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Unformatted text preview: Solutions to Homework 4 May 1, 2010 1. We will prove this via induction: The first term in the Upper Central Series is { e } , which is clearly a characteristic group. Suppose, for proof by mathematical induction, we have shown that { e } = Z Z 1 Z k are all characteristic subgroups of G . Let us now consider Z k +1 : Let : G G/Z k be the obvious mapping (note that Z k is normal in G ). Notice that since Z k is characteristic, we have that will map the kernel of into itself; in other words, ( x ) = e ( )( x ) = e. (1) Let Z = Z ( G/Z k ) Then, Z is normal in G/Z k , and its inverse image via is normal in G , and is what we call Z k +1 . Now suppose z Z k +1 . We will now attempt to show that ( )( z ) commutes with everything in G/Z k , which would prove ( )( z ) Z , and therefore ( z ) Z k +1 ; applying the same argument using 1 , which is also an automorphism of G , we will get ( Z k +1 ) =...
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This note was uploaded on 10/23/2011 for the course MATH 4108 taught by Professor Staff during the Spring '10 term at Georgia Institute of Technology.
 Spring '10
 Staff
 Algebra, Mathematical Induction

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