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Problem session solutions 1. Let a = (1 2 3 4) ∈ S4 and consider the subgroup H = hai. Determine the right cosets of H in S4 . Solution We have H = {(1), a, a2 , a3 } = {(1), (1 2 3 4), (1 3)(2 4), (1 4 3 2)}, and |S4 | = 24. The right cosets partition S4 , so there are 24/4 = 6 of them. They are: H(1) = H = {(1), (1 2 3 4), (1 3)(2 4), (1 4 2 3)} H(1 2) = {(1 2), (1 3 4), (1 4 2 3), (2 4 3)} H(1 3) = {(1 3), (1 4)(2 3), (2 4), (1 2)(3 4)} H(1 4) = {(1 4), (2 3 4), (1 2 4 3), (1 3 2)} H(2 3) = {(2 3), (1 2 4), (1 3 4 2), (1 4 3)} H(3 4) = {(3 4), (1 2 3), (1 3 2 4), (1 4 2)}. 2. Let G be a group, let a ∈ G and let H be a subgroup of G. Recall that the right coset of H by a is Ha = {ha : h ∈ H}, and let us also define the left coset of H by a to be the set aH = {ah : h ∈ H}. (a) Prove that Ha = H ⇐⇒ a ∈ H ⇐⇒ aH = H. (b) Suppose that G is a finite group and |G| = 2|H|. If a ∈ G, prove that: (i). G = H ∪ Ha = H ∪ aH for any a ∈ G with a 6∈ H. (ii). aH = Ha for any a ∈ G. Solution (a) If a ∈ H then ha ∈ H for all h ∈ H, so a ∈ H =⇒ Ha ⊆ H. If a ∈ H and h ∈ H then ha−1 ∈ H and so h = (ha−1 )a ∈ Ha. So a ∈ H =⇒ H ⊆ Ha. So a ∈ H = H ⊆ Ha. Similarly, a ∈ H =⇒ H = aH. If H = Ha then, since e ∈ H, we have a = ea ∈ Ha = H, so H = Ha =⇒ a ∈ H. If aH = H then, since e ∈ H, we have a = ae ∈ aH = H, so aH = H =⇒ a ∈ H. (b) (i) By (a), there are at most 2 different right cosets of H. Since |H| < |G| and the right cosets partition G, there must be exactly 2, and by (a) these must be H and Ha for some a 6∈ H. Moreover, by (a) we have Ha = Ha′ for every a′ 6∈ H. By symmetry, G = H ∪ aH for any a 6∈ H. (ii) If a ∈ H then aH = H = Ha. If a 6∈ H, then by (i) and the fact that the right cosets partition G, we have Ha = G \ H. It’s not hard to show (if you like, by symmetry) that the left cosets also partition G, so aH = G \ H, too.