Adv Alegbra HW Solutions 68

Adv Alegbra HW Solutions 68 - 68 Solution. Suppose, on the...

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68 Solution. Suppose, on the contrary, that G / Z ( G ) is cyclic, say, with gen- erator aZ ( G ) . Since G is not abelian, a / Z ( G ) . On the other hand, we now show that a commutes with every element in g G . By hypothesis, there is an integer i and an element z Z ( G ) with g = a i z . Therefore, ag = aa i z = a i az = a i za = ga , and this contradicts a / Z ( G ) . 2.99 Let G be a f nite group, let p be a prime, and let H be a normal subgroup of G . Prove that if both | H | and | G / H | are powers of p , then | G | is a power of p . Solution. If | H |= p h and | G / H p m , then | G |=| G / H || H p h + m . 2.100 Call a group G fnitely generated if there is a f nite subset X G with G =h X i . Prove that every subgroup S of a f nitely generated abelian group G is itself f nitely generated. Solution. Use induction on n 1, where X ={ a 1 ,..., a n } . The inductive step should consider the quotient group G / h a n + 1 i . 2.101 (i) Let π : G H be a surjective homomorphism with ker π = T . Let H X i , and, for each x X , choose an element g x G with π( g x ) = x . Prove that G is generated by T ∪{ g x : x X } . Solution. Absent. (ii) Let G be a group and let T C G . If both T and G / T are f nitely generated, prove that G is f nitely generated. Solution. Absent. 2.102 Let
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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