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I'm having a lot of trouble with this problem. If anyone could help me out it would be much appreciated:

Suppose $ G $ is a group and $|G : Z(G)|=4$. Prove that $G/Z(G)\approx \mathbb Z_2 \oplus \mathbb Z_2 $.

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There's a general result you can use here (you should prove it, first): If $G/Z(G)$ is cyclic, then $G$ is abelian, and so $|G:Z(G)|=1$. Thus, since the only groups of order $4$ are $\Bbb Z_4$ and $\Bbb Z_2\oplus\Bbb Z_2$, and since the first one is cyclic, then we really have only one possibility for the isomorphism class of the quotient group $G/Z(G)$.

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Thanks. This makes perfect sense. –  mkeachie Nov 17 '12 at 0:09
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Probably you've already observed this, but $H := G/Z(G)$ has to be a group of order 4, and the only isomorphism types of groups of order 4 are $\mathbb{Z}_4$ and $\mathbb{Z}_2\oplus \mathbb{Z}_2$. So if you could show that $H$ can't be a cyclic group, you would be done.

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