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Course 111: Algebra, 17th November 2006 1. Suppose G is a cyclic group of order 4, G = {e, a, a2 , a3 } and a4 = e. Find the order of each element of G. Find all the subgroups of G. List the distinct left cosets of the subgroup H = {e, a2 } in G. Solution o(e) = 1 since e1 = e o(a) = 4 since a.a.a.a = a4 = e o(a2 ) = 2 since a2 .a2 = a4 = e o(a3 ) = 4 since a3 .a3 .a3 .a3 = (a4 )3 = e The only proper subgroup of G is {e, a2 }. ie it contains the group identity element, e, it is closed under the group operation, a2 .e = a2 , a2 .a2 = a4 = e, e.e = e and each element has an inverse since e.e = e so e−1 = e and a2 .a2 = e so (a2 )−1 = a2 . In addition, {e}, {e, a, a2 , a3 } are also (improper) subgroups. Given H = {e, a2 } the left cosets are {e, a2 }, {a, a3 }, {a2 , a4 = e}, {a3 , a5 = a} so the two distinct left cosets are {e, a2 } and {a, a3 }. 2. Prove that for H and K both subgroups of a group G. The product, HK is itself a subgroup of G iff HK = KH. Note: HK = {x ∈ G|x = hk, h ∈ H, k ∈ K} Proof: Begin by supposing that HK = KH - then we must prove that HK is a subgroup. HK = KH means that if h ∈ H and k ∈ K then hk = k1 h1 for some k1 ∈ K and h1 ∈ H. Note this doesn’t require that k1 = k or that h1 = h. To prove that HK a subgroup we have to show it is closed and that every element in HK has an inverse in HK. • closure: suppose x = hk ∈ HK and y = h0 k 0 ∈ HK. Then xy = hkh0 k 0 . But now since kh0 ∈ KH = HK we have kh0 = h2 k2 with h2 ∈ H and k2 ∈ K. Hence xy = h(h2 k2 )k 0 = (hh2 )(k2 k 0 ) ∈ HK and so HK is closed. • inverses: x−1 = (hk)−1 = k −1 h−1 ∈ KH = HK. So, x−1 ∈ HK. Then, HK a subgroup. Now suppose HK is a subgroup - then we must show HK = KH. If HK a subgroup of G then for any h ∈ H and k ∈ K, h−1 k −1 ∈ HK and so kh = (h−1 k −1 )−1 ∈ HK. Thus KH ⊂ HK. Now if x is any element of HK then x−1 = hk ∈ HK and so x = (x−1 )−1 = (hk)−1 = k −1 h−1 ∈ KH and so HK ⊂ KH. Thus, HK = KH. 3. Recall the properties of cosets listed in your notes. From this list prove the following: For a subgroup H of a group G, • x ∈ xH for all x ∈ G. Proof: Consider x ∈ G. Then x = xe for the identity element, e of G. But e ∈ H. Therefore x ∈ xH. • if x and y ∈ G and if y = xa for some a ∈ H then xH = yH. Proof: Let x, y ∈ G where y = xa and a ∈ H. Then yh = x(ah) and xh = y(a−1 h) for all h ∈ H. Also, ah ∈ H and a−1 h ∈ H for all h ∈ H since H a subgroup of G. Therefore yH ⊂ xH and xH ⊂ yH and hence xH = yH. • each left coset of H in G has the same number of elements as H. Proof: Suppose that xH ∩ yH is non-empty for x, y ∈ G. And let z ∈ (xH ∩ yH). Then z = xa for some a ∈ H and z = yb for some b ∈ H. It follows from the second proof that zH = xH and zH = yH. Therefore xH = yH.