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Math Help - Nonabelian group of order p^3

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    Nonabelian group of order p^3

    [Hungerford, p99 Q7]

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    If G is a nonabelian group of order p^{3} (p prime), then the center of G is the subgroup generated by all elements of the form aba^{-1}b^{-1} (a,b \in G).
    Last edited by aliceinwonderland; November 18th 2008 at 08:10 PM.
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    Quote Originally Posted by aliceinwonderland View Post
    If G is a nonabelian group of order p^{3} (p prime), then the center of G is the subgroup generated by all elements of the form aba^{-1}b^{-1} (a,b \in G).
    Let C be the subgroup generated by elements of the form aba^{-1}b^{-1} i.e. commutator subgroup. By Burnside's lemma Z(G), the center, is non-trivial. Therefore, |Z(G)|\geq p. But then the group |G/Z(G)| has order p\text{ or }p^2 and therefore it is abelian. Thus, we must have that C\subseteq Z(G). We are told that G is not abelian therefore, C\not = \{ 1 \} and Z(G) \not = G. Suppose that |Z(G)|=p^2 and choose a\in G - Z(G). Then C(a), the centralizer, contains Z(G) properly, and is a subgroup of G. Thus, Lagrange's theorem forces |C(a)| = p^3. But that would mean the index of C(a) in G is 1. This means that the conjugacy class of a is just \{ a \} and so a\in Z(G) - a contradiction! Thus, we see that |Z(G)| \not = p^2. The only remaining possibility is that |Z(G)| = p. However, C is a non-trivial subgroup of Z(G) which forces C = Z(G).
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