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Math Help - Abstract Algebra: Suppose that G is a group with |G|=p3. Prove that |Z(G)| != p2.

  1. #1
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    Question Abstract Algebra: Suppose that G is a group with |G|=p3. Prove that |Z(G)| != p2.

    Suppose that G is a group with |G|=p^3 where p is prime. Prove that |Z(G)|\ne p^2.
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  2. #2
    Senior Member Tinyboss's Avatar
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    If so, then |G/Z(G)|=p, and so G/Z(G) is cyclic. But G/Z(G) cyclic implies G is abelian, a contradiction.
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