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# aug22 - prime order are cyclic we deduce from the previous...

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Math 5125 Monday, August 22 August 22, Ungraded Homework Exercise 3.1.36 on page 89 Prove that if G / Z ( G ) is cyclic, then G is abelian. Write Z = Z ( G ) . If G / Z is cyclic, then G / Z = h xZ i for some x G . This means that the left cosets of Z in G are of the form x a Z for some a Z . Therefore every element of G is of the form x a z for some z Z . Suppose we have another such element, say x b w . Then ( x a z )( x b w ) = zwx a x b = wzx b x a = ( wx b )( zx a ) , consequently any two elements of G commute and we have proven that G is abelian. Exercise 3.2.4 on page 95 Show that if G = pq for some primes p and q (not necessarily distinct), then either G is abelian or Z ( G ) = 1. By Lagrange’s theorem, | Z ( G ) | = 1 , p , q or pq . If | Z ( G ) | = pq , then G is abelian, so we may assume without loss of generality that | Z ( G ) | = p . But then | G / Z ( G ) | = q . Since groups of
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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 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 iso-morphism 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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