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I need to show that, if I have a non-abelian group G of order 12 with only one element has order 2, then G is soluble and the center Z(G) is such that

$Z(G)\cong \mathbb{Z}_2$ and $\frac{G}{Z(G)}\cong S_3$

I don't know how to solve this problem, and any help would be most welcome.

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    Doesn't one need $|G| = |Z(G)| |\frac{G}{Z(G)}|$?2012-05-15
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    Conjugation preserves order, so if you only have one element of order $2$ then it must be central. Now, there are only two groups of order $6$, $S_3$ and the cyclic group of order $6$. It is not too difficult to show that if $G/Z(G)$ is cyclic then $G$ is abelian. So, here you must have $G/Z(G)\cong S_3$.2012-05-15
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    @DylanMoreland, you are correct, that was a typo. I have edited my text.2012-05-15
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    @DylanMoreland, but the order of $S_3$ is $3!$, so everything is fine2012-05-16
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    @DylanMoreland I see. I didn't noted that the question was edited.2012-05-16
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    Reposting: [This KCd handout](http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/group12.pdf) is useful for all of your groups of order 12 needs.2012-05-16

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