0,0)$ | $(0,1)$ | $(0,2)$ | $(0,3)$ | $(1,0)$ | $(1,1)$ | $(1,2)$ | $(1,3)$ |
|---|---|---|---|---|---|---|---|---|
| $(0,0)$ | $(0,0)$ | $(0,1)$ | $(0,2)$ | $(0,3)$ | $(1,0)$ | $(1,1)$ | $(1,2)$ | $(1,3)$ |
| $(0,1)$ | $(0,1)$ | $(0,2)$ | $(0,3)$ | $(0,0)$ | $(1,1)$ | $(1,2)$ | $(1,3)$ | $(1,0)$ |
| $(0,2)$ | $(0,2)$ | $(0,3)$ | $(0,0)$ | $(0,1)$ | $(1,2)$ | $(1,3)$ | $(1,0)$ | $(1,1)$ |
| $(0,3)$ | $(0,3)$ | $(0,0)$ | $(0,1)$ | $(0,2)$ | $(1,3)$ | $(1,0)$ | $(1,1)$ | $(1,2)$ |
| $(1,0)$ | $(1,0)$ | $(1,1)$ | $(1,2)$ | $(1,3)$ | $(0,0)$ | $(0,1)$ | $(0,2)$ | $(0,3)$ |
| $(1,1)$ | $(1,1)$ | $(1,2)$ | $(1,3)$ | $(1,0)$ | $(0,1)$ | $(0,2)$ | $(0,3)$ | $(0,0)$ |
| $(1,2)$ | $(1,2)$ | $(1,3)$ | $(1,0)$ | $(1,1)$ | $(0,2)$ | $(0,3)$ | $(0,0)$ | $(0,1)$ |
| $(1,3)$ | $(1,3)$ | $(1,0)$ | $(1,1)$ | $(1,2)$ | $(0,3)$ | $(0,0)$ | $(0,1)$ | $(0,2)$ |
Given a group \(G\) with subgroups \(H,K\leq G\), the map
is an isomorphism — in which case \(G\) is called the internal direct product of the subgroups \(H\) and \(K\) — if and only if
-
\(G=HK\);
-
\(H\cap K=\set{e}\);
-
\(H\unlhd G\) and \(K\unlhd G\).
Derives Proposition 4.70.
Throughout, \(H\times K\) carries the operation of Definition 4.65, whose neutral element is \((e,e)\).
(\(\Leftarrow\)) Assume (1), (2) and (3). We first show that every element of \(H\) commutes with every element of \(K\). Let \(h\in H\), \(k\in K\) and put
Read as \(c=\left(hkh^{-1}\right)k^{-1}\), the normality of \(K\) gives \(hkh^{-1}\in K\), so \(c\in K\); read as \(c=h\left(kh^{-1}k^{-1}\right)\), the normality of \(H\) gives \(kh^{-1}k^{-1}\in H\), so \(c\in H\). Hence \(c\in H\cap K=\set{e}\) by (2), that is,
Now \(\varphi\) is a homomorphism: for \(h,h'\in H\) and \(k,k'\in K\),
It is surjective by (1), since \(G=HK\) is exactly the set of products \(hk\). It is injective because its kernel is trivial: if \(\varphi(h,k)=e\) then \(h=k^{-1}\), and \(k^{-1}\in K\), so \(h\in H\cap K=\set{e}\); thus \(h=e\) and then \(k=e\), that is, \(\ker\varphi=\set{(e,e)}\), and Proposition 4.17 applies. A bijective homomorphism is an isomorphism (Definition 4.44).
(\(\Rightarrow\)) Assume \(\varphi\) is an isomorphism. Surjectivity says that every \(g\in G\) can be written \(g=hk\) with \(h\in H\) and \(k\in K\), which is \(G=HK\): this is (1). For (2), let \(x\in H\cap K\). Then \(x^{-1}\in K\), and
so injectivity forces \(\left(x,x^{-1}\right)=(e,e)\) and \(x=e\).
For (3), note that \(H\times\set{e}\) is a normal subgroup of \(H\times K\), since
and that \(\varphi\left(H\times\set{e}\right)=H\), because \(\varphi(h,e)=he=h\). Let \(g\in G\) and write \(g=\varphi(u)\) with \(u\in H\times K\), which is possible by surjectivity. Since \(\varphi\) is a homomorphism, \(\varphi(u)^{-1}=\varphi\left(u^{-1}\right)\) (Proposition 4.47), and therefore
so \(H\unlhd G\). The same argument with \(\set{e}\times K\) gives \(K\unlhd G\).
∎It is essential that the proposition names the map Equation (4.42). The weaker reading — that \(G\) be isomorphic to \(H\times K\) by some isomorphism or other — does not imply (1)–(3), and the difference is not a technicality. Take \(G=\Z_2\times\Z_2\) and let \(H=K=\set{(0,0),(1,0)}\leq G\). Each of \(H\) and \(K\) is a two-element group, hence isomorphic to \(\Z_2\) by the uniqueness of the two-element Cayley table established in Section 4.5.1, so \(H\times K\) is isomorphic to \(\Z_2\times\Z_2=G\); yet \(HK=H\neq G\) and \(H\cap K=H\neq\set{e}\). An abstract isomorphism knows nothing about how \(H\) and \(K\) sit inside \(G\), and it is exactly that placement which conditions (1)–(3) describe.
Semidirect product
The direct product is symmetric in its two factors: Proposition 4.70 demands that both \(H\) and \(K\) be normal in \(G\), and Equation (4.43) then forces every element of one to commute with every element of the other. That is a strong condition, and most of the groups physics actually uses fail it. In the group of rigid motions of space, a rotation and a translation do not commute; the translations form a normal subgroup, and the rotations do not. What is needed is a product in which one factor is normal and the other merely acts on it. That is the semidirect product, and it is the shape of the Euclidean and Poincaré groups alike.
Semidirect product of groups; characterization of the semidirect product
The action of one group on another is by automorphisms, so we first note that the automorphisms of a group form a group in their own right.
Let \(N\) be a group. The set \(\operatorname{Aut}(N)\) of all automorphisms of \(N\) is a subgroup of the symmetric group \(S_N\), and in particular a group under composition. Rests on Definitions 4.36, 4.45 and 4.48.
Derives Proposition 4.72. An automorphism of \(N\) is in particular a bijection of \(N\) onto itself, so \(\operatorname{Aut}(N)\subseteq S_N\). The identity map is an automorphism, so the neutral element of \(S_N\) lies in \(\operatorname{Aut}(N)\). If \(\alpha,\beta\in\operatorname{Aut}(N)\) then for all \(n_1,n_2\in N\)
so \(\alpha\circ\beta\) is a homomorphism, and it is a bijection as a composite of bijections. Finally, since \(\alpha\) is a homomorphism,
and applying \(\alpha^{-1}\) to both sides gives \(\alpha^{-1}(n_1)\,\alpha^{-1}(n_2)=\alpha^{-1}(n_1n_2)\), so \(\alpha^{-1}\in\operatorname{Aut}(N)\). The three conditions of Definition 4.36 are met.
∎Let \(N\) and \(H\) be groups and let \(\varphi:H\longrightarrow\operatorname{Aut}(N)\) be a group homomorphism, written \(h\longmapsto\varphi_h\). The external semidirect product \(N\rtimes_{\varphi}H\) is the Cartesian product \(N\times H\) equipped with the internal binary operation
Rests on Proposition 4.72, Definition 4.43 and Definition 4.1.
The first slot is not multiplied blindly: the second factor of the left element is allowed to act on the first slot of the right element before the two are combined. Setting \(\varphi_h=\id_N\) throughout removes the action and returns Definition 4.65, as Proposition 4.77 makes precise.
\(\left(N\rtimes_{\varphi}H,\ast\right)\) is a group, with neutral element \((e_N,e_H)\) and with
Derives Proposition 4.74. We use twice over that \(\varphi\) is a homomorphism into \(\operatorname{Aut}(N)\): each \(\varphi_h\) is an automorphism of \(N\), and \(\varphi_{h_1h_2}=\varphi_{h_1}\circ\varphi_{h_2}\). In particular \(\varphi_{e_H}=\id_N\) and \(\varphi_h(e_N)=e_N\), by Proposition 4.46 applied to \(\varphi\) and to \(\varphi_h\) respectively.
Closure. \(\varphi_{h_1}(n_2)\in N\) and \(h_1h_2\in H\), so the right-hand side of Equation (4.44) lies in \(N\times H\).
Associativity. On the one hand,
and on the other,
the third step because \(\varphi_{h_1}\) is a homomorphism of \(N\) and the fourth because \(\varphi\) is a homomorphism into \(\operatorname{Aut}(N)\). The two agree.
Neutral element. Using \(\varphi_h(e_N)=e_N\) and \(\varphi_{e_H}=\id_N\),
Inverses. With the candidate Equation (4.45),
and in the other order
All four axioms of Definition 4.21 hold.
∎Write \(G=N\rtimes_{\varphi}H\), \(\widetilde{N}=N\times\set{e_H}\) and \(\widetilde{H}=\set{e_N}\times H\). Then \(\widetilde{N}\unlhd G\), \(\widetilde{H}\leq G\), \(\widetilde{N}\cap\widetilde{H}=\set{(e_N,e_H)}\) and \(G=\widetilde{N}\widetilde{H}\); moreover \(n\longmapsto(n,e_H)\) and \(h\longmapsto(e_N,h)\) are isomorphisms of \(N\) onto \(\widetilde{N}\) and of \(H\) onto \(\widetilde{H}\), and conjugation inside \(G\) reproduces \(\varphi\):
Rests on Proposition 4.74, Definition 4.60 and Definition 4.67.
Derives Proposition 4.75. By Equation (4.44), \((n_1,e_H)\ast(n_2,e_H)=\left(n_1\varphi_{e_H}(n_2),e_H\right) =(n_1n_2,e_H)\) and \((e_N,h_1)\ast(e_N,h_2)=\left(\varphi_{h_1}(e_N),h_1h_2\right) =(e_N,h_1h_2)\), so both maps are homomorphisms; each is evidently bijective onto its image, hence an isomorphism (Definition 4.44), and each image contains the neutral element and the inverses of its elements, so both are subgroups. Their intersection is \(\set{(e_N,e_H)}\) because a pair lying in both has \(h=e_H\) and \(n=e_N\), and \((n,e_H)\ast(e_N,h)=\left(n\varphi_{e_H}(e_N),h\right)=(n,h)\) shows \(G=\widetilde{N}\widetilde{H}\).
For normality, let \((m,k)\in G\) and \((n,e_H)\in\widetilde{N}\). Then
which lies in \(\widetilde{N}\); hence \(\widetilde{N}\unlhd G\). Finally, \((e_N,h)^{-1}=\left(\varphi_{h^{-1}}(e_N),h^{-1}\right)=(e_N,h^{-1})\), so
which is Equation (4.46).
∎The construction is therefore faithful to its data, and the converse holds too: whenever a group contains a normal subgroup and a complement to it, it is a semidirect product. Note that the complement is not required to be normal — that is the whole difference from Proposition 4.70.
Let \(G\) be a group, \(N\unlhd G\) and \(H\leq G\) with \(G=NH\) and \(N\cap H=\set{e}\). Then
defines a homomorphism \(\varphi:H\longrightarrow\operatorname{Aut}(N)\), and
is an isomorphism. Conversely, by Proposition 4.75, every external semidirect product is of this form. Rests on Definitions 4.60, 4.67 and 4.73.
Derives Theorem 4.76. \(\varphi_h\) is an automorphism of \(N\). Since \(N\unlhd G\) and \(h\in G\), we have \(hnh^{-1}\in N\), so \(\varphi_h\) maps \(N\) into \(N\); it is a homomorphism because
and \(\varphi_{h^{-1}}\) is a two-sided inverse for it, so it is bijective.
\(\varphi\) is a homomorphism. For \(h_1,h_2\in H\) and \(n\in N\),
using Equation (4.16) for \((h_1h_2)^{-1}\); hence \(\varphi_{h_1h_2}=\varphi_{h_1}\circ\varphi_{h_2}\).
\(\psi\) is a homomorphism. Using Equation (4.44) and then Equation (4.47),
\(\psi\) is bijective. It is surjective because \(G=NH\) is precisely the set of products \(nh\). Its kernel is trivial: if \(nh=e\) then \(n=h^{-1}\), which lies in \(N\cap H=\set{e}\), so \(n=e\) and then \(h=e\). By Proposition 4.17, \(\psi\) is injective, and a bijective homomorphism is an isomorphism.
∎Let \(G=N\rtimes_{\varphi}H\) and let \(\widetilde{N}\), \(\widetilde{H}\) be as in Proposition 4.75. The following are equivalent:
-
\(\widetilde{H}\unlhd G\);
-
\(\varphi_h=\id_N\) for every \(h\in H\);
-
the operation Equation (4.44) is the operation of Definition 4.65, so that \(N\rtimes_{\varphi}H\) is the direct product \(N\times H\).
Rests on Proposition 4.75, Definition 4.65 and Definition 4.60.
Derives Proposition 4.77. \((2)\implies(3)\): substituting \(\varphi_{h_1}=\id_N\) into Equation (4.44) gives \((n_1,h_1)\ast(n_2,h_2)=(n_1n_2,h_1h_2)\), which is Definition 4.65.
\((3)\implies(1)\): in the direct product,
so \(\widetilde{H}\unlhd G\).
\((1)\implies(2)\): let \(h\in H\) and \(n\in N\). Since \((n,e_H)^{-1}=\left(\varphi_{e_H}\!\left(n^{-1}\right),e_H\right) =(n^{-1},e_H)\),
If \(\widetilde{H}\) is normal this must lie in \(\set{e_N}\times H\), that is, \(n\,\varphi_h\!\left(n^{-1}\right)=e_N\), or \(\varphi_h\!\left(n^{-1}\right)=n^{-1}\). As \(n\) runs over \(N\) so does \(n^{-1}\) (Proposition 4.26), hence \(\varphi_h=\id_N\) for every \(h\).
∎Let \(N=\left(\R^{3},+\right)\), the translations of Euclidean space, and let \(H=\SO(3)\), the rotations about a fixed origin. A rotation \(R\) is a linear bijection of \(\R^{3}\), so \(\varphi_R(\vect{a})=R\vect{a}\) is an automorphism of the additive group \(\R^{3}\), and \(\varphi_{R_1R_2}=\varphi_{R_1}\circ\varphi_{R_2}\) because matrix multiplication is associative. The semidirect product
therefore exists, and by Equation (4.44) its operation is
which is exactly the composition of the rigid motions \(\vect{x}\longmapsto R\vect{x}+\vect{a}\) of \(\R^{3}\): applying \(\left(\vect{a}_2,R_2\right)\) and then \(\left(\vect{a}_1,R_1\right)\) sends \(\vect{x}\) to \(R_1R_2\vect{x}+R_1\vect{a}_2+\vect{a}_1\). The same construction with \(\Ogrp(3)\) in place of \(\SO(3)\) adjoins the reflections and gives the full group of distance-preserving maps of \(\R^{3}\), the Euclidean group proper,
of which \(\mathrm{SE}(3)\) is the orientation-preserving subgroup of index two; the isometries of the equilateral triangle in Proposition 4.55 are the planar analogue. The naming follows the convention Equation (4.51) uses below and Definition 39.10 fixes: an inhomogeneous group \(\ISO(p,q)\) carries the full orthogonal group of its form, reflections included, and the prefix \(\mathrm{S}\) marks the orientation-preserving subgroup.
The product is genuinely semidirect and not direct. Conjugating a rotation by a translation gives, by the computation in Proposition 4.77,
whose translation part vanishes only when \(R\vect{a}=\vect{a}\). The rotations therefore do not form a normal subgroup — a rotation about the origin, viewed from a translated origin, is a rotation about a different point — while the translations do, by Proposition 4.75. This is the algebraic content of the familiar fact that rotating and then translating is not the same as translating and then rotating. Rests on Definition 4.73 and Proposition 4.77.
The same construction with the translations of Minkowski space and the Lorentz group in place of the rotations gives the symmetry group of special relativity. Take \(N=\left(\R^{4},+\right)\) and for \(H\) the Lorentz group \(L\) of Definition 39.1, whose elements are invertible linear maps and therefore automorphisms of the additive group \(\R^{4}\), with \(\varphi_{\Lambda}(a)=\Lambda a\). Then
and Equation (4.44) reads
which is the composition law Equation (39.17) of the Poincaré transformations, defined and used in Minkowski Space and Its Symmetries; restricting \(L\) to its proper subgroup \(\SO(3,1)\) gives the proper Poincaré group in the same way. That the translations are normal and the Lorentz transformations are not is precisely why Equation (39.17) is not the componentwise product of a direct product, by Proposition 4.77. The infinitesimal counterpart of Equation (4.51) is the semidirect sum Equation (14.99) of Lie Groups, Lie Algebras, and Fibre Bundles, in which the translation generators span an abelian ideal on which the Lorentz generators act. Rests on Definition 4.73 and Proposition 4.75.