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.

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