## Burnside’s Basis Theorem and some basic consequences

Let $p$ be a prime and let $P$ be a finite $p$-group.

1. Prove that $P/\Phi(P)$ is an elementary abelian $p$-group. [Hint: Show that $P^\prime \leq \Phi(G)$ and that $x^p \in \Phi(P)$ for all $x \in P$.]
2. Prove that if $N \leq P$ is a normal subgroup such that $P/N$ is elementary abelian then $\Phi(P) \leq N$. State this (universal) property in terms of homomorphisms and commutative diagrams.
3. Let $P/\Phi(P)$ be elementary abelian of order $p^r$ (by part (a)). Deduce from this previous exercise that if $x_i \Phi(P)$, $1 \leq i \leq r$, is a basis for the $r$-dimensional vector space $P/\Phi(P)$ over $\mathbb{F}_p$, then $P = \langle x_1, \ldots, x_r \rangle$. Show conversely that if $y_1, \ldots, y_s$ is any set of generators for $P$, then $s \geq r$. (You may assume that every minimal generating set for an $r$-dimensional vector space has $r$ elements; i.e., every basis has $r$ elements.) Deduce Burnside’s Basis Theorem: a set $y_1, \ldots, y_s$ is a minimal generating set for $P$ if and only if $y_1 \Phi(P), \ldots, y_s \Phi(P)$ is a basis for $P/\Phi(P)$. Deduce that any minimal generating set for $P$ has $r$ elements.
4. Prove that if $P/\Phi(P)$ is cyclic then $P$ is cyclic. Deduce that if $P/P^\prime$ is cyclic then so is $P$.
5. Let $\sigma$ be any automorphism of $P$ with prime order $q$ and $q \neq p$. Show that if $\sigma$ fixes the coset $x \Phi(P)$ then $\sigma$ fixes some element of this coset. [Hint: Every automorphism of $P$ induces an automorphism of $P/\Phi(P)$. Use the observation that $\sigma$ acts as a permutation of order 1 or $q$ on the $p^a$ elements in $x \Phi(P)$.]
6. Use parts 5 and 3 to deduce that every nontrivial automorphism of $P$ of order prime to $p$ induces a nontrivial automorphism of $P/\Phi(P)$. Deduce that any group of automorphisms of $P$ which has order prime to $p$ is isomorphic to a subgroup of $\mathsf{Aut}(P/\Phi(P)) = GL_r(\mathbb{F}_p)$.

1. [With hints from PlanetMath.]

We begin with some lemmas.

Lemma 1: If $G$ is a finite elementary abelian $p$-group, then $\Phi(G) = 1$. Proof: Let $I = \{1,2,\ldots,n\}$. Then we have $G = \times_I Z_p$. Now the subgroup $M_i = G_{I \setminus \{i\}}$ (as defined in this previous exercise) is maximal, and clearly $\bigcap_I M_i = 1$. Thus $\Phi(G) \leq 1$, so that $\Phi(G) = 1$. $\square$

Lemma 2: If $G$ is a group and $N \leq G$ such that $N \leq \Phi(G)$, then $\Phi(G/N) = \Phi(G)/N$. Proof: Note that $xN \in \Phi(G/N)$ if and only if for all maximal $M/N \leq G/N$, $xN \in M/N$, if and only if for all maximal $M \leq G$, there exists $m \in M$ such that $xm^{-1} \in N$, if and only if $xN \in \Phi(G)/N$. $\square$

Lemma 3: If $P$ is a finite $p$-group, then $[P,P]P^p \leq \Phi(P)$, where $P^p = \{x^p \ |\ x \in P \}$. Proof: Let $Q \leq P$ be maximal. By Sylow’s Theorem, $[P:Q] = p$ is prime, so that $P/Q \cong Z_p$. Since $P/Q$ is abelian, $[P,P] \leq Q$. Since every element of $P/Q$ satisfies $(xQ)^p = Q$, $x^p \in Q$ for all $x \in P$. Thus $P^p \leq Q$. Now since $[P,P] \leq P$ is characteristic, hence normal, $[P,P]P^p \leq Q$ is a subgroup for all maximal $Q \leq P$. Thus $[P,P]P^p \leq \Phi(P)$. $\square$

Lemma 4: $\Phi(P) = [P,P] P^p$. Proof: By Lemma 3, $H = [P,P]P^p \leq \Phi(P)$. Consider $P/H$. Since $[P,P] \leq H$, $P/H$ is abelian, and since $P^p \leq H$, $P/H$ is elementary abelian. By the Lattice Isomorphism Theorem and the definition of $\Phi(P)$, the maximal subgroups of $P$ are in bijective correspondence with the maximal subgroups of $P/H$. By Lemma 1 and Lemma 2, we have $\Phi(P)/H = \Phi(P/H) = 1$. Thus $\Phi(P) = [P,P]P^p$. $\square$

Lemma 3 implies that $P^\prime \leq \Phi(P)$.

2. Let $N$ be such a subgroup. Since $P/N$ is abelian, $[P,P] \leq N$, and since $x^p \in N$ for all $x \in P$, $P^p \leq N$. Thus $\Phi(P) = [P,P] P^p \leq N$.

We will give the “universal property” of the Frattini subgroup of a $p$-group in terms of short exact sequences. A short exact sequence of groups is a pair of homomorphisms $(\iota, \pi)$ such that $\iota$ is injective, $\pi$ is surjective, and $\mathsf{cod}(\iota) = \mathsf{dom}(\pi)$. We will often abbreviate short exact sequences with a diagram as follows.

Short Exact Sequence of Groups

If $(\iota_1,\pi_1)$ and $(\iota_2,\pi_2)$ are short exact sequences, a homomorphism (of short exact sequences) is a triple $(\alpha,\beta,\gamma)$ such that the following diagram commutes.

Homomorphism of Short Exact Sequences

Now if $P$ is a $p$-group, then letting $\Phi(P)$ be the Frattini subgroup and $\iota$ and $\pi$ the injection and natural projection, respectively, $(\iota,\pi)$ is a short exact sequence of groups whose last term is elementary abelian. Moreover, this SES has the following “universal property”: If $(\iota_2,\pi_2)$ is any other short exact sequence whose middle term is $P$ and whose last term is elementary abelian, then there exists an injective homomorphism $\alpha : \Phi(P) \rightarrow N$ and a surjective homomorphism $\gamma : P/\Phi(P) \rightarrow P/N$ such that $(\alpha, \mathsf{id}_P, \gamma)$ is a homomorphism of short exact sequences. Moreover, the homomorphisms are unique (since injective and surjective maps are left- and right-cancellable, respectively).

3. Let $P/\Phi(P)$ have rank $r$ and let $x_i \Phi(P)$, $1 \leq i \leq r$, be a minimal generating set of $P/\Phi(P)$. Let $H = \langle x_1, \ldots, x_r \rangle$, and suppose $H \leq P$ is proper. Let $S$ be a minimal set such that $\langle H, S \rangle = P$. Note that each $s \in S$ has the form $wz$ where $w \in H$ and $z \in \Phi(P)$; thus, without loss of generality, we may assume that $S \subseteq \Phi(P)$. Now $H \leq P$ is proper and $\langle H, S \rangle = P$; since $P$ is finite, there must exist an element $s \in S$ and a subset $T \subseteq S$ such that $\langle H, T \rangle \leq P$ is finite but $\langle H, T, s \rangle = P$. This is a contradiction because $\Phi(P)$ is the set of nongenerators. Thus $P = H$.

Now let $P = \langle y_1, \ldots, y_s \rangle$. Clearly none of the $y_i$ are in $\Phi(P)$, so that $P/\Phi(P) = \langle y_1 \Phi(P), \ldots, y_s \Phi(P) \rangle$. Since $P/\Phi(P)$ is a dimension $r$ vector space, $s \geq r$.

We now approach Burnside’s Basis Theorem. $(\Rightarrow)$ Suppose $P = \langle y_1, \ldots, y_s \rangle$ is a minimal generating set of $P$. Now $y_i \Phi(P)$ is a generating set of $P/\Phi(P)$; if this generating set is not minimal, say $y_s \Phi(P)$ is redundant, then by the above argument $P = \langle y_1, \ldots, y_{s-1} \rangle$, a contradiction since the generating set of $P$ is minimal. Thus $y_i \Phi(P)$ is a basis for $P/\Phi(P)$. $(\Leftarrow)$ Suppose $y_i \Phi(P)$ is a basis for $P/\Phi(P)$; by the first argument above, $y_i$ is a generating set for $P$ of size $r$. By the second argument above, this generating set is minimal.

Suppose now that $y_i$ is a minimal generating set for $P$. Then $y_i \Phi(P)$ is a minimal generating set (basis) for $P/\Phi(P)$, and thus has cardinality $r$.

4. If $P/\Phi(P)$ is cyclic, then there is a generating set for $P/\Phi(P)$ with one element. By Burnside’s Basis Theorem, there is a corresponding generating set for $P$ with one element; thus $P$ is cyclic. Suppose now that $P/P^\prime$ is cyclic. By the Third Isomorphism Theorem and part (1), we have $P/\Phi(P) \cong (P/P^\prime)/(\Phi(P)/P^\prime)$, so that $P/\Phi(P)$ is cyclic, hence $P$ is cyclic.
5. Now $\langle \sigma \rangle$ acts on the set $x \Phi(P)$ by application. By the orbit-stabilizer theorem, we have $[\langle \sigma \rangle, \mathsf{stab}(x)] = |\langle \sigma \rangle \cdot x|$, so that each orbit of this action has order 1 or $q$. If no orbit has order 1, then we have $qk = p^a$, where $p^a = |\Phi(P)|$. This is a contradiction since $p \neq q$; thus some orbit $\{z\}$ has order 1, and we have $\sigma(z) = z$.
6. Let $\alpha$ be a nontrivial automorphism of $P$ such that $|\alpha| = k$ and $p$ are coprime. Now $\alpha$ induces an automorphism $\overline{\alpha}$ of $P/\Phi(P)$ by $\overline{\alpha}(x\Phi(P)) = \alpha(x) \Phi(P)$ since $\Phi(P)$ is characteristic. Suppose $\overline{\alpha}$ is trivial, and choose a power $\beta = \alpha_m$ of prime order $q$; note that $q \neq p$. Let $\overline{\beta}$ be the automorphism of $P/\Phi(P)$ induced by $\beta$. By Part 5 above, for each $x_i \Phi(P) \in P/\Phi(P)$, there exists an element $y_i \in x_i \Phi(P)$ which is fixed by $\beta$. Moreover, by part 3, $y_i$ is a generating set for $P$. Since $\beta(y_i) = y_i$ for each $y_i$, $\beta$ is trivial – a contradiction since $\beta$ has prime order in $\mathsf{Aut}(P)$. Thus every nontrivial automorphism of $P/\Phi(P)$ of order prime to $p$ induces a nontrivial automorphism of $P/\Phi(P)$.

Now let $K \leq \mathsf{Aut}(P)$ be a subgroup such that $\mathsf{gcd}(p,|K|) = 1$. For every $\varphi \in K$, $|\varphi|$ is necessarily coprime to $p$ by Lagrange. Thus every $\varphi \in K$ induces a homomorphism $\overline{\varphi} \in \mathsf{Aut}(P/\Phi(P))$; define $\theta : K \rightarrow \mathsf{Aut}(P/\Phi(P))$ by $\theta(\psi)(x \Phi(P)) = \theta(x) \Phi(P)$. This mapping is clearly a homomorphism. Moreover, if $\theta(\varphi) = \theta(\psi)$, then for all $x \Phi(P)$, we have $\varphi(x)\psi(x)^{-1} \in \Phi(P)$. If $x$ is a generator of $P$, then (since $\Phi(P)$ is characteristic in $P$) $\varphi(x)$ and $\psi(x)$ are also generators, a contradiction. Thus $\varphi(x) \psi(x)^{-1} = 1$, so that $\varphi(x) = \psi(x)$. Thus $\theta$ is injective, and the desired result follows.