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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 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 nite, we see that | HK | / | K | = | H | / | H K | and hence | HK | / | H | = | K | / | K H | . The result follows....
View Full Document

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.

Ask a homework question - tutors are online