10Ma5aHw5Sol

# 10Ma5aHw5Sol - M a 5a HOMEWORK 5 SOLUTION FALL 2010 The exercises are taken from the text Abstract Algebra(third edition by Dummitt and Foote Page

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Ma 5 a HOMEWORK 5 SOLUTION FALL 2010 The exercises are taken from the text, Abstract Algebra (third edi- tion) by Dummitt and Foote. Page 106, 11 . We will ﬁrst show that if H is a normal subgroup of a solvable group G . Then H is also solvable. Let 1 = G 0 G 1 · · · G n G be a normal series with G i +1 /G i abelian. Consider H i = H G i , then each H i is clearly normal in H i +1 , and we have a normal series 1 = H 0 H 1 · · · H n = H . Also, take g, h H i +1 = G i +1 H , then since G i +1 /G i is abelian, ghg - 1 h - 1 G i H = H i . Therefore H i +1 /H i is abelian, and H is solvable. Now H has a derived series H = H (0) H (1) · · · H ( n ) = 1. We claim that H ( k ) is normal in G for any 0 6 k 6 n . Since H is trivial, the base case holds. Assume H ( k - 1) is normal. Take hkh - 1 k - 1 H ( k ) with h , k H ( k - 1) . For any g G , ghkh - 1 k - 1 g - 1 = ghg - 1 gkg - 1 ( ghg - 1 ) - 1 ( gkg - 1 ) - 1 . Since H ( k - 1) is normal,

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## This note was uploaded on 03/03/2011 for the course MATH 5A taught by Professor Dinakarramakrishnan during the Fall '10 term at Caltech.

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10Ma5aHw5Sol - M a 5a HOMEWORK 5 SOLUTION FALL 2010 The exercises are taken from the text Abstract Algebra(third edition by Dummitt and Foote Page

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