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21 (77) Assume that Z(G) 6= {e} but G is not abelian, then Z(G) = p or q, and therefore G/Z (G) = p or q But then G/Z (G) is cyclic By Theorem 738, G is abelian Contradiction 11 (78) Define f R∗ → R∗∗ by f (x) = x Then f is a surjective homomorphism, and Kerf = {1,−1}/ , 1 ) 3 7 ?Title INBC2RUpdf Author kkasprzak Created Date 8/19/14 449 PM O C N P C ƒ}ƒCƒNƒ‰ ƒX[ƒp[ƒtƒ‰ƒbƒg ƒvƒŠƒZƒbƒg

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