x 2 = xn xn = x 2 N = N = 0


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1 Potpourri. Spring 2010 Problem 2 Let G be a finite group with commutator subgroup G. Let N be the subgroup of G generated by the set {x 2 : x G}. Then N is a normal subgroup of G and N contains G. Proof. Let h N. Then h = x 2 1 x 2 n for some elements x i G. Let g G and note that ghg 1 = (gx 1 g 1 ) 2 (gx n g 1 ) 2 N. So N is normal. Now, let x, y G. Note that (xyn)(xyn) = (xy) 2 N = N. On the other hand, (xn)(xn) = N = (yn)(yn). So xn = x 1 N and yn = y 1 N. So (xyn)(xyn) = (xyn)(x 1 y 1 N) = xyx 1 y 1 N = N. Hence, G N. Spring 2007 Problem 2. Let G be a group with commutator subgroup [G : G]. Let N be the subgroup of G generated by {x 2 : x G}. N is a normal subgroup of G containing [G, G]. Proof. Note that gx 2 g 1 = gxg 1 gxg 1 = (gxg 1 ) 2 N. Indeed, gxyg 1 = gxg 1 gyg 1. And so since conjugation by an arbitrary element g is multiplicative, every element of N is a product of squares, and conjugation by an arbitrary element g of a square is a square, we have that gng 1 N, that is, N is normal. Consider G/N. Let x G/N. Then x 2 = xn xn = x 2 N = N = 0 So every element of G/N has order 2. So xȳ xȳ = 0, that is xȳ = ( xȳ) 1, that is, xȳ = ȳ x. So G/N is abelian. So xyx 1 y 1 N = N, that is, [G, G] N. Spring 2009 Problem 2. Suppose a finite group G has normal subgroups A, B G such that A B = {e}. Assume that (i) AB is a subgroup. (ii) ab = ba for all a A, b B. 1
2 2 (iii) Every g AB can be expressed uniquely as a product g = ab for some a A and b B. (iv) The four subgroups {e}, A, B, and AB are normal in AB. Suppose A and B are simple groups, and there exists a normal subgroup N AB that is not equal to one of the four subgroups listed in (iv) above. Then A B. Proof. Note that N A and N B must be normal in A and B, respectively. Since N isn t A, B, or AB, this implies that N A = N B = {e} since A and B are simple. Suppose ab 1, ab 2 N. Then ab 1 (ab 2 ) 1 = b 1 b 1 2 N. So it must be that b 1 b 1 2 = e, that is, b 1 = b 2. Likewise, if a 1 b, a 2 b N, then a 1 = a 2. Let π : AB A be defined by ab a. This map is welldefined by (iii), and is easily verified to be a group homomorphism in lieu of (ii). By the isomorphism theorems, π(n) is a normal subgroup of A. Hence π(n) is either all of A or {e}. But π(n) {e} since this would imply N B {e} or N = {e}. So π(n) = A. We have proven the following: for each a A, there is a unique b(a) = b B such that ab N. Let ϕ : A B be defined by a b(a). It s easy to verify that ϕ is a group homomorphism with trivial kernel. Hence A B. The above arguments are completely symmetric in A and B, so we conclude that B A. Hence A B. Autumn 2008 Problem 2. Which of the following groups are isomorphic? (a) the multiplicative group of the integers modulo 13 (b) the multiplicative group of the integers modulo 28 (c) the alternating group on 4 letters (d) the group of symmetries of the regular hexagon. Proof. First, (a) and (b) are abelian while (c) and (d) are not. So the only possibilities for isomorphisms are from (a) to (b) and from (c) to (d). But (a) Z/12Z is cyclic while (b) (Z/7Z Z/4Z) Z/6Z Z/2Z is not. So (a) is not isomorphic to (b). What s more, (d) has a element of order 6 while (c) does not. So (c) is not isomorphic to (d). Autumn 2007 Problem 1.
3 3 Let G be a group, and suppose that G has a normal subgroup N of order p (where p is prime). Let g be an element of G. Then N is contained in the centralizer of g p 1. Proof. Since N = p a prime, we have that N is cyclic, generated by an element, say x. Since N is normal, gxg 1 = x j for some j < p. Since conjugation by g is an automorphism of G, we have that g p 1 xg 1 p = x jp 1. Since j < p, p j. So by Fermat s Little theorem, j p 1 1 (mod p). Hence g p 1 xg 1 p = x, that is, x 1 g p 1 x = g p 1. Since x generates N, it follows that y 1 g p 1 y = g p 1 for all y N, that is, N C G (g p 1 ). Autumn 2004 Problem 5. Let G be a group such that G/Z is cyclic where Z is the center of G. Then G is abelian. Proof. Let z = G/Z. Let g, h G, and let c Z. Then there exists z 1, z 2 Z and n, m Z >0 such that hg = (hc)(gc 1 ) = (z n z 1 )(z m z 2 ) = (z n z m )(z 1 z 2 ) = (z m z n )(z 2 z 1 ) = (z m z 2 )(z n z 1 ) = gh. Spring 2003 Problem 2. Let G be a group. For a, b G put a b = b 1 ab, [a, b] = a 1 b 1 ab. (a) [a, bc] = [a, c][a, b] c for a, b, c G. Proof. Note that [a, c][a, b] c = a 1 c 1 ac(a 1 b 1 ab) c = a 1 c 1 acc 1 a 1 b 1 abc = a 1 c 1 b 1 abc = [a, bc]. (b) Suppose that G = AB, where A, B are subgroups of G. Let [A, B] denote the subgroup of G generated by all [a, b], where a A, b B. Then [A, B] is a normal subgroup of G.
4 4 Proof. Let a, α A and b B. Then [a, b] α = α 1 a 1 b 1 abα = ( (aα) 1 b 1 (aα)b ) ( b 1 α 1 bα ) = [aα, b] [α, b] 1 [A, B]. Since [a, b] β = [a, β] 1 [a, bβ] by part (a), we have that [a, b] β [A, B] for all β B. Since G = AB and [A, B] is generated by elements of the form [a, b] (and considering that conjugation is an automorphism of G), it follows that [A, B] is normal. Alternatively, note that for a, α A and b B, we have [a, b] α = ([b, a] 1 ) α = ([b, a] α ) 1 = ([b, α] 1 [b, aα]) 1 = [b, aα] 1 [b, α] = [aα, b][α, b] 1 [A, B]. The fact that [a, b] β [A, B] follows from part (a), and the rest follows identically. (c) Suppose, in addition, that A and B are abelian. abelian and [G, G] = [A, B]. Then G/[A, B] is Proof. Let a 1, a 2 A and b 1, b 2 B. Since a 1 1 b 1 1 a 1b 1 [A, B] we have that a 1 b 1 [A, B] = b 1 a 1 [A, B], in G/[A, B]. So the fact that g 1 g 2 [A, B] = g 2 g 1 [A, B] follows from the fact that G = AB, A commutes with B, and A and B are internally commutative. Hence G/[A, B] is abelian. Now, it s obvious that [A, B] [G, G]. On the other hand, let g 1 = a 1 b 1 and g 2 = a 2 b 2. So g 1 1 g 1 2 g 1g 2 = b 1 1 a 1 1 b 1 2 a 1 2 a 1b 1 a 2 b 2 = b 1 1 a 1 1 b 1 2 a 1a 1 2 b 1a 2 b 2 = (b 1 b 2 ) 1 [a 1, b 1 2 ] 1 [a 2, b 1 1 ](b 1b 2 ). [A, B], since [A, B] is a normal subgroup of G. Hence [G, G] [A, B], so [G, G] = [A, B]. Alternatively, since G/[A, B] is abelian, we must have that [A, B] contains the commutator subgroup [G, G].
5 5 Autumn 2010 Problem 1. Suppose that G is a group, and H is a nontrivial subgroup such that H J for every nontrivial subgroup J of G. Then H is contained in the center of G. Proof. Suppose G is finite. Since H is contained in every nontrivial subgroup, H is contained in every Sylowp subgroup of G. Since H is nontrivial, it must be that G is a pgroup. If G has a trivial center, then the class equation implies that p n = G = 1 + p c i 1 mod p where the sum is over the distinct nontrivial conjugacy classes of G and p c i is the order said conjugacy classes. This is clearly absurd, so G does not have a trivial center. By assumption, H then must be contained in the center. Suppose G is not finite. Then H is contained in the centralizer of every element of G. So H is contained in the intersection of all the centralizers, so H is contained in the center of G. Spring 2001 Problem 2. Let G be a finite group and P a Sylowp subgroup of G. Let K be a subgroup of G which contains N G (P), the normalizer of P in G. Then N G (K) = K. Proof. Note that P N G (P) K. Let g N G (K). Then gkg 1 = K gpg 1 = P. So P is a Sylowp subgroup of K. Since the Sylowp subgroups of K are conjugate in K, let k K such that kp k 1 = P. Then kgpg 1 k 1 = P, so kg N G (K) K. Since k K, it follows that g K. So N G (K) K, and the reverse inclusion is obvious. Autumn 2000 Problem 1. Let G be a finite group and p a prime dividing the order of G. Assume that a Sylowp subgroup P of G is cyclic. (a) P has a unique subgroup of order p. Proof. Cyclic groups have unique subgroups of all orders dividing the order of the group. To be precise, let P = x and suppose P = p n. Then x pn 1 is a subgroup of order p. Let H be another subgroup of order p. Then H = y for some y P. Write y = x k for some positive integer k. Since y p = x kp = 1, we have that p n pk, so p n 1 k. So H x pn 1. Since H = x pn 1, we get H = x pn 1. (b) Suppose P x 1 Px {1} for all x G. Then G possesses a nonidentity normal p subgroup.
6 6 Proof. Let H be the unique order p subgroup of P. For any x G, we have that P xpx 1 is a nontrivial pgroup. All the Sylowp subgroups of G are conjugate in G, hence isomorphic. So xpx 1 is cyclic too. So P xpx 1 is a cyclic pgroup, thus, by part (a), it contains a subgroup of order p, say H. But H P, hence H = H. So H P xpx 1 for all x G. Since conjugation is an automorphism, it must be that H = xhx 1. So xhx 1 = H for all x G. Autumn 2000 Problem 2 Let Q + be the additive group of rational numbers. (a) Suppose M is a maximal subgroup of Q +. Then Q + /M is a group of prime order. Proof. Let x Q + /M be nonzero. Then Q + /M = x. Suppose Q + /M =. Then Q + /M Z. But this is impossible since Z has (many) proper subgroups implying via a pullback proper subgroups containing M in Q + not equal to M. So Q + /M = m <. Let p m where p is prime. Then Q + /M contains a subgroup of order p (by Cauchy). If this subgroup is proper, then, again, via a pullback we would have a proper subgroup of Q + containing M not equal to M. So this subgroup must not be proper, that is, p = m. (b) Q + does not possess maximal subgroups. Proof. Let a/b Q +. By part (a), for any x Q +, we have that px M. Set x = a/(pb). Then a/b M. Since a/b was arbitrary, it follows that M = Q +, a contradiction. Autumn 1999 Problem 2. Let G be a finite group with a unique maximal subgroup. Show that G is a cyclic pgroup. Proof. Let M be the unique maximal subgroup of G. If G is composite, then M contains every Sylow subgroup from which we infer M = G, a contradiction. So G is a pgroup. Let g G \ M. If g G, then g M by the uniqueness property of M, a contradiction. So g = G, i.e., G is cyclic.
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