p-Groups and Inductive methods. Spring 2010 Let p be a prime, and let P be a finite p-group. Suppose that Q 6 P is a proper subgroup. Then N(Q) 6= Q. Proof. Let jPj = pn. We proceed by induction on n. Suppose n = 1. Then Q = feg and N(Q) = P. Now suppose the claim is true for all positive integers less or equal n+ to some integer n 2 Z. Suppose jPj = p 1 and let Q be a proper subgroup. Let Z be the center 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 order p where k 6 n, so the induction hypothesis applies. Let Q¯ be the image 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 isomorphism theorems 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 class equation in conjunction with the orbit stabilizer theorem. To be precise, conjugation partitions G into orbits where two elements belong to the same orbit if they are conjugate in G. The singleton 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 element (save the singleton orbit feg 6= G, of course). The size of these orbits is -1 -1 determined by the orbit stabilizer theorem: jGxG j = jGj=CG(x) where GxG = -1 -1 fgxg : g 2 Gg, the orbit of x, and CG(x) = fg 2 G : gxg = xg, the centralizer of k(x) x in G. Since G is a p-group, we get that jGj=CG(x) = p for some k(x) > 1 for every x belonging to a non-singleton orbit. Let X be a set of unique representatives from each non-singleton orbit. Then by what has already been shown we have that jGj = jZj + pk(x) x2X X Hence jZj ≡ 0 mod p. This implies that p j jZj, hence Z is non-trivial, as claimed. Spring 2002 Problem 5(b). Let M be a maximal subgroup 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 2 G=M be 1 2 of order p. Let Mf = π-1(hxi) where π : G ! G=M is the natural map. Since M is maximal, it must be that Mf = G, so p = jhxij = [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 jGj = 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 subgroups in G. Let G act on M and its conjugates by conjugation. This induces a homomorphism ' : G ! Sp, the symmetric group on p letters. Since p itself is the highest power of p dividing jSpj, it follows that j ker 'j = jMj. On the other hand, ker ' ⊆ NG(M) = M where NG(M) is the normalizer of M in G. So ker ' = 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 normal subgroup 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 2 N, let CP(x) = fg 2 P : gxg = xg. Let -1 orbP(x) = fgxg : g 2 Pg. By the orbit-stabilizer theorem we know that j orbP(x)j = jPj=jCP(x)j. Let X be a set of unique representatives from each non-singleton orbit of N. Then jNj = jZ \ Nj + jPj=jCP(x)j. x2X X Note that jNj and jPj=jCP(x)j are non-trivial p-powers. It follows that jZ \ Nj ≡ 0 mod p. Since jZ \ Nj > 1, it must be that jZ \ Nj is non-trivial, as claimed. .