Let be a prime and let be a finite -group.

- Prove that is an elementary abelian -group. [Hint: Show that and that for all .]
- Prove that if is a normal subgroup such that is elementary abelian then . State this (universal) property in terms of homomorphisms and commutative diagrams.
- Let be elementary abelian of order (by part (a)). Deduce from this previous exercise that if , , is a basis for the -dimensional vector space over , then . Show conversely that if is any set of generators for , then . (You may assume that every minimal generating set for an -dimensional vector space has elements; i.e., every basis has elements.) Deduce
*Burnside’s Basis Theorem*: a set is a minimal generating set for if and only if is a basis for . Deduce that any minimal generating set for has elements. - Prove that if is cyclic then is cyclic. Deduce that if is cyclic then so is .
- Let be any automorphism of with prime order and . Show that if fixes the coset then fixes some element of this coset. [Hint: Every automorphism of induces an automorphism of . Use the observation that acts as a permutation of order 1 or on the elements in .]
- Use parts 5 and 3 to deduce that every nontrivial automorphism of of order prime to induces a nontrivial automorphism of . Deduce that any group of automorphisms of which has order prime to is isomorphic to a subgroup of .

- [With hints from PlanetMath.]
We begin with some lemmas.

Lemma 1: If is a finite elementary abelian -group, then . Proof: Let . Then we have . Now the subgroup (as defined in this previous exercise) is maximal, and clearly . Thus , so that .

Lemma 2: If is a group and such that , then . Proof: Note that if and only if for all maximal , , if and only if for all maximal , there exists such that , if and only if .

Lemma 3: If is a finite -group, then , where . Proof: Let be maximal. By Sylow’s Theorem, is prime, so that . Since is abelian, . Since every element of satisfies , for all . Thus . Now since is characteristic, hence normal, is a subgroup for all maximal . Thus .

Lemma 4: . Proof: By Lemma 3, . Consider . Since , is abelian, and since , is elementary abelian. By the Lattice Isomorphism Theorem and the definition of , the maximal subgroups of are in bijective correspondence with the maximal subgroups of . By Lemma 1 and Lemma 2, we have . Thus .

Lemma 3 implies that .

- Let be such a subgroup. Since is abelian, , and since for all , . Thus .
We will give the “universal property” of the Frattini subgroup of a -group in terms of short exact sequences. A short exact sequence of groups is a pair of homomorphisms such that is injective, is surjective, and . We will often abbreviate short exact sequences with a diagram as follows.

If and are short exact sequences, a homomorphism (of short exact sequences) is a triple such that the following diagram commutes.

Now if is a -group, then letting be the Frattini subgroup and and the injection and natural projection, respectively, is a short exact sequence of groups whose last term is elementary abelian. Moreover, this SES has the following “universal property”: If is any other short exact sequence whose middle term is and whose last term is elementary abelian, then there exists an injective homomorphism and a surjective homomorphism such that is a homomorphism of short exact sequences. Moreover, the homomorphisms are unique (since injective and surjective maps are left- and right-cancellable, respectively).

- Let have rank and let , , be a minimal generating set of . Let , and suppose is proper. Let be a minimal set such that . Note that each has the form where and ; thus, without loss of generality, we may assume that . Now is proper and ; since is finite, there must exist an element and a subset such that is finite but . This is a contradiction because is the set of nongenerators. Thus .
Now let . Clearly none of the are in , so that . Since is a dimension vector space, .

We now approach Burnside’s Basis Theorem. Suppose is a minimal generating set of . Now is a generating set of ; if this generating set is not minimal, say is redundant, then by the above argument , a contradiction since the generating set of is minimal. Thus is a basis for . Suppose is a basis for ; by the first argument above, is a generating set for of size . By the second argument above, this generating set is minimal.

Suppose now that is a minimal generating set for . Then is a minimal generating set (basis) for , and thus has cardinality .

- If is cyclic, then there is a generating set for with one element. By Burnside’s Basis Theorem, there is a corresponding generating set for with one element; thus is cyclic. Suppose now that is cyclic. By the Third Isomorphism Theorem and part (1), we have , so that is cyclic, hence is cyclic.
- Now acts on the set by application. By the orbit-stabilizer theorem, we have , so that each orbit of this action has order 1 or . If no orbit has order 1, then we have , where . This is a contradiction since ; thus some orbit has order 1, and we have .
- Let be a nontrivial automorphism of such that and are coprime. Now induces an automorphism of by since is characteristic. Suppose is trivial, and choose a power of prime order ; note that . Let be the automorphism of induced by . By Part 5 above, for each , there exists an element which is fixed by . Moreover, by part 3, is a generating set for . Since for each , is trivial – a contradiction since has prime order in . Thus every nontrivial automorphism of of order prime to induces a nontrivial automorphism of .
Now let be a subgroup such that . For every , is necessarily coprime to by Lagrange. Thus every induces a homomorphism ; define by . This mapping is clearly a homomorphism. Moreover, if , then for all , we have . If is a generator of , then (since is characteristic in ) and are also generators, a contradiction. Thus , so that . Thus is injective, and the desired result follows.