551, Fall '00                   A1b.    Isomorphism                   p. 1

Math 551 – Algebra – Fall 2000

A. Groups

1. Definition and examples.

Definition. A group is a set G together with a binary operation G × G → G, written as multiplication, such that

(1) The operation is associative: g(hk) = (gh)k for all g, h, k ∈ G;

(2) There exists 1 ∈ G such that g1 = 1g = g for all g ∈ G;

(3) For every g ∈ G there is g^{−1} ∈ G such that gg^{−1} = g^{−1}g = 1.

1a. Symmetric groups Ex. A. Let X be any set. A permutation of X is a bijection σ : X → X, i.e., a mapping on X

which is one-to-one and onto X. Let

Σ_{X} = {σ : X → X | σ is a permutation of X}.

For σ, τ ∈ Σ_{X} define στ = σ ∘ τ, the composite of σ and τ. Also let 1_{X} be the identity mapping 1_{X}(x) = x, x ∈ X, and for each σ ∈ Σ_{X}, let σ^{−1} be the inverse mapping: σ^{−1}(x) = y ⟺ σ(y) = x. Then Σ_{X} is a group.

Remarks: Composition of mappings is always associative, whether or not the mappings are injective or surjective. If X is finite, then |Σ_{X}| = |X|!. We write mappings on the left, so σ ∘ τ(x) = σ(τ(x)).

1b. Isomorphism

Definition. Let G and H be groups. An isomorphism from G to H is a mapping ϕ : G → H such that

1) ϕ(xy) = ϕ(x)ϕ(y) for all x, y ∈ G.

2) ϕ is a bijection.

We write G ≅ H if and only if there exists an isomorphism from G to H. In 1), the products xy and ϕ(x)ϕ(y) are in G and H, respectively. The relation ≅ is then reflexive, symmetric and transitive on the class of all groups. Indeed one sees quickly from the definition of isomorphism that

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1) For any G, id_{G} : G → G is an isomorphism;

2) If ϕ : G → H is an isomorphism, then ϕ^{−1} : H → G is an isomorphism; and

3) If ϕ : G → H and ψ : H → K are isomorphisms, then ψ ∘ ϕ : G → K is an

isomorphism. and these facts imply respectively that ≅ is reflexive, symmetric and transitive. Consequently (ignoring set-theoretic difficulties) the class of all groups is partitioned into “isomorphism classes” or “isomorphism types”. As an example of isomorphisms, let X and Y be sets and suppose that there exists a bijection f : X → Y. Define ϕ_{f} : Σ_{X} → Σ_{Y} by ϕ_{f}(σ) = f ∘ σ ∘ f^{−1}. Notice that the right side is a composite of bijections so is a bijection, and maps Y to Y. Then ϕ_{f} is an isomorphism. Indeed ϕ_{f}(στ) = (fσf^{−1})(fτf^{−1}) = fστf^{−1}. Moreover, ϕ_{f}ϕ_{f^{−1}} : Σ_{Y} → Σ_{X} → Σ_{Y} takes σ to ϕ_{f}(f^{−1}σf^{−1}^{−1}) = ff^{−1}σf^{−1}^{−1}f^{−1} = σ, so ϕ_{f}ϕ_{f^{−1}} = 1_{Σ_{Y}}, and similarly ϕ_{f^{−1}}ϕ_{f} = 1_{Σ_{X}}. Hence ϕ_{f} is a bijection (with inverse ϕ_{f^{−1}}). Thus if X and Y are two sets of the same cardinality, then Σ_{X} ≅ Σ_{Y}. If ϕ : G → H is an isomorphism, and ϕ(1_{G}) = h, then h^{2} = ϕ(1_{G})ϕ(1_{G}) = ϕ(1_{G}^{2}) = ϕ(1_{G}) = h, so h = 1_{H}. Likewise for any x ∈ G, ϕ(x)ϕ(x^{−1}) = ϕ(xx^{−1}) = ϕ(1_{G}) = 1_{H} and vice-versa, so ϕ(x^{−1}) = ϕ(x)^{−1}. We shall consider isomorphic groups to be the “same”, and group theory is the study of those properties of groups which are invariant under isomorphism. We write Σ_{n} for Σ_{{1,2,...,n}}.

1c. Rudiments Note that 1_{X} is the only mapping such that 1_{X}σ = σ for all σ ∈ Σ_{X}, and f^{−1} is the unique inverse of f.

Proposition. Let G be a group. Then

1) 1_{G} is the only identity element of G; indeed it is the unique right identity element of G and the unique left identity element.

2) For each x ∈ G, x^{−1} is the unique inverse of x; indeed it is the unique right inverse and the unique left inverse.

3) If g, h ∈ G then the equation gx = h has a unique solution in G, namely x = g^{−1}h.

4) If n ≥ 3 and x_{1}, …, x_{n} ∈ G, then any two associations of x_{1} … x_{n} represent the same element of G.

5) If x, y ∈ G then (xy)^{−1} = y^{−1}x^{−1} and (x^{−1})^{−1} = x.

Proof. 1) If e is a right identity then 1_{G} = 1_{G}e = e, and similarly for left identity elements.

2) If y is a right inverse then x^{−1} = x^{−1}1_{G} = x^{−1}(xy) = (x^{−1}x)y = 1_{G}y = y. Similarly for left inverses.

3) Trivial

 THIS IS A PRIVATE DRAFT, NOT FOR PUBLIC CONSUMPTION. AUTHOR IS NOT RESPONSIBLE FOR ERRORS!!

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4) Induction on n, starting with n = 3, where it is the associative law. Show that the element represented by any association of x_{1} … x_{n} equals the element represented by the standard association [x_{1} ⋯ x_{n}] = x_{1}(x_{2}(x_{3}(⋯ (x_{n−1}x_{n}) ⋯)). Namely any association equals a product of associations of x_{1}, …, x_{k} and x_{k+1}, …, x_{n}, so equals [x_{1} ⋯ x_{k}][x_{k+1} ⋯ x_{n}] by induction. This equals

(x_{1}[x_{2} ⋯ x_{k}])[x_{k+1} ⋯ x_{n}] = x_{1}([x_{2} ⋯ x_{k}][x_{k+1} ⋯ x_{n}]) = x_{1}[x_{2} ⋯ x_{n}] = [x_{1} ⋯ x_{n}],

the three steps by associativity, induction and definition, respectively.

5) xy(y^{−1}x^{−1}) = (y^{−1}x^{−1})xy = 1 and xx^{−1} = x^{−1}x = 1. QED

Ex. B. Let G be a group. A subset H of G which is itself a group with respect to the same operation is a subgroup of G. (Notation: H ≤ G.)

For instance, if G = Σ_{X} and x ∈ X, then G_{x} = {σ ∈ Σ_{X} | σ(x) = x is a subgroup of G. It is isomorphic to Σ_{X−{x}}, via the isomorphism G_{x} → Σ_{X−{x}} taking σ ↦ σ|X − {x}.

Proposition. If G is a group and H ⊆ G, then H ≤ G if and only if H is nonempty and closed under multiplication and inversion. In that case 1_{H} = 1_{G} and for x ∈ H, x^{−1} has the same meaning in both H and G.

Proof. Left to reader. QED

Theorem. (Cayley) Let G be a group. For each g ∈ G let λ_{g} ∈ Σ_{G} be defined by λ_{g}(h) = gh. Then G^{G} = {λ_{g} | g ∈ G} is a subgroup of Σ_{G} and is isomorphic to G.

Proof. Notice that λ_{g′}(λ_{g}(h)) = λ_{g′}(gh) = g′(gh) = (g′g)h = λ_{g′g}(h). So

λ_{g′} ∘ λ_{g} = λ_{g′g}.

Similarly,

λ_{1_{G}} = id_{G}, whence λ_{g}λ_{g^{−1}} = λ_{gg^{−1}} = λ_{1_{G}} = id_{G} = λ_{g^{−1}}λ_{g}.

It follows that each λ_{g} is bijective with inverse λ_{g^{−1}}, so λ_{g} ∈ Σ_{G}. Closure, identity and inverse follow easily. Define λ : G → G^{G} by λ(g) = λ_{g}. The first displayed equation is the first condition for an isomorphism. λ is surjective by definition of G^{G}. And if λ(g) = λ(g′), then g = λ_{g}(1_{G}) = λ_{g′}(1_{G}) = g′, so λ is bijective and hence an isomorphism. QED

This is generally not very useful, except for philosophical reasons: groups are inherently groups of permutations, or can be viewed as such.

1d. More examples

Ex. C. Let X be a set “with structure”. A mapping σ ∈ Σ_{X} is an automorphism of X if and only if it “preserves” the structure. Aut(X) is the set of all automorphisms of X. Then Aut(X) is a group, indeed a subgroup of Σ_{X}.

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Ex. C′. Let G be a group. An automorphism of G is an isomorphism from G to G. Then Aut(G), the set of automorphisms of G under composition, is a group. (Check this. Need the inverse of an automorphism is an automorphism.)

Ex. C′′. Let R be Rubik’s cube; more precisely the set {s_{1}, …, s_{54}} of all 54 positions which can be held by a colored square on the surface of the cube. Let G(R) be the set of all σ ∈ Σ_{54} such that by legitimate moves one may transform a given configuration to a new one in which the square at position i has been moved to position σ(i), i = 1, …, 54. Then G(R) is a group. (It has order 2^{10}3^{7} · 8! · 12!, if memory serves.)

Ex. D. Let n be an integer and G = GL_{n}(C) the set of n × n nonsingular matrices over C, the complex field. Then G is a group (under matrix multiplication). A subgroup is SL_{n}(C) = {g ∈ GL_{n}(C) | det g = 1}.

The field C can be replaced by R or any commutative ring here. Also Cayley’s theorem has an analogue here: any finite group G is isomorphic to a subgroup of GL_{|}G|(C) (consisting of the permutation matrices, rows and columns indexed by G, corresponding to the permutations λ_{g}, g ∈ G.

Ex. D′. An example similar to, but different from, GL_{2}(C) is the group of fractional linear transformations of the complex plane, that is, the group LF (2, C) of all permutations σ of the extended complex plane (Riemann sphere) C ∪ {∞} of the form

σ(z) = \frac{az + b}{cz + d}

such that a, b, c, d are complex constants for which ad − bc ≠ 0. This last condition prevents σ from being a constant mapping, so is actually redundant. We may write σ = σ_{A}, where A is the “matrix of coefficients”

A = [ \begin{matrix}a & b \\ c & d \end{matrix} ] .

It is then easy to check that

LF (2, C) = {σ_{A} | A ∈ GL_{2}(C)}, and σ_{A}σ_{B} = σ_{AB}∀A, B ∈ GL_{2}(C).

However, the mapping

ϕ : GL_{2}(C) → LF (2, C) defined by ϕ(A) = σ_{A},

is not an isomorphism, because it fails to be injective. In fact σ_{cA} = σ_{A} for every A and for every c ∈ C^{×}.

Exercise. Show that for any A, B ∈ GL_{2}(C), we have σ_{A} = σ_{B} if and only if B = cA for some c ∈ C^{×}.

1e. Powers; cyclic groups

Ex. E. Let G be any group and g ∈ G. Define g^{n} for n ∈ Z as follows: g^{0} = 1_{G} and g^{n+1} = g^{n}g for n ≥ 0; g^{n} = (g^{−n})^{−1} for n < 0.

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Proposition. Let g ∈ G and m, n ∈ Z. Then g^{m}g^{n} = g^{m+n}, and (g^{m})^{−1} = g^{−m}. Moreover (g^{m})^{n} = g^{mn}.

This is a consequence of 4) of the Proposition above, when m and n have the same sign; otherwise, there are several cases, e.g. if n ≥ |m| with m < 0, then g^{n} = g^{−m}g^{m+n}, so g^{m}g^{n} = g^{m}g^{−m}g^{m+n} = g^{m+n}. The remaining assertions are left to the reader to prove.

The set {g^{n} | n ∈ Z} is a subgroup of G, and is denoted by ⟨g⟩. A group arising in this way–as the powers of a single element–is called cyclic, and g is called a generator.

Examples of cyclic groups: Z, the set of integers, with respect to addition. 1 is a generator, 0 = 1_{Z}, and usually one writes g + h for gh and −g for g^{−1}. Also Z_{n}, the integers mod n (here n is a positive integer), with elements [i], i ∈ Z, and [i] = [j] if and only if i ≡ j mod n;

[i] + [j] = [i + j]. A generator is [1].

Proposition. Every cyclic group G is isomorphic to exactly one of the groups Z, Z_{n}, n = 1, 2, . . .. Moreover if G = ⟨g⟩, then there is a unique isomorphism Z → G or Z_{n} → G, as the case may be, taking 1 ↦ g or [1] ↦ g.

Proof. If g^{n} ≠ 1 for all n ≠ 0, then the elements g^{n}, n ∈ Z are all different (for g^{n} = g^{m} would imply g^{n−m} = g^{n}(g^{n})^{−1} = 1 and hence n − m = 0). Map Z → G by n ↦ g^{n}. This is clearly injective, surjective, and multiplicative.

If g^{n} = 1 for some n ≠ 0, then g^{−n} = (g^{n})^{−1} = 1, so we may take n > 0. We may also assume that n is the smallest positive integer such that g^{n} = 1. Then the elements g^{i}, 0 ≤ i < n, are all distinct, by a similar argument, and g^{i} = g^{j} ⟺ i ≡ j mod n. The mapping Z_{n} → G taking [m] ↦ g^{m}, 0 ≤ m < n is then well-defined and bijective, and g^{m}g^{m′} = g^{m+m′}, providing the isomorphism.

The uniqueness of the isomorphism is clear since if 1 or [1] maps to g, then m = 1 + ⋯ + 1 or [m] = [1] + ⋯ + [1] maps to g ⋯ g = g^{m} by repeated use of the multiplicativity of the isomorphism. QED

Exercise. The distinct subgroups of Z are the subgroups ⟨n⟩, n > 0. The distinct subgroups of Z_{n} are the subgroups ⟨[m]⟩, m > 0, m dividing n. Consequently every subgroup of a cyclic group is cyclic; and a cyclic group of order n has a subgroup of order m if and only if m divides n.

As a corollary we can define the order of an element.

Definition. Let G be a group and g ∈ G. The order of g is |g| = |⟨g⟩|, the cardinality of ⟨g⟩.

Thus the order of g is a positive integer or ∞. In a finite group, every element has finite order.

Exercise. If g ∈ G and g^{n} = 1 for some integer n ≠ 0, then the order of g is finite and divides n.

1f. Generation

Ex. F. For this example we need a lemma.

Lemma. The intersection of any collection of subgroups of a group G is a subgroup of G.

Proof. Left to reader. QED

Definition. Let G be a group and S any subset of G. The intersection of all subgroups of G containing S (G is one of them) is a subgroup of G called ⟨S⟩, the subgroup of G generated by S.

Proposition. ⟨S⟩ = {s_{1}^{ϵ_{1}}s_{2}^{ϵ_{2}} ⋯ s_{n}^{ϵ_{n}} | n ≥ 0, s_{i} ∈ S and ϵ_{i} = ±1 ∀i}.

The expressions on the right are called “words in S ∪ S^{−1}”. Here if n = 0, the empty product is interpreted as 1_{G}.

Proof. Since (xy)^{−1} = y^{−1}x^{−1} and (x^{−1})^{−1} = x, the right side H is obviously a subgroup of G, and S ⊆ H via words of length 1. Therefore ⟨S⟩ ≤ H. On the other hand ⟨S⟩ contains each s^{±1}, s ∈ S, and hence contains all the elements of H, as it is closed under inversion and products. QED

Though the situation is nice for cyclic groups, those generated by one element, it quickly becomes intractable for groups generated by two or more elements.

Exercise. Σ_{n} is generated by two elements.

Hint: Every permutation is the composition of “transpositions”, i.e., interchanges of pairs of objects. The transposition (12) and the cycle (12 ⋯ n) generate a subgroup containing all transpositions and hence generate Σ_{n}.

Furthermore, it can be shown that lim_{n→∞} P (⟨x, y⟩ = Σ_{n}) = 1, for x and y chosen (uniformly) randomly and independently in Σ_{n}.

Another example of this is

SL_{2}(Z) = ⟨[ \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ] , [ \begin{matrix}0 & 1 \\ −1 & −1 \end{matrix} ]⟩ .

Taking matters to larger extremes, Burnside proposed the following problem a hundred years ago.

Let m and n be positive integers. Let G be a group such that g^{n} = 1 for all g ∈ G, and G = ⟨S⟩ for some set of cardinality m. Is G necessarily finite?

The answer turns out to be “no” in general, although for some very small values of n, it is “yes”. Of course it is “yes” for m = 1. It is not known for (m, n) = (2, 5).

1g. Abelian groups

Ex. G. However, the situation is better controlled for abelian groups, those which satisfy the commutative law.

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Definition. A group G is commutative (or abelian) if and only if xy = yx for all x, y ∈ G.

Exercise. If G is abelian then (xy)^{m} = x^{m}y^{m} for all x, y ∈ G and m ∈ Z. Moreover if G is abelian and generated by elements x_{1}, …, x_{r} and x_{i}^{n} = 1 for all i, then G is finite and its cardinality is at most n^{r}.

Examples of abelian groups are cyclic groups, the additive group of the rationals, reals, complexes (or any field), the multiplicative group of nonzero rationals, reals, complexes (or any field), the multiplicative group of all roots of unity in C (or in any field), and the matrix group

U = \{[ \begin{matrix}1 & c \\ 0 & 1 \end{matrix} ] | c ∈ C\} .

Exercise. If G is an abelian group, then any subgroup of G is abelian.

Definition. A periodic group (torsion group) is one all of whose elements have finite order. The exponent of a torsion group is the least common multiple of the orders of its elements (or ∞, if there is no such common multiple). The exponent of a non-torsion group is ∞.

Exercise. Suppose that G = ⟨S⟩. If G is abelian, then the exponent of G is the least common multiple of the orders of the elements of S.

Exercise. A group of exponent 2 is necessarily abelian. There exist nonabelian groups of exponent 3.

1h. Dihedral groups

Ex. H. Let n > 2 be an integer and let D_{2n} be the group of symmetries of the regular n-gon. Thus D_{2n} contains rotations through integer multiples of 2π/n, which form a cyclic subgroup of cardinality n, generated by rotation σ through 2π/n. In addition there are the reflections about an axis of symmetry; there are n of these as well. Notice that if τ is one such reflection, then σ^{i}τ is also a reflection. There are no other symmetries, so D_{2n} = {σ^{1}, σ^{2}, …, σ^{n}, σ^{1}τ, …, σ^{n}τ} = {σ^{i}τ^{j} | 0 ≤ i < n, 0 ≤ j ≤ 1}. The multiplication is determined by the following rules:

σ^{n} = 1, τ^{2} = 1, τστ = σ^{−1}.

For then

σ^{i}τ^{j}σ^{k}τ^{ℓ} = \{ \begin{matrix}σ^{i+k}τ^{ℓ} & if j = 0 \\ σ^{i−k}τ^{1+ℓ} & if j = 1 \end{matrix}

Notice however that D_{2n} = ⟨τ, στ⟩, and both of these elements have order 2.

1i. Direct products Ex. I. Let G and H be groups. The (external) direct product G × H is the group based on

the set which is the Cartesian product {(g, h) | g ∈ G, h ∈ H} and with multiplication (g, h)(g′, h′) = (gg′, hh′). Then G × H has subgroups G_{1} = {(g, 1) | g ∈ G} isomorphic to G and H_{1} = {(1, h) | h ∈ H} isomorphic to H, every element of G is the product of an element of G_{1} and one of H_{1}, and g_{1}h_{1} = h_{1}g_{1} for all g_{1} ∈ G_{1}, h_{1} ∈ H_{1}. But any nonabelianness of G and H is preserved.

Indeed if {G_{i} | i ∈ I} is a family of groups indexed by the set I, then ∏_{i∈I} G_{i} is the set

of all functions f : I → ∪G_{i} such that f(i) ∈ G_{i} for all i; multiplication is pointwise: (ff′)(i) = f(i)f′(i). Within this is the restricted direct product, the subgroup consisting of all elements f such that f(i) = 1_{G_{i}} for all but finitely many i ∈ I.

We shall see that every finite (or even finitely generated) abelian group is isomorphic to the direct product of finitely many cyclic groups. But nothing like this is true for infinitely generated groups or for finite nonabelian groups.

Ex. J. Let G = Σ_{X} and let Y be a subset of X, ∅ ≠ Y ≠ X. Let G_{[Y]} = {σ ∈ G | σ(Y) = Y }. Then G_{[Y]} ≤ G. Moreover, the mapping

G_{[Y]} ≅ Σ_{Y} × Σ_{X−Y}

via the isomorphism σ ↦ (σ|Y, σ|(X − Y)). Checking this is a matter of checking that a) composition of mappings is preserved upon restriction to a subset (left invariant by the mappings in question); b) every permutation of X leaving Y invariant is made up of a permutation of Y and one of X − Y, and every such pair conversely together form a permutation of X.