551, Fall '00                A5a.    Some Simple Groups                p. 1

Math 551 – Algebra – Fall 2000

Richard Lyons Rutgers University New Brunswick, New Jersey, USA

A. Groups

5. The Analysis of Groups.

We consider here the question of how properties of a group G may be related to properties of its normal subgroups N and quotients G/N. We also give some natural ways to attempt to break up a group into simpler parts. We also give some examples of simple groups.

5a. Some Simple Groups

Of course the cyclic groups Z_{p} of prime order are simple, and they are the only simple abelian groups (If G is simple abelian then every subgroup is normal, so is either 1 or G, so G = ⟨g⟩ for any 1 ≠ g ∈ G. Let p be a prime divisor of |G| (or any prime if G is infinite). Then ⟨g^{p}⟩ < G so g^{p} = 1.)

One classic family of nonabelian simple groups is the family of alternating groups.

Theorem. Suppose that n ≥ 5. Then A_{n} is simple.

Corollary. Suppose that n ≥ 5. Then the only nontrivial normal subgroup of Σ_{n} is A_{n}.

A standard approach to proving this theorem is a) show that A_{n} is generated by all its 3-cycles (note that we can’t use 2-cycles since they are odd permutations); b) show that all 3-cycles in A_{n} are conjugate in A_{n}; c) show that if 1 < N ◁ A_{n}, then an element of N whose support has minimum size is a 3-cycle. Then by c) and b), N contains all 3-cycles, so equals G, as desired. The dirtiest part of this is c), where one produces from an element g ∈ A_{n} which is not a 3-cycle another non-identity element h ∈ A_{n} which has smaller support, but is the product of powers of g and their conjugates.

Instead we give an inductive proof, still using the action of A_{n} on Ω = {1, …, n} of course. This is a transitive action for which the stabilizer of a point is isomorphic to A_{n−1}.

To start the induction, observe that A_{5} has 15 “Klein” elements (those of shape (ab)(cd)e), and they are all conjugate in A_{5}. A_{5} also has 20 3-cycles, and they too are all conjugate in A_{5}. A_{5} has 24 5-cycles, each of which is self-centralizing and so lies in a conjugacy class of size |A_{5}|/5 = 12. The class equation for A_{5} is therefore

60 = 1 + 15 + 20 + 12 + 12.

p. 2             A5.    The Analysis of Groups.               551, Fall '00

Now a normal subgroup N of A_{5} must consist of the identity and some of the other conjugacy classes, so |N| is a subsum of the above sum, containing the term 1. But no such subsum divides 60, other than the full sum or 1. Therefore N = 1 or N = A_{5}.

For the induction step, we need to check that for n ≥ 6, the action of A_{n} on {1, …, n} is 4-transitive in the following sense.

Definition. Let G act on Ω. Then G acts 4-transitively if and only if for any quadruple (α, β, γ, δ) ∈ Ω × Ω × Ω × Ω such that α, β, γ, δ are pairwise distinct, and for any other such quadruple (α′, β′, γ′, δ′), there is g ∈ G such that gα = α′, gβ = gβ′, gγ = γ′, and gδ = δ′.

One may similarly define n-transitivity, using n-tuples instead of quadruples.

Lemma. If G acts n-transitively on Ω, then for each ω ∈ Ω, G_{ω} acts n − 1-transitively on Ω − {ω}.

Proof Left to reader.

Lemma. For n ≥ 6, A_{n} acts 4-transitively on {1, …, n}.

Proof Σ_{n} certainly does, and if g is taken in Σ_{n} and is odd, then g′ = gτ works just as well, where τ is a transposition fixing α, β, γ, δ.

A group action of G on Ω is called faithful if and only if the kernel of the corresponding mapping G → Σ_{Ω} is trivial. The action of A_{n} on {1, …, n} is certainly faithful.

Lemma. Suppose that G acts n-transitively and faithfully on Ω, n ≥ 2. Let 1 ≠ N ◁ G. Then N acts transitively on Ω. Moreover, if N_{α} = 1 for some α ∈ Ω, then the action of G on N − {1} by conjugation is n − 1-transitive.

Proof For any ω ∈ Ω, the N-orbit N · ω is mapped by g ∈ G as follows:

g · (N · ω) = (gN) · ω = Ng · ω = N(gω)

Thus each g ∈ G permutes the set of N-orbits on Ω. Suppose that there are two or more orbits Ω_{1}, Ω_{2}, . . .. Then |Ω_{i}| > 1 since otherwise N would act trivially on Ω, so N = 1 by faithfulness, contrary to assumption. Choose α, β ∈ Ω_{1} and γ ∈ Ω_{2}. There is g ∈ G such that gα = α and gβ = γ. Therefore gΩ_{1} = Ω_{1} and gΩ_{1} = Ω_{2}, a contradiction. Therefore there’s only one orbit and N is transitive.

Next fix α ∈ Ω and suppose that N_{α} = 1. We map

ϕ : N → Ω, n ↦ nα,

and claim that ϕ is a bijection. Since N is transitive ϕ is onto. If nα = n′α then n^{−1}n′ ∈ N_{α} = 1 so n = n′. We also claim that

551, Fall '00                 A5b.    Solvable Groups                 p. 3

the actions of G_{α} on N (by conjugation) and Ω are isomorphic via ϕ, that is, ϕ(^{g}n) = gϕ(n) for all g ∈ G_{α} and n ∈ N. Namely, ϕ(^{g}n) = ϕ(gng^{−1}) = gng^{−1}α = gnα = gϕ(n).

But the action of G_{α} on Ω − {α} is n − 1-transitive, and so its action on N − {1} is n − 1-transitive as well. QED

Now we complete the proof of the theorem. Suppose that n ≥ 6 and let G = A_{n}. Proceeding by contradiction suppose that N ◁ G, with N ≠ 1 and N ≠ G. We reach a contradiction. Let α ∈ {1, …, n} = Ω. Then G_{α} ∩ N ◁ G_{α} by the parallelogram law. But G_{α} ≅ A_{n−1} is simple by induction. Therefore G_{α} ∩ N = G_{α} or 1. Also N is transitive on Ω, so |N : N_{α}| = n.

If G_{α} ∩ N = 1, then N_{α} = 1 and so |N| = n. The previous lemma shows that G_{α} acts 4 − 1-transitively on N − {1} by conjugation, since G is 4-transitive on Ω. Choose distinct elements x, y ∈ N − {1} and let z = xy, so that z is distinct from x and y. As n ≥ 6 there is another element w ∈ N − {1}. By the 3-transitivity of G_{α} there is g ∈ G_{α} such that

gxg^{−1} = x, gyg^{−1} = y, gzg^{−1} = w.

However, gzg^{−1} = gxg^{−1}gyg^{−1} = xy = z ≠ w, a contradiction.

Therefore G_{α} ∩ N = G_{α} so G_{α} = N_{α}. Then |G : G_{α}| = n = |N : N_{α}| = |N : G_{α}| and so N = G. QED

Proof of Corollary. Suppose that n ≥ 5 and N ◁ Σ_{n}. Then N ∩ A_{n} ◁ A_{n} so N ∩ A_{n} = 1 or A_{n}. In the first case N ≅ N/N ∩ A_{n} ≤ Σ_{n}/A_{n} ≅ Z_{2}, so |N| ≤ 2. If |N| = 2, then N consists of a single odd element which must be the unique element of its cycle shape (as N ◁ G). But clearly for each cycle shape (except for the identity element) there is more than one element of that cycle shape, contradiction. In the second case A_{n} ≤ N and the Correspondence Theorem implies N = A_{n} or Σ_{n}.

5b. Solvable Groups

The following result is easy:

Proposition. All subgroups and quotients of an abelian group are abelian.

However the “converse” is false; if N ◁ G and N and G are both abelian, then G need not be abelian. Example: G = Σ_{3}, N = A_{3}, or G = D_{2n}, N ≅ Z_{n}. To obtain a property which “persists under extensions” in the most economical way we are let to the following notion.

Definition. A finite group is solvable if and only if all its composition factors are abelian (hence cyclic of prime order). In general a group is solvable if and only if there exists a normal series (by definition of finite length!) all of whose factors are abelian.

The two definitions coincide for finite groups, since if a finite group has such a normal series then it can be refined to a composition series, whence all composition factors are abelian.

Example: Σ_{4} is solvable. For any v ∈ V − {1}, the series

1 ◁ ⟨v⟩ ◁ V ◁ A_{4} ◁ Σ_{4}

is a composition series.

Proposition. All subgroups and quotients of solvable groups are solvable. If N ◁ G, and both N and G/N are solvable, then G is solvable.

Proof If K ◁ H ≤ G, and A ≤ G, then A ∩ K ◁ A ∩ H and A ∩ H/A ∩ K is isomorphic to a subgroup of H/K. This follows from the parallelogram law applied to K and A ∩ H. Now if

1 = G_{n} ◁_{≠} G_{n−1} ◁_{≠} ⋯ ◁_{≠} G_{2} ◁_{≠} G_{1} ◁_{≠} G_{0} = G

is a normal series of G with abelian factors, then

1 = G_{n} ∩ H ◁_{≠} G_{n−1} ∩ H ◁_{≠} ⋯ ◁_{≠} G_{2} ∩ H ◁_{≠} G_{1} ∩ H ◁_{≠} G_{0} ∩ H = H  (5A)

is a normal series of H, and G_{i−1} ∩ H/G_{i} ∩ H embeds in the abelian group G_{i−1}/G_{i} so is abelian. The statement for quotients similarly follows from the following fact: If K ◁ H ≤ G and N ◁ G, then KN ◁ HN and HN/KN is isomorphic to a quotient of H/K. The parallelogram law applid to KN and H yields in fact that HN/KN ≅ H/H ∩ KN; but K ≤ H ∩ KN so H/H ∩ KN is a quotient of H/K. In this situation the above series gives the series

N = NG_{n} ◁_{≠} NG_{n−1} ◁_{≠} ⋯ ◁_{≠} NG_{2} ◁_{≠} NG_{1} ◁_{≠} NG_{0} = G  (5B)

and so

1 = N/N ◁_{≠} NG_{n−1}/N ◁_{≠} ⋯ ◁_{≠} NG_{2}/N ◁_{≠} NG_{1}/N ◁_{≠} NG_{0}/N = G/N  (5C)

the factors of which are quotients of the original abelian factors so are again abelian.

Conversely suppose that N and G/N are solvable. Then G/N has a normal series with abelian factors, and replacing each term by its inverse image in G and using the correspondence theorem we get a series starting with N and going to G, with all factors abelian. We are assuming that N possesses such a series, and it can be attached to this one to give the desired series for G.

One useful way to analyze a solvable group is by its action on a minimal normal subgroup.

Theorem. Let G be a finite solvable group. Let 1 ≠ N ◁ G and suppose that no proper subgroup of N is normal in G. Then N ≅ Z_{p} × ⋯ × Z_{p} for some prime p.

A group is called elementary abelian if it is the direct product of groups of the same prime order. The structure theorem for finite abelian groups (see below) implies that if N is a finite abelian group and p a prime such that x^{p} = 1 for all x ∈ N, then N is elementary abelian. We shall use this fact in the proof, and aim to prove that N is abelian and has exponent p.

The proof uses the notion of characteristic subgroup:

Definition. A subgroup H ≤ G is characteristic in G if and only if α(H) = H for all α ∈ Aut(G). We write H char G.

If H char G, then Int(g)(H) = H for all g ∈ G, so gHg^{−1} = H, i.e., H ◁ G. The converse is false.

It also uses the notion of commutator subgroup.

Definition. Let x, y ∈ G. Then [x, y] = xyx^{−1}y^{−1}. Moreover, [G, G] = ⟨[x, y] | x, y ∈ G⟩.

A group is abelian if and only if [G, G] = 1.

Lemma. The following subgroups of any group G are characteristic subgroups:

a) [G, G];

b) Z(G);

c) G^{n} = ⟨x^{n} | x ∈ G⟩, for an integer n. Moreover, if G is finite and has a normal Sylow p-subgroup P for some prime p, then P char G.

Proof ^{g}[x, y] = [^{g}x,^{g} y] so conjugation by an element of G leaves the set of commutators invariant; therefore it leaves invariant the subgroup generated by them. The proof for c) is similar, and b) is left to the reader. For the final statement, P is the only Sylow psubgroup, being normal, so is characteristic since automorphisms carry Sylow p-subgroups to Sylow p-subgroups. QED

Lemma. If H char N ◁ G, then H ◁ G.

Proof Let g ∈ G. Then Int(g) : x ↦ gxg^{−1} is an automorphism of G, and it leaves N invariant since N ◁ G. Therefore Int(g)|N is an automorphism of N, so it leaves the characteristic subgroup H invariant. Therefore gHg^{−1} = H. QED

Here is the universal property of [G, G], or really the quotient G/[G, G], or really the projection G → [G, G].

Proposition. If N ≤ G, then N ≥ [G, G] if and only if N ◁ G and G/N is abelian.

Proof Suppose [G, G] ≤ N ≤ G. Then for any g ∈ G and n ∈ N, gng^{−1} = [g, n]n ∈ N, so N ◁ G. Now under the projection G → G/N, [x, y] maps to [xN, yN]. But [x, y] = 1 for all x, y since [G, G] ≤ N. Therefore [G/N, G/N] = 1. Conversely if N is a normal subgroup and G/N is abelian, the same reasoning shows that [x, y] ∈ N for all x, y ∈ G. Therefore [G, G] ≤ N.

Corollary. [G, G] < G if and only if G has a nontrivial abelian quotient. If G is a solvable group and G ≠ 1, then [G, G] < G.

Proof of Theorem Let G and N be as in the theorem. Then N is solvable since G is solvable. So [N, N] < N. But [N, N] char N ◁ G so [N, N] ◁ G. Therefore [N, N] = 1 and N is abelian. Let p be a prime divisor of |N|. The mapping N → N defined by x ↦ x^{p} is then a homomorphism, with image N^{p}. Its kernel is nontrivial by Sylow, and so N^{p} < N. As with the commutator subgroup we get N^{p} = 1, that is, x^{p} = 1 for all x ∈ N. Now we quote the structure theorem for finite abelian groups (see below) to complete the proof.

5c. N/C-Theorem

The fact that a solvable group has a normal subgroup which is elementary abelian suggests a way to analyze the group: consider the series

1 ≤ N ≤ C_{G}(N) ≤ G.

Here the centralizer of a subgroup N is defined as

C_{G}(N) = {g ∈ G | gn = ng ∀n ∈ N}.

It is easily checked that C_{G}(N) is a subgroup of G. It is not always true that N ≤ C_{G}(N); indeed this condition is equivalent to N being abelian.

Thus we have the three factors N, C_{G}(N) and G/C_{G}(N). Of these we have little control over C_{G}(N), but since N ◁ G, G/C_{G}(N) can be regarded as a subgroup of Aut(N) as follows.

Theorem. Let N ◁ G. The mapping G → Aut(N) defined by g ↦ Int(g)|N is a homomorphism, and its kernel is C_{G}(N). Consequently C_{G}(N) ◁ G and G/C_{G}(N) is isomorphic to a subgroup of Aut(N).

The assertions of the theorem are easy to check and are left to the reader.

When N is not necessarily normal, we salvage at least this much:

Corollary. Let H ≤ G. Then the mapping N_{G}(H) → Aut(H) defined by g ↦ Int(g)|H is a homomorphism with kernel C_{G}(H). Consequently C_{G}(H) ◁ N_{G}(H) and N_{G}(H)/C_{G}(H) is isomorphic to a subgroup of Aut(H).

Furthermore,

5d. More Simple Groups

Another source for simple groups are matrix groups, i.e. certain subgroups and quotients (and subgroups of quotients) of GL_{n}(K) for various fields K. We define

P GL_{n}(K) = GL_{n}(K)/Z and P SL_{n}(K) = SL_{n}(K)/Z ∩ SL_{n}(K)

where Z is the group of all scalar matrices (scalar multiples of the identity matrix) in GL_{n}(K). Notice that Z ≤ Z(GL_{n}(K)).

Exercise. Z = Z(GL_{n}(K)).

Also by the parallelogram law, P SL_{n}(K) ≅ ZSL_{n}(K)/Z ◁ P GL_{n}(K). So P GL_{n}(K) has a copy of P SL_{n}(K) as a normal subgroup.

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Theorem. Let K be a field and n ≥ 2 an integer. If n = 2 assume that |K| ≥ 4. Then P SL_{2}(K) is simple.

The “P ” in P GL and P SL is for “projective”. Another source for simple groups are matrix groups, i.e. certain subgroups and quotients (and subgroups of quotients) of GL_{n}(K) for various fields K. We define

P GL_{n}(K) = GL_{n}(K)/Z and P SL_{n}(K) = SL_{n}(K)/Z ∩ SL_{n}(K)

where Z is the group of all scalar matrices (scalar multiples of the identity matrix) in GL_{n}(K). Notice that Z ≤ Z(GL_{n}(K)).

Exercise. Z = Z(GL_{n}(K)).

Also by the parallelogram law, P SL_{n}(K) ≅ ZSL_{n}(K)/Z ◁ P GL_{n}(K). So P GL_{n}(K) has a copy of P SL_{n}(K) as a normal subgroup.

Theorem. Let K be a field and n ≥ 2 an integer. If n = 2 assume that |K| ≥ 4. Then P SL_{2}(K) is simple.

The “P ” in P GL and P SL is for “projective”.

We sketch a proof. Let V be a 2-dimensional vector space over the field K. Define GL(V) to be the group of nonsingular linear transformations: V → V; SL(V) to be the subgroup of GL(V) which is the kernel of the determinant homomorphism SL(V) → K^{×}; Z to be the subgroup of GL(V) consisting of all scalar mappings (for each α ∈ K^{×} there is unique corresponding scalar mapping s_{α} ∈ GL(V), namely s_{α}(v) = αv for all v ∈ V). Also define P GL(V) = GL(V)/Z and P SL(V) = SL(V)/Z ∩ SL(V).

Choosing a (ordered) basis for V leads to an isomorphism P SL(V) ≅ P SL_{2}(K). We may then think of P SL(V) acting on P(V), or equivalently, we may think of P SL_{2}(K) acting on the set of 1-dimensional spaces of 2 × 1 column vectors.

The group GL(V) acts naturally on the set P(V) of 1-dimensional subspaces of V, so SL(V) does as well. Moreover Z acts trivially on P(V), so the action GL(V) → Σ_{P(V)} lifts to an action GL(V)/Z → Σ_{P(V)}, i.e., P GL(V) acts on P(V). Likewise P SL(V) acts on P(V).

Exercise. The action of P SL(V) on P(V) is 2-transitive.

We also can show:

Lemma. The action of P SL(V) on P(V) is faithful.

This amounts to showing that if T : V → V is a linear transformation and for each v ∈ V there exists a scalar α_{v} (depending on v, perhaps) such that T v = α_{v}v, then α_{v} = α_{w} for all v, w ∈ V − {0}. (Look at T (v + w) to see that this is true, if v and w are linearly independent.)

The structure of a point stabilizer is also important, and this is easily seen concretely in P SL_{2}(K).

Lemma. The stabilizer in P SL_{2}(K) of the subspace ω spanned by [ \begin{matrix}1 \\ 0 \end{matrix} ] is the image \overline{B}

in P SL_{2}(K) of the group

B = \{[ \begin{matrix}α & β \\ 0 & α^{−1} \end{matrix} ] | α ∈ K^{×}, β ∈ K\} .

under the canonical homomorphism SL_{2}(K) → P SL_{2}(K). (B is a subgroup of SL_{2}(K); check it.)

This merely states that

[ \begin{matrix}α & β \\ γ & δ \end{matrix} ] [ \begin{matrix}1 \\ 0 \end{matrix} ] = [ \begin{matrix}ζ \\ 0 \end{matrix} ]

for some ζ if and only if γ = 0.

Lemma. B and \overline{B} are solvable.

Proof. Define ϕ : B → K^{×} by

ϕ ([ \begin{matrix}α & β \\ 0 & α^{−1} \end{matrix} ]) = α.

It is easily checked that ϕ is a homomorphism. Let u(β) = [ \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ] for any β ∈ K, and

U = ker ϕ = {u(β) | β ∈ K} .

Then U is abelian and B/U is isomorphic to a subgroup of K^{×} so is abelian. Therefore the normal series B > U > 1 has abelian factors and so B is solvable. Since \overline{B} is the image of B, it is isomorphic to a quotient of B and so is solvable.

We need one additional fact about P SL_{2}(K); to motivate it we now give the main logic of the proof.

Main Logic Let G = P SL_{2}(K), which acts on Ω = P(V). Suppose that N ◁ G, but 1 < N < G. We derive a contradiction.

Since G acts faithfully and 2-transitively on Ω, and N ≠ 1, N is transitive on Ω (the same fact was used in the proof that A_{n} is simple, n ≥ 5). Therefore G = NG_{ω} = \overline{B}N. (For any g ∈ G there is n ∈ N with gω = nω and so g = gn^{−1}n ∈ G_{ω}N. Then G/N = \overline{B}N/N ≅ \overline{B}/\overline{B} ∩N is a quotient of the solvable group \overline{B} so is solvable. Since N ≠ G, G/N is a nontrivial (solvable) group. Therefore it has a nontrivial abelian quotient. Therefore G has a nontrivial abelian quotient. Therefore SL_{2}(K) has a nontrivial abelian quotient. Therefore SL_{2}(K) ≠ [SL_{2}(K), SL_{2}(K)]. This completes the proof as it contradicts the following fact:

551, Fall '00          A5e.    Inner and Outer Automorphisms          p. 9

Lemma. If |K| ≥ 4, and H = SL_{2}(K), then H = [H, H].

The lemma is proved by

a) checking that H = SL_{2}(K) is generated by all the elements u(β) defined above and all their transposes v(β) = u(β)^{T}. (This amounts to showing that any matrix of determinant 1 can be reduced to I by certain types of row and column operations.)

b) checking that each u(β) and v(β) lies in [H, H]. (It is here that the hypothesis |K| ≥ 4 is used, to find an element α ∈ K such that α ≠ 0 and α ≠ ±1. Fix α and let h(α) be the diagonal matrix with diagonal entries α, α^{−1}; one computes

[h(α), u(β)] = u((α^{2} − 1)β).

As β varies over K, so does (α^{2} − 1)β, since α^{2} ≠ 1. Therefore every u(β) is in [H, H], and similarly so is every v(β).)

Then as the u’s and v’s generate H, [H, H] = H. QED

A similar argument, slightly more complicated, can be used to prove:

Theorem. For any field K and any n ≥ 3, P SL_{n}(K) is simple.

There are no exceptions here; the larger size of matrices makes it possible to express certain critical matrices as commutators, regardless of the size of the field.

5e. Inner and Outer Automorphisms

This is one of the “natural ways” referred to above. For any group G, and any g ∈ G, the mapping

Int(g) : G → G, h ↦ ghg^{−1}

is an automorphism of G. Any automorphism of G arising from some element g ∈ G in this fashion is called an inner automorphism of G. Furthermore, the mapping

I : G → Aut(G), g ↦ Int(g)

is then a homomorphism: Int(gg′) = Int(g) Int(g′). (Check it!) The

image of I consists of all inner automorphisms, which therefore constitute a subgroup Inn(G) ≤ Aut(G).

Theorem. ker I = Z(G), and G/Z(G) ≅ Inn(G) ◁ Aut(G).

Proof z ∈ ker I ⟺ zgz^{−1} = g for all g ∈ G ⟺ z ∈ Z(G). The first isomorphism theorem then gives the isomorphism. The normality of Inn(G) follows from the following identity:

If α ∈ Aut(G) and g ∈ G, then α Int(g)α^{−1} = Int(α(g)).

Namely α Int(g)α^{−1}(h) = α(gα^{−1}(h)g^{−1}) = α(g)hα(g)^{−1}.

Elements of Aut(G)/ Inn(G) ≅ Out(G) are called “outer automorphisms”. The following conjecture is a consequence of the classification of finite simple groups, but still awaits a straightforward proof.

Schreier Conjecture. If G is a finite simple group, then Out(G) is solvable.

Not every group can arise as Inn(G) for some G. For instance:

Theorem. If G/Z(G) is cyclic, then G is abelian (and so G/Z(G) = 1).

Proof. Some element gZ(G) ∈ G/Z(G) generated G/Z(G). Then G = ∪g^{i}Z(G). Consequently G = ⟨S⟩ where S = Z(G) ∪ {g}. But any two elements of S commute, hence any two elements of G commute, since they are represented by words in S.

Corollary. If |G| = p^{2} for some prime p, then G ≅ Z_{p} × Z_{p} or Z_{p^{2}}.

Proof. Since |G| = p^{2} we know that Z(G) ≠ 1. Therefore G/Z(G) has order 1 or p, so is cyclic. By the theorem, G is abelian. If G has an element of order p^{2}, then G ≅ Z_{p^{2}}. Otherwise every element of G has order p (except the identity element). Choose g ∈ G with g ≠ 1, so ⟨g⟩ ≅ Z_{p}. Then ⟨g⟩ ≠ G and we may choose h ∈ G with h ∉ ⟨g⟩. Construct the abstract group ⟨g⟩ × ⟨h⟩ and define

ϕ : ⟨g⟩ × ⟨h⟩ → G

by ϕ((g^{i}, h^{j})) = g^{i}h^{j}. Since G is abelian this turns out to be a homomorphism. Its image clearly contains both ⟨g⟩ and ⟨h⟩ so by Lagrange’s Theorem its image is all of G. Since |⟨g⟩ × ⟨h⟩| = p^{2} = |G| and ϕ is surjective, ϕ must be an isomorphism. QED