551, Fall '02      A1a.    Definition and rudimentary properties      p. 1

Math 551 – Algebra – Fall 2002

Richard Lyons Rutgers University New Brunswick, New Jersey, USA

A. Groups

1. Definition and examples.

1a. Definition and rudimentary properties

Definition. A group (G, ·) is a set G together with a binary operation · : G × G → G, written for simplicity 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.

(If (3) is dropped, we have the definition of “monoid”. If (2) and (3) are dropped, we have the definition of “semigroup”. If gh = hg for all g, h ∈ G, then G is said to be abelian. In this case, and only in this case, additive notation g + h may be used for the product gh, and 0 instead of 1, and −g instead of g.)

Proposition. Let G be a group. Then

1) 1 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; likewise xg = h has the unique solution x = hg^{−1}.

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.

p. 2            A1.    Definition and examples.              551, Fall '02

3) Trivial 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) (y^{−1}x^{−1})xy = 1 and x^{−1}x = 1; now use the first statement of 3). QED

1b. Some examples

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)).

Ex. B. Let V be a vector space (of any dimension) over a field k. Then GL(V), the set of all invertible linear transformations from V → V, is a group, under composition of mappings.

Related to this, for a given natural number n and field (indeed any commutative ring) k, is the group GL_{n}(k) of all invertible n × n matrices with entries in k, under the operation of matrix multiplication.

Ex. C. 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; if also H ≠ G we write 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.

Also if n is a natural number and k is a commutative ring, then we define SL_{n}(k) = {A ∈ GL_{n}(k) | det(A) = 1} and have that SL_{n}(k) ≤ GL_{n}(k).

Trivially, in any group G, {1} (written 1) and G itself are both subgroups.

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.

551, Fall '02   A1c.    Isomorphisms, homomorphisms, representations   p. 3

Proof. Left to reader. QED Ex. D. Let X be any set, and Φ a structure on X. An automorphism of Φ is a bijection X → X

preserving Φ. The set Aut(Φ) of all automorphisms of Φ is a group, indeed a subgroup of Σ_{Φ}.

1c. Isomorphisms, homomorphisms, representations

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.

In 1), the products xy and ϕ(x)ϕ(y) are in G and H, respectively.

If 2) is omitted, we have the definition of homomorphism.

Proposition. Let ϕ : G → H be a homomorphism. Then im(ϕ) = ϕ(G) ≤ H, and ker(ϕ) = ϕ^{−1}(1) ≤ G. Moreover ϕ is injective if and only if ker(ϕ) = 1. In this case ϕ induces an isomorphism from G to ϕ(G).

Proof. Left to reader. QED

We write G ≅ H if and only if there exists an isomorphism from G to H.

The relation ≅ is then reflexive, symmetric and transitive on the class of all groups. Indeed one sees quickly from the definition of isomorphism that

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 isomorphism, let X be a set, x ∈ X, G = Σ_{X} and G_{x} the stabilizer of x in G, as above. Then G_{x} is isomorphic to Σ_{X−{x}}, via the isomorphism G_{x} → Σ_{X−{x}} taking σ ↦ σ|(X − {x}).

For another 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}.

We write Σ_{n} for Σ_{{1,2,...,n}}. Thus for any finite X, Σ_{X} ≅ Σ_{n} where n = |X|. However, there are (as long as n > 2) many such isomorphisms, for a given X.

Yet another example of isomorphism is obtained by taking an n-dimensional vector space V over a field k, and choosing an ordered basis B of V. For each linear transformation T : V → V, let [T]_{B} be the matrix of T with respect to B. The mapping

GL(V) → GL_{n}(k), T ↦ [T]_{B}

is an isomorphism. Thus there are many isomorphisms between these two groups, (at least) one for each choice of B.

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}. Indeed these properties hold for any homomorphism ϕ.

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. Representation theory, on the other hand, is the study, for a given group G, of structures Φ and homomorphisms G → Aut(Φ).

If G is a group, then Aut(G), the set of all isomorphisms from G to itself, is again a group.

Exercise. If G ≅ H then Aut(G) ≅ Aut(H).

Representations of any group G are not hard to find. We give three important examples.

1. For each g ∈ G define Int(G) : G → G by Int(g)(x) = gxg^{−1}. Then Int(g) ∈ Aut(G), and the mapping

i : G → Aut(G), g ↦ Int(g)

is a homomorphism. In general this homomorphism is not surjective; its image is called Inn(G) and an automorphism is called inner if and only if it lies in Inn(G). Likewise in general this homomorphism is not injective; its kernel is

Z(G) = {g ∈ G | gxg^{−1} = x ∀x ∈ G} = {g ∈ G | gx = xg ∀x ∈ G}.

At one extreme, if G is abelian, i.e., satisfies the commutative law

gh = hg ∀g, h ∈ G

then Z(G) = G and Inn(G) = 1. At the other extreme, if Z(G) = 1, then G ≅ Inn(G).

Exercise. If X is a set, what is Z(Σ_{X})?

2. If G is a group, consider the symmetric group Σ_{G} (the set of all permutations of G, disregarding their effect on the group structure, i.e. whether or not they are automorphisms). For each g ∈ G define λ_{g} : G → G

λ_{g}(x) = gx, ∀x ∈ G.

Then

λ_{g} ∘ λh(x) = g(hx) = (gh)x = λ_{gh}(x) ∀x ∈ G  (1A)

and of course λ_{1} = id_{G} is the identity function on G.

Consequently λ_{g}λ_{g^{−1}} = λ_{1} = id_{G} = λ_{g^{−1}}λ_{g}, so each λ_{g} ∈ Σ_{G} and then (1A) shows that the mapping

λ : G → Σ_{G}, g ↦ λ_{g}

is a homomorphism. If λ_{g} = λ_{h} then g = λ_{g}(1) = λ_{h}(1) = h so λ is injective. We have proved Cayley’s Theorem:

Theorem. Every group G is isomorphic to a subgroup of Σ_{G} (via λ).

This is generally not very useful, except for philosophical reasons: groups are inherently groups of permutations, or can be viewed as such. Moreover one can try to “decompose” this representation.

One can also define ρ_{g} : G → G by x ↦ xg, and we get the possibly different representation ρ : G → Σ_{G} defined by ρ(g) = ρ_{g^{−1}}.

Exercise. Let G be a group. For any subgroup H ≤ G, define C_{G}(H) = {g ∈ G | gh = hg ∀h ∈ H}. Show that C_{G}(H) ≤ G. Show also that C_{Σ_{G}}(ρ(G)) = λ(G) and C_{Σ_{G}}(λ(G)) = ρ(G).

3. For any group G, let V be a vector space over R (or any field k) with basis G. Thus V is the set of formal linear combinations ∑_{x∈G} α_{x}x such that α_{x} ∈ k for all x ∈ G and almost all α_{x} are 0 (i.e. α_{x} ≠ 0 for only finitely many elements x of G). The vector space operations on V are the obvious formal operations. For each g ∈ G there is a linear transformation Λ_{g} : V → V defined by

Λ_{g} (∑_{x} α_{x}x) = ∑_{x} α_{x}gx = ∑_{y∈G} α_{g^{−1}y}y,

and the mapping Λ : G → GL(V), g ↦ Λ_{g} is a homomorphism. It is easily checked that Λ is injective, so we get a matrix version of Cayley’s Theorem above. In particular: Every group G of finite cardinality |G| = n is isomorphic to a subgroup of GL_{n}(k) for any field k.

The decomposition of this “regular” representation is crucial for the representation theory of finite groups.

Remark. An element ∑ α_{x}x of V is really just an array of coefficients (α_{x}) indexed by G, that is, a function f : G → k (namely the one such that f(x) = α_{x} for every x ∈ G). Regarding elements of V this way we find that

Λ_{g}(f)(y) = f(g^{−1}y), f ∈ V, g, y ∈ G.

Note the g^{−1} (not g), so that Λ_{g}Λ_{h} = Λ_{gh} (not Λ_{hg}).

Ex. E. 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 homomorphism

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^{×}. Explain why this assertion is essentially the same as the assertion that ker ϕ = {cI | c ∈ C^{×}}, where I is the 2 × 2 identity matrix.

Exercise. Let G be a group and α ∈ Aut(G). Set

C_{G}(α) = {g ∈ G | α(g) = g} and I_{G}(α) = {g ∈ G | α(g) = g^{−1}}.

Show that C_{G}(α) ≤ G, but that I_{G}(α) need not be a subgroup of G. Show that if I_{G}(α) = G, then G is abelian. Show that if G is finite and |C_{G}(α)| > |G|/2, then C_{G}(α) = G. Show that if G is finite and |I_{G}(α)| > |G|/2, then α^{2} = 1. Finally show that if G is finite and |I_{G}(α)| > 3|G|/4, then G is abelian and I_{G}(α) = G.

1d. Powers; cyclic groups

Definition. 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. (In an abelian group with additive notation, we write ng instead of g^{n}.)

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, as is the following elementary fact:

551, Fall '02              A1d.    Powers; cyclic groups              p. 7

Proposition. If ϕ : G → H is a homomorphism and g ∈ G, then ϕ(g^{n}) = (ϕ(g))^{n} for every n ∈ Z.

Given g ∈ G, 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 is the identity element, −g is the inverse of g. Additive notation is customary. 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].

Theorem. 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} implies 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 i ≡ j mod n implies g^{i} = g^{j}. The mapping ϕ : Z_{n} → G taking [m] ↦ g^{m}, 0 ≤ m < n is then well-defined and injective. For any i ∈ Z we may write i = qn + r with 0 ≤ r < n, so g^{i} = g^{r} and ϕ is surjective. Finally g^{m}g^{m′} = g^{m+m′}, so ϕ is an 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

One corollary of this is that we can compute the structure of the automorphism group of a cyclic group, which we do below. As a simpler 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 ∞. The identity element has order 1. In a finite group, every element has finite order.

Theorem. Subgroups and homomorphic images of cyclic groups are cyclic.

Proof. The second statement means that if G is cyclic and ϕ : G → H is a homomorphism, then ϕ(G) is cyclic. But if G = ⟨g⟩, then every element of ϕ(G) has the form ϕ(g^{n}) = (ϕ(g))^{n} for some n ∈ Z, so ϕ(G) = ⟨ϕ(g)⟩ is cyclic.

For the first statement, we know that G ≅ Z or Z_{n} for some n > 0, so it is sufficient to prove the statement for Z and Z_{n}. We prove a stronger statement:

Theorem. 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; the subgroup ⟨[m]⟩ has order n/m if m divides n. Consequently a cyclic group of order n has a subgroup of order m if and only if m divides n.

Obviously the listed subgroups are indeed subgroups. In ⟨n⟩ ≤ Z, n is the least positive element. Moreover if m > 0 divides n, then [m] has order n/m in Z_{n}. These facts imply that the claimed subgroups are distinct.

It remains to show that every subgroup is one of them. Let H ≤ Z. Then either H = 0 or H contains a positive integer m (since H is closed under inversion). Similarly if H ≤ Z_{n}, then either H = 0 or H contains [m] for some 0 < m < n. Let m be the least such integer, in either case. Then H contains ⟨m⟩ or ⟨[m]⟩ respectively, and we argue that H in fact equals that cyclic group. For if k ∈ H (or [k] ∈ H) we may write k = qm + r with 0 ≤ r < m. Then r = k − qm ∈ H (or [r] ∈ H). By minimality of m, r = 0 so k = qm ∈ H (or [k] = q[m] ∈ H). Hence H = ⟨m⟩ or ⟨[m]⟩. QED

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

Exercise. If g ∈ G has finite order n, and ϕ : G → H is a homomorphism, then ϕ(g) has finite order dividing n.

To determine the automorphism group of a cyclic group G, observe first that G ≅ Z or Z_{n}, so Aut(G) ≅ Aut(Z) or Aut(Z_{n}). Thus it suffices to consider the groups G = Z and Z_{n}, n > 0. Accordingly let g = 1 or [1]. If ϕ ∈ Aut(G) then G = ϕ(G) = ⟨ϕ(g)⟩, so ϕ(g) is also a generator of G. Indeed for each generator h of G there exists by the above theorem a unique ϕ ∈ Aut(G) such that ϕ(g) = h. Hence the mapping

ϕ ↦ ϕ(g)

is a bijection between Aut(G) and the set of all generators of G.

Now if G = Z it is easy to see that the only generators of G are 1 and −1. Hence Aut(Z) consists of two elements, the identity mapping and inversion, and Aut(Z) ≅ Z_{2}.

On the other hand if G = Z_{n} then (exercise) the only generators of G are the elements [m] such that gcd(m, n) = 1. Thus for each such [m] there exists a unique ϕ_{[m]} ∈ Aut(Z_{n}) such that ϕ_{[m]}([1]) = [m]. Moreover

ϕ_{[m]} ∘ ϕ_{[m′]}([1]) = ϕ_{[m]}([m′]) = ϕ_{[m]}(m′[1]) = m′ϕ_{[m]}([1]) = m′[m] = [mm′],

so ϕ_{[m]} ∘ ϕ_{[m′]} = ϕ_{[mm′]}. Thus if we let S be the set of all generators of G, and use the multiplication in the ring Z/nZ as the operation on S, then Aut(G) ≅ S (and S is a group).

Exercise. S is the group of multiplicative units in the ring Z/nZ.

These results imply:

551, Fall '02                    A1e.    Generation                    p. 9

Theorem. Aut(Z) ≅ Z_{2}, and for n > 0, Aut(Z_{n}) is isomorphic to the multiplicative group of units in the ring Z/nZ.

1e. Generation

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, which remained unknown for 50 years, 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 all values of m and n, and in particular the case (m, n) = (2, 5) remains open.

On the other hand, some non-finitely-generated groups can be very tractable. The additive group Q^{+} of rational numbers, for instance, is “locally cyclic”: every finitely generated subgroup is cyclic, even though Q^{+} itself is not cyclic (or finitely generated). The better behavior is largely due to the fact that Q^{+} is abelian.

1f. Abelian groups

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. Subgroups and homomorphic images of abelian groups are 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. Moreover, the hypothesis of commutativity cannot be dropped.

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

1g. Dihedral groups

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 (say) rotation σ through 2π/n. In addition there are the reflections about an axis of symmetry; there are exactly 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 compact rules:

σ^{n} = 1, τ^{2} = 1, τστ = σ^{−1}.

551, Fall '02                  A1i.    Free products                  p. 11

For these imply that τσ^{n}τ = Int(τ)(σ^{n}) = Int(τ)(σ)^{n} = σ^{−n}, and 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.

1h. Direct products

Definition. 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.

A naturally occurring example of direct products is as follows:

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 boils down to the following evident statements: 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.

Exercise. Develop an analogue of the previous example for GL(V) and GL_{n}(k).

Exercise. Show that Z_{6} ≅ Z_{2} × Z_{3}, but Z_{4} ≅ Z̸_{2} × Z_{2}. Generalize.

1i. Free products

If G and H are groups then G × H is generated by the isomorphic copies G_{1} = {(g, 1) | g ∈ G} of G and H_{1} = {(1, h) | h ∈ H} of H. Moreover, every element of G_{1} commutes with every element of H_{1}. Given G and H we may similarly search for a group containing – and generated by – two subgroups which are isomorphic copies of G and H, but with no additional equations (or “relations”) holding between elements of these subgroups. The free product G ∗ H of G and H has the desired property. To describe it let us assume for simplicity that the groups G and H are disjoint except that they share an identity element (this assumption can easily be satisfied by replacing H by a suitable isomorphic copy of itself). Then the elements of G∗H are (formal) finite words a_{1} ⋯ a_{n}, n ≥ 0, in which either all the odd a_{i} lie in G − {1} and all the even a_{i} lie in H − {1}, or vice-versa. We include the “empty word” of length n = 0. Multiplication is accomplished simply by juxtaposing words and carrying out the obvious reduction to a reduced word, using the multiplication in G (or H) if two elements of G (or H) end up adjacent, and removing identity elements whenever they appear in this reduction process. It is clear that multiplication is welldefined, the empty word is an identity element, and that inverses exist (reverse a word and replace every letter by its inverse). It also seems clear that associativity holds, although some thought shows that there is something to check: a triple product has two possible locations for reduction, and one must show that the result of reduction is independent of the choice of location at which to begin reduction. We shall verify this in a later section.

Free products have the following important mapping property:

Proposition. Let ϕ : G → K and ψ : H → K be group homomorphisms. Then there exists a unique group homomorphism χ : G ∗ H → K such that χ(g) = ϕ(g) and χ(h) = ψ(h) for all g ∈ G and h ∈ H.

Here we identify g ∈ G with the one-letter word g ∈ G ∗ H, and identify h ∈ H with the one-letter word h ∈ G ∗ H.

The proposition has a trivial proof: for a word a_{1}a_{2} ⋯ a_{n} ∈ G ∗ H with a_{1} ∈ G, the only possibility is to define

χ(a_{1}a_{2} ⋯ a_{n}) = χ(a_{1})χ(a_{2}) ⋯ χ(a_{n}) = ϕ(a_{1})ψ(a_{2})ψ(a_{3}) ⋯;

then χ is well-defined and it is easy to check that it preserves multiplication. The fact that ϕ and ψ are homomorphisms comes into play when checking χ(xy) = χ(x)χ(y) in the case that the juxtaposition of x and y requires reduction.

Exercise. Formulate and prove a similar proposition for the direct product G×H instead of G ∗ H. An extra hypothesis about the homomorphisms ϕ and ψ will have to be added.

Exercise. If G and H are subgroups of a group K and K is generated by them (i.e., K = ⟨G ∪ H⟩, usually written K = ⟨G, H⟩), then there is a surjective homomorphism G ∗ H → K.

Exercise. Let ⟨a⟩ and ⟨b⟩ be groups of order 2 and 3, respectively. Show that there is an injective homomorphism

ϕ : ⟨a⟩ ∗ ⟨b⟩ → LF (2, C) such that ϕ(a) : z ↦ \frac{−1}{z} and ϕ(b) : z ↦ \frac{1}{1 − z} .

551, Fall '02          A1j.    Cosets and Lagrange's Theorem          p. 13

Conclude that Z_{2} ∗ Z_{3} is isomorphic to the image of SL_{2}(Z) in LF (2, C). (This image is usually called P SL_{2}(Z).)

1j. Cosets and Lagrange’s Theorem

Let G be a group and H a subgroup of G. We define an equivalence relation ∼_{H} on G as follows:

a ∼_{H} b ⟺ a^{−1}b ∈ H.

The facts that 1 ∈ H, H is closed under inversion, and H is closed under multiplication translate directly into the reflexivity, symmetry and transitivity of ∼_{H}. (Check this!) The equivalence classes of G are called the left cosets of H in G, and the set of all left cosets of G is denoted G/H. It is a partition of G, as is the set of equivalence classes for any equivalence relation. The cardinality of G/H is the “index” of H in G and is written |G : H|. Thus by definition, |G : H| = |G/H|.

Lemma. The left coset of H containing the element a ∈ G is the set aH = {ah | h ∈ H}. Moreover |H| = |aH| for any a ∈ G.

Proof. If a ∼_{H} b, then b = a(a^{−1}b) ∈ aH. Conversely if b ∈ aH, then b = ah for some h ∈ H, and a^{−1}b = h, so a ∼_{H} b. Finally the mapping h ↦ ah is a bijection between H and aH. QED

Lagrange’s Theorem. If H ≤ G, then |G| = |H||G : H|. If G is finite, then |G : H| = |G|/|H|.

The first statement is actually true even if G is infinite, with cardinal multiplication. In any case for G finite, it is a trivial consequence of the previous lemma.

Corollary. If H ≤ G and G is finite, then |H| divides |G|, and |G : H| divides |G|.

Corollary. If g ∈ G and G is finite, then |g| divides |G|.

Corollary. A group of prime order is cyclic.

Proof. Let G have order p and choose g ∈ G with g ≠ 1. Then |g| > 1, so |g| = p by the previous corollary, and hence |⟨g⟩| = p. Therefore G = ⟨g⟩. QED

Proposition. If K ≤ H ≤ G, then

|G : K| = |G : H||H : K|.

If G is finite this is immediate from Lagrange’s theorem, by which |G : K| = |G|/|K| (and similarly for the other terms). But in general it remains true and can be proved as follows. Choose a “transversal” T for H in G, that is, a subset T of G containing exactly one element from each coset in G/H. Thus |T | = |G : H|. Similarly choose a transversal U for K in H. Then we argue that

the elements tu, t ∈ T, u ∈ U are all distinct and form a transversal for K in G.

Assuming this, we conclude that |G : K| = |T × U| = |T ||U| = |G : H||H : K| as desired.

To prove the displayed statement, let g ∈ G. Then gH = tH for some t ∈ T since T is a transversal. Hence g = th for some h ∈ H, and then hK = uK for some u ∈ U as U is a transversal. Therefore gK = thK = tuK. So every coset in G/K contains one of the listed elements. On the other hand if tuK = t′u′K for some t, t′ ∈ T and u, u′ ∈ U (for example if tu = t′u′), then tH = tuH = tuKH = t′u′KH = t′u′H = t′H. Therefore t = t′ since T is a transversal. But then tuK = t′u′K implies uK = u′K, so u = u′ since U is a transversal, completing the proof.

Exercise. Show that if H ≤ G, K ≤ G and both |G : H| and |G : K| are finite, then |G : H ∩K| is finite. (Hint. Show that if h, h′ ∈ H, then h ∼_{H∩K} h′ if and only if h ∼_{K} h′.)

The definition of |G : H| may seem left-handed, but it is even-handed.

Exercise. If H ≤ G, let H\G be the set of all right cosets Hx, x ∈ G (i.e. the set of equivalence classes under the relation a ∼ b ⟺ ab^{−1} ∈ H). Then |G/H| = |H\G|. (Hint. Use the anti-isomorphism G → G defined by x ↦ x^{−1}.)

1k. Normal subgroups and quotients

In general if H ≤ G, then the coset space G/H is just a set (endowed with an action of G, see the next section), but G/H cannot be given the structure of a group in a reasonable way†. The obvious attempt (xH)·(yH) = (xy)H to define a binary operation on G/H fails to be well-defined. (Example: In G = Σ_{3}, let H = ⟨(12)⟩. Then (23)H · (13)H = (123)H according to this definition and (132)H · (13)H = (12)H. Unfortunately, the left sides are equal but the right sides are not: (23)H = (132)H but (12)H = H ≠ (123)H.)

Definition-Lemma. Let H ≤ G. Then H ◁ G (H is a normal subgroup of G) if and only if the following equivalent conditions hold:

a) Hx = xH for all x ∈ G;
b) xHx^{−1} = H for all x ∈ G;
c) xHx^{−1} ≤ H for all x ∈ G.

Proof. a) and b) are equivalent by right multiplication by x^{−1}. Trivially b) implies c), and the converse holds by applying c) to x^{−1} in place of x and conjugating by x.

† “Reasonable” means that the mapping G → G/H, g ↦ gH, should be a homomorphism.

PRIVATE DRAFT, NOT FOR PUBLIC CONSUMPTION. NO WARRANTY MADE OR IMPLIED.    9/11/2002

551, Fall '02         A1k.    Normal subgroups and quotients         p. 15

Example. If ϕ : G → K is any homomorphism, then ker ϕ ◁ G. (By c): if ϕ(g) = 1 then ϕ(xgx^{−1}) = ϕ(x)ϕ(x^{−1}) = 1 for all x ∈ G. By a): for g ∈ G, g^{−1}x ∈ H if and only if ϕ(g) = ϕ(x) if and only if gx^{−1} ∈ H. So xH = ϕ^{−1}(ϕ(x)) = Hx.)

Theorem. If H ◁ G, then the definition (xH) · (yH) = xyH makes G/H a group, and the mapping x ↦ xH is a surjective homomorphism π_{H} : G → G/H with kernel H.

Proof. For any X, Y ⊆ G set XY = {xy | x ∈ X, y ∈ Y }. This is clearly a well-defined associative binary operation on the set of all subsets of G. If H ◁ G, then (xH)(yH) = x(Hy)H = x(yH)H = (xy)(HH) = (xy)H, which proves that the operation · is welldefined. That the operation is associative, H is an identity element and (xH)^{−1} = x^{−1}H is easy to check, and the multiplicative property of π_{H} is trivial.

Corollary. Let H ≤ G. Then H ◁ G if and only if there is a group K and a homomorphism ϕ : G → K such that ker ϕ = H.

Examples. a) Any subgroup of an abelian group is normal. b) Any subgroup of Z(G) is normal in G. c) ⟨(123)⟩ ◁ Σ_{3}, but ⟨(12)⟩ ◁̸ Σ_{3}. d) SL_{n}(k) = {g ∈ GL_{n}(k) | det g = 1} ◁ GL_{n}(k), since det(gh) = det(g) det(h), i.e., det : GL_{n}(k) → k^{×} is a homomorphism. Here k^{×} is the group of all nonzero elements of k, under multiplication. e) If |G : H| = 2, then H ◁ G. (Proof: For x ∈ H, xH = H = Hx. For x ∉ H, G = H ∪ Hx = H ∪ xH so xH = G − H = Hx.)

Warning: It is not the case that H ◁ K ◁ G implies H ◁ G. The group G = Σ_{4} has a normal abelian subgroup V of order 4 (consisting of 1 and the three elements (ab)(cd), and V has several normal subgroups H of order 2, but Σ_{4} has no normal subgroup H of order 2. (Such a subgroup would have to be contained in Z(Σ_{4}), which is trivial.)