Unformatted text preview: prime order are cyclic, we deduce from the previous exercise that G is abelian and the result follows. Let G be a ﬁnite group, let H ≤ G and let K ± G . Prove that  HK : H  =  K : K ∩ H  . Note that HK and K ∩ H are subgroups of G (the former by, for example, the second isomorphism theorem), so the result to be proven makes sense. By the second isomorphism theorem, HK / K ∼ = H / H ∩ K . Since everything in sight is ﬁnite, we see that  HK  /  K  =  H  /  H ∩ K  and hence  HK  /  H  =  K  /  K ∩ H  . The result follows....
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 Fall '07
 PALinnell
 Algebra, Sets, Prime number, Cyclic group, Coset, form xa

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