1. Consider the dihedral group D14. (a) Identify the elements of the subgroup hsri. {1, sr} (b) Identify all cosets of this subgroup (as sets). The left cosets are S = {1, sr}, rS = {r, s}, r2S = {r2, sr6}, r3S = {r3, sr5}, r4S = {r4, sr4}, r5S = {r5, sr3}, r6S = {r6, sr2}. I only meant to ask for left cosets, but some people noticed (correctly!) that I actually asked for all cosets, so it’s better to include the right cosets as well: S = {1, sr}, Sr = {r, sr2}, Sr2 = {r2, sr3}, Sr3 = {r3, sr4}, Sr4 = {r4, sr5}, Sr5 = {r5, sr6}, Sr6 = {r6, s}.
2. Give two non-isomorphic groups of order 18. (Prove that they are non-isomorphic.) There are three non-isomorphic groups of order 18 that are easy to produce based on what we’ve seen: Z18, D18, and Z3 × Z6. D18 is easy to distinguish because it is non-abelian, and the others are abelian. Distinguishing Z18 from Z3 × Z6 is harder; the easiest way is probably to notice that Z18 has an element of order 18, while every element of Z3 × Z6 has order at most 6.
3. A group G = hx, y | x4 = y4 = 1, yxy−1 = x3i. This group has 16 elements, which can be written xayb for 0 ≤ a, b < 4. Note that it is not abelian. Exactly one of hxi and hyi is normal; which? (Prove both that the one you pick is normal and that the other is not.) a −1 −1 a Consider NG(hxi); certainly x ∈ NG(hxi). For any 1 ≤ a < 4, yx y = (yxy ) = 3 a 3a (x ) = x ∈ hxi. Since x, y ∈ NG(hxi) and x, y generate G, hxi is normal in G. (Many people forgot that we know the normalizer is a subgroup, and solved this correctly but with much more work by checking every element of G.) On the other hand, yx−1y = (yxy−1)−1 = (x3)−1 = x, so yx−1 = xy−1, so xyx−1 = x2y−1 = x2y3 6∈ hyi. Therefore hyi is not normal in G.
4. Suppose four people, 1, 2, 3, and 4, are going to divide into two pairs. There are three ways of doing this: 1 is paired with 2, with 3, or with 4, and the other two most be paired. We name these pairing 12/34, 13/24, and 14/23, and let U = {12/34, 13/24, 14/23} be the set of these three possibilities. There is a natural action of S4 on the set U: σ · ab/cd is the version where σ(a) is paired with σ(b) and σ(c) is paired with σ(d). (a) Identify (12) · u for each u ∈ U. (12) · 12/34 = 12/34, (12) · 13/24 = 14/23, and (12) · 14/23 = 13/24. (b) Explain how this gives a homomorphism φ : S4 → S3 and illustrate by identifying φ((12)) and φ((123)). Any action on a three element set gives a homomorphism into S3: if we identify 12/34 with 1, 13/24 with 2, and 14/23 with 3, φ(σ) is the permutation mapping 1 to σ·1, 2 to σ·2, and 3 to σ·3. (On this part of the problem, I was really hoping to see both an awareness of the general theorem that actions correspond to homomorphisms, and an understanding of the proof which made it easy to produce the isomorphism.) φ((12)) = (23), φ((123)) = (132) (Note that different identifications will give different homomorphisms, which of course differ by conjugation.) (c) Identify the kernel of this action.
1 of 2 The kernel does not contain (12), since (12)·13/24 = 14/23. By normality, the kernel contains no transposition. The kernel does not contain (123) since (123)·13/24 = 12/34, and by normality, does not contain any 3-cycle. The kernel does not contain (1234) since (1234) · 12/34 = 14/23, so does not contain any 4-cycle. On the other hand, the kernel does contain the double transpositions, since (12)(34)·12/34 = 12/34, (12)(34) · 13/24 = 13/24, and (12)(34) · 14/23 = 14/23, and by normality, contains all double transpositions. So the kernel is K = {1, (12)(34), (13)(24), (14)(23)}. (d) Identify the quotient of S4 by the kernel. The kernel has size 4 and |S4| = 24, so the quotient has size 6. The quotient is a subgroup of S3, so must be S3. (Many people got answers where the quotient wasn’t a subgroup of S3, which should have been a big warning sign.)
5. (a) Suppose |G| = 15, |H| = 28, and φ : G → H is a homomorphism. What is |ker(φ)|? Since G/ker(φ) is a subgroup of H, |G/ker(φ)| divides |H| = 28. But also |G/ker(φ)| divides |G| = 15. The only common factor is 1. (b) Let H ≤ G, G be finite, N E G, |H| = 15, and |G : N| = 28. Show that H ≤ N. By the second isomorphism theorem, HN is a subgroup of G, H ∩N is a subgroup of H, and HN/N is isomorphic to H/H ∩ N. In particular 28 = |G : N| = |G : HN| · |HN : N|, so |HN : N| is a factor of 28. Also |HN : N| = |H : H ∩ N|, so |HN : N| is also a factor of 15. Again, the only common factor is 1.
6. Suppose {a | a3 = 1} is a subgroup of G. Prove that it is a normal subgroup. Let g ∈ G and a3 = 1. Then (gag−1)3 = gag−1gag−1gag−1 = ga3g−1 = g1g−1 = 1, so gag−1 belongs to this subgroup as well. Suppose N and M are both normal subgroups of G. Show that if n ∈ N and m ∈ M then nmn−1m−1 ∈ N ∩ M. Since M is normal, nmn−1 ∈ M, so nmn−1m−1 ∈ M as well. Also n−1 ∈ N, so mn−1m−1 ∈ N, so nmn−1m−1 ∈ N. So nmn−1m−1 ∈ N ∩ M. Suppose N and M are both normal subgroups of G and N ∩ M = {1}. Show that if n ∈ N and m ∈ M then nm = mn. Since nmn−1m−1 ∈ N ∩ M, nmn−1m−1 = 1. Therefore nm = mn.
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