p-Groups and Inductive methods.

Spring 2010 Let p be a prime, and let P be a finite p-. Suppose that Q 6 P is a proper . Then N(Q) 6= Q.

Proof. Let |P| = pn. We proceed by induction on n. Suppose n = 1. Then Q = {e} and N(Q) = P. Now suppose the claim is true for all positive less or equal n+ to some n ∈ Z. Suppose |P| = p 1 and let Q be a proper subgroup. Let Z be the of P. Note that Z 6 N(Q) and Z is non-trivial. If Z 6 Q, then N(Q) 6= Q, as desired. k If Z 6 Q, consider P/Z. This group has p where k 6 n, so the induction hypothesis applies. Let Q¯ be the of Q in P/Z. Then there exists some subgroup N¯ of P/Z such that Q¯ is normal in N¯ and Q¯ 6= N¯ . The then guarantee that there exists some subgroup N of P such that Q is normal in N and Q 6= N. Hence N(Q) 6= Q. This proves the claim.  Spring 2002 Problem 5(a). If G is a nontrivial p-group, then the center of G is nontrivial.

Proof. This follows from the equation in conjunction with the stabilizer . To be precise, conjugation partitions G into orbits where two elements belong to the same orbit if they are conjugate in G. The orbits constitute those elements that belong to the center. Let Z denote the subgroup consisting of these elements. Note that no orbit is all of G since no orbit contains the identity (save the singleton orbit {e} 6= G, of course). The size of these orbits is −1 −1 determined by the orbit stabilizer theorem: |GxG | = |G|/CG(x) where GxG = −1 −1 {gxg : g ∈ G}, the orbit of x, and CG(x) = {g ∈ G : gxg = x}, the centralizer of k(x) x in G. Since G is a p-group, we get that |G|/CG(x) = p for some k(x) > 1 for every x belonging to a non-singleton orbit. Let X be a of unique representatives from each non-singleton orbit. Then by what has already been shown we have that

|G| = |Z| + pk(x) x∈X X Hence |Z| ≡ 0 mod p. This implies that p | |Z|, hence Z is non-trivial, as claimed.  Spring 2002 Problem 5(b). Let M be a of a non-trivial p-group G. Then [G : M] = p.

Proof. Let Z be the center of G, and let M be a maximal subgroup of G. Then either MZ = M or MZ = G. If MZ = G, then M is normal in G. Let x ∈ G/M be 1 2 of order p. Let Mf = π−1(hxi) where π : G → G/M is the natural . Since M is maximal, it must be that Mf = G, so p = |hxi| = [Mf : M] = [G : M]. Now, suppose MZ = M. Then Z ⊆ M. Suppose for all p-groups of order less than G we have that the claim holds. Let π : G → G/Z be the natural map. Then, by the correspondence theorem, π(M) must be a maximal subgroup of G/Z. So, by our assumption (and by part(a)), p = [G/Z : π(M)] = [G : M]. Now, if |G| = p, there’s nothing to prove. Hence, by induction, the claim holds.  Spring 2002 Problem 5(c). A maximal subgroup of a non-trivial p-group G is normal in G.

Proof. Let M be a maximal subgroup of G and suppose M is not normal. By part (b), [G : M] = p. By the orbit stabilizer theorem, we have that M has p conjugate in G. Let G act on M and its conjugates by conjugation. This induces a ϕ : G → Sp, the on p letters. Since p itself is the highest power of p dividing |Sp|, it follows that | ker ϕ| = |M|. On the other hand, ker ϕ ⊆ NG(M) = M where NG(M) is the normalizer of M in G. So ker ϕ = M, that is, M is normal. This is absurd. It follows that if M is maximal, then M must be normal.  Autumn 2001 Problem 1. Let N be a non-identity of a finite p-group P. Then N intersects the center of P non-trivially.

Proof. Let P act on N by conjugation. This action partitions N into orbits. The singleton orbits consist of those elements in N contained in the center of P. Let Z −1 be the center of P. Now, for any x ∈ N, let CP(x) = {g ∈ P : gxg = x}. Let −1 orbP(x) = {gxg : g ∈ P}. By the orbit-stabilizer theorem we know that

| orbP(x)| = |P|/|CP(x)|. Let X be a set of unique representatives from each non-singleton orbit of N. Then

|N| = |Z ∩ N| + |P|/|CP(x)|. x∈X X Note that |N| and |P|/|CP(x)| are non-trivial p-powers. It follows that |Z ∩ N| ≡ 0 mod p. Since |Z ∩ N| > 1, it must be that |Z ∩ N| is non-trivial, as claimed.