Homework7 - Homework 7 Solutions 1. Let |G| = pn for n 0....

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Homework 7 Solutions 1. Let | G | = p n for n 0. ( ) Suppose G is simple. Since G is simple, n 6 = 0. In addition, we know that G has normal subgroups of order p k for k n . Since G is simple, n must equal 1. Thus | G | = p . ( ) Suppose | G | = p . By Lagrange’s theorem, since p is prime, the only subgroups of G are { e } and G itself. Thus G has no non-trivial, proper normal subgroups. 2. (a) C G ( x ) H = H ∩ { g G | gxg - 1 = x } = { g H | gxg - 1 = x } = C H ( x ) (b) [ G : H ] = 2 H C G . By the second isomorphism theorem we know C G / ( H C G ) = HC G /H and HC G G . In addition, since [ G : H ] = 2 then HC G = H or G . By part (a) we have C G / ( H C G ) = C G /C H = HC G /H . Since | x H | = | H | / | C H | and | C G /C H | = | HC G /H | we have | x H | = | HC G | / | C G | . Since HC G = H or G and | G | = 2 | H | we have | x H | = 1 2 | G | / | C G | or | G | / | C G | and thus | x H | = 1 2 | x G | or | x G | . 3. (a) Assume there exists
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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.

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Homework7 - Homework 7 Solutions 1. Let |G| = pn for n 0....

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