ALGEBRA HW 1 CLAY SHONKWILER 1 Define the center of a group G to be Z = {g ∈ G|(∀h ∈ G)gh = hg}. (a) Is Z a subgroup? Is it normal? Answer: Yes to both questions. To demonstrate that Z is a subgroup, we must demonstrate that it is closed under the group operation and that each element in the center has an inverse in the center (since associativity will be inherited and the identity is clearly in the center). To that end, let a, b ∈ Z and let h ∈ G. Then h(ab)h−1 = hah−1hbh−1 = ab, so Z is closed under the group operation. Now, let a ∈ Z. Then a has an inverse a−1 in G; we want to show that a−1 ∈ Z. Let h ∈ G. Then (ha−1h−1)−1 = hah−1 = a, which implies that ha−1h−1 = a−1, so a−1 ∈ Z. Therefore, we conclude that Z is, in fact, a subgroup of G. Furthermore, the calculation we did above to demonstrate that Z is closed under the group operation is precisely the calculation necessary to demonstrate that Z is normal in G. ♣ (b) Find the center of Cn, Dn, Sn, An, Q, Z, GL2(R). Answer: Let a ∈ Cn. Then, for all b ∈ Cn, ab = ba, since Cn is abelian. Since our choice of a was arbitrary, we see that the center of Cn is Cn itself. n 2 −1 −1 Consider Dn under the presentation hx, y|x = y = 1, yxy = x i. Then if σ, τ ∈ Dn, there exist i, j ∈ {0, . , n − 1}, a, b ∈ {0, 1} such that σ = xiya, τ = xjyb. Then if a = b = 1, στ = xiyaxjyb = xix−jyayb = xi−jya+b and τσ = xjybxiya = xjx−iybya = xj−iya+b. i−j j−i If σ in the center of Dn, then it must be the case that x = x for all choices of j; Clearly this is impossible. On the other hand, if a = 0, then στ = xixjyb = xi+jyb and xj−iyb if b = 1 τσ = xjybxi = xj+i if b = 0. 1 2 CLAY SHONKWILER If σ in the center of Dn, then it must be the case that i + j ≡ j − i( mod n) for all choices of j. This, in turn, implies that 2i ≡ 0(modn), which is to say that i = n/2, which is only possible if n is even. Therefore, we conclude n/2 that the center of Dn is trivial when n is odd and is {1, x } when n is even. Since S2 = C2, the center of S2 is itself. If n ≥ 3, let σ ∈ Sn. We can write σ as the product of a unique decomposition of disjoint cycles, σ = ζ1 ··· ζj. If j = 1 and σ is just a transposition, then σ = ζ1 = (ab) for some a, b ∈ {1, . , n}. Then (bc)σ = (bc)(ab) = (acb) 6= (abc) = (ab)(bc) = σ(bc), so σ is not in the center of Sn. If j > 1, then we know that ζ1 = (a1a2 . am) for m ≤ n, ai ∈ {1, . , n}. Then, since a1 is fixed by ζi for i > 1 (because the cycle decomposition of σ is into disjoint cycles), the product (a1a2)σ = (a1a2)ζ1 ··· ζj = (a1a2)(a1a2 . am)ζ2 ··· ζj fixes a1. On the other hand, σ(a1a2) = (a1a2 . am)ζ2 ··· ζj(a1a2) = (a1a3 . d)ξ1 ··· ξk, where d ∈ {1, . , n} and the ξl are disjoint cycles. Specifically, a1 is not fixed by σ(a1a2), so we see that σ is not in the center of Sn. Since our choice of σ was arbitrary, we see that the center of Sn is trivial. When n ≤ 3, An is cyclic and so its center is all of An. If n ≥ 4, let σ ∈ Sn. Then we can write σ as the product of disjoint cycles σ = ζ1 ··· ζj. Suppose ζ1 is at least a 3-cycle. Then σ = ζ1 ··· ζj = (a1a2a3 . am)ζ2 ··· ζj for m ≥ 3. Then (a1a2a3) = (a1a3)(a1a2) ∈ An and (a1a2a3)σ = (a1a2a3)(a1a2a3 . am)ζ1 ··· ζj = (a1a3 . d)ξ1 ··· ξk for d ∈ {1, . , n} and disjoint cycles ξl. On the other hand, σ(a1a2a3) = (a1a2a3 . am)ζ1 ··· ζj(a1a2a3) fixes a1, so (a1a2a3)σ 6= σ(a1a2a3) and so σ is not in the center of An. Therefore, if σ is to be in the center of An, it must be the case that ζ1 is a transposition; specifically, σ = ζ1 ··· ζj = (b1b2)(b3b4 ··· br)ζ3 ··· ζj for r ≥ 2. Now, again, (b1b2b3) = (b1b3)(b1b2) ∈ An and (b1b2b3)σ = (b1b2b3)(b1b2)(b3b4 . br)ζ3 ··· ζj = (b1b3 . c)ξ1 ··· ξk for c ∈ {1, . , n} and disjoint cycles ξl. On the other hand, σ(b1b2b3) = (b1b2)(b3b4 . br)ζ3 ··· ζj(b1b2b3) ALGEBRA HW 1 3 fixes b1, meaning (b1b2b3)σ 6= σ(b1b2b3) and so σ is not in the center of An. Since ζ1 must either be a transposition or a cycles of length ≥ 3 and we’ve just demonstrated that in both cases σ∈ / Z, we conclude that the center of An is trivial. Suppose a ∈ Q. If a = ±1, the fact that ag = ga for all g ∈ Q follows directly from the definition of Q. Otherwise, since ij = k ji = −k ik = −j ki = j we see that a cannot be in the center of Q. Thus, the center of Q is simply {1, −1}. Since Z is abelian, the same argument we gave to show that the center of Cn is the entire group suffices to show that the center of Z is Z itself. a b To characterize the center of GL ( ), let A = be in the center 2 R c d e f of GL ( ) and let B = ∈ GL ( ). Then it must be the case that 2 R g h 2 R a b e f ae + bg af + bh = c d g h ce + dg cf + dh is equal to e f a b ae + cf be + df = . g h c d ag + ch bg + dh Hence, it must be the case that ae + bg = ae + cf, or bg = cf. Since our choice of B was arbitrary, we see that this relationship must hold for any choice of g and f. The only values of b and c for which this is true are if b = c = 0. Using this fact, then the equation af + bh = be + df reduces to af = df or a = d. Hence, elements in the center of GL2(R) must be of the form a 0 . Furthermore, if a 6= 0, then 0 a a 0 1 0 1 0 a 0 C = aC = a C = C 0 a 0 1 0 1 0 a 4 CLAY SHONKWILER so every matrix of this form is in the center. Hence, the center of GL2(R) is the set a 0 |a 6= 0 . 0 a ♣ 2 If H is a subgroup of G, define the normalizer of H by N(H) = {a ∈ G|aHa−1 = H}. Is N(H) a subgroup of G? Is H a subgroup of N(H)? Is H C N(H)? Is N(H) C G? Answer: To demonstrate that N(H) is a subgroup of G, we need to demonstrate that it is closed under the group operation, that the identity is in N(H) and that the inverse of each element in N(H) is in N(H). To that end, let a, b ∈ N(H) and let h ∈ H. Then we know that bhb−1 = h0 for some h0 ∈ H. Therefore, (ab)h(ab)−1 = abhb−1a−1 = ah0a−1 ∈ H. Therefore, ab ∈ N(H). The identity e is clearly in N(H), since ehe−1 = ehe = eh = h ∈ H. Now, suppose a ∈ N(H) and h ∈ H. Then a has an inverse a−1 in G. Furthermore, since a ∈ N(H), aHa−1 = H, so there exists h0 ∈ H such that ah0a−1 = h. Multiplying both sides of this equation by a−1 on the left and a on the right, we see that h0 = a−1ha = a−1h(a−1)−1, so a−1 ∈ N(H). Therefore, we conclude that N(H) is a subgroup of G. To see that H is a subgroup of N(H), it suffices to show that H is con- tained in N(H), since both H and N(H) are subgroups of G. However, this is clear, since for any a, b ∈ H, bab−1 ∈ H. This implies that b ∈ N(H). Since our choices of a and b were arbitrary, we see that every element of H is in N(H), and so we conclude that H is a subgroup of N(H). The fact that H C N(H) follows directly from how we have defined N(H). For any a ∈ N(H), aHa−1 = H, which is precisely what it means to say that H C N(H). To see that it is not necessarily the case that N(H) C G, consider the case where G = S3 and H = h(12)i = {1, (12)}. Now, (13)(12)(13) = (23) (23)(12)(23) = (13) (123)(12)(132) = (23) (132)(12)(123) = (13), ALGEBRA HW 1 5 so we see that N(H) = H. However, the above calculations demonstrate that H is not normal in G, so it’s clear that N(H) is not normal in G. ♣ 3 (a) If H is a subgroup of G and H 6= G, we say that H is a maximal subgroup if the only subgroups containing H are itself and G. Show that if H is maximal then so is aHa−1, for any a ∈ G.