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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 nite 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 nite, we see that  HK  /  K  =  H  /  H K  and hence  HK  /  H  =  K  /  K H  . The result follows....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.
 Fall '07
 PALinnell
 Algebra, Sets

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