4
$\begingroup$

During reading a book, I have faced to this problem telling:

$G$ is a group of order $12$ such that $Z(G)$ has no element of order $2$ . Then $G≅A_4$.

Obviously, this group is not abelian and I think some information about $S_4$ is involved here because of the desired deduction. Can we say $|\frac{G}{Z(G)}|≠3$? And if so, is it useful for the problem?

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    If $|G / Z(G)|$ were 3, then $Z(G)$ would be a group of four elements and by Cauchy's theorem would have an element of order 2.2012-06-14
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    You can deduce that the center of $G$ has to be trivial. I'm not sure if this is useful for this problem though.2012-06-14
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    Up to isomorphism, there are 5 groups of order 12. There are exactly 2 abelian groups by the theory of finite abelian groups, and the rest can be found by using semi-direct products.2012-06-14

1 Answers 1

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If $H$ is a subgroup of order $3$, then $[G:H] = 4$ and there is a homomorphism from $G$ to $S_4$ with the kernel contained in $H$. Show that the kernel is trivial, implying that $G$ is isomorphic to a subgroup of $S_4$ of order $12$, which implies that $G \cong A_4$.

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    I do know using the notation $|G:H|$ doesn't imply $H$ is a normal subgroup but, may I ask you is this subgroup normal in &G&?2012-06-14
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    every subgroup of index 2 is normal2012-06-14
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    @BabakSorouh: No, this $G$ does not have a normal subgroup of order $3$. You can then embed $G$ into $S_4$ using the coset representation, which is the main thing to realize here.2012-06-14
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    @m.k.: Thanks for your complete answer. I 've got more than I had expected. Thanks2012-06-14
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    More detail at http://math.stackexchange.com/questions/162332/g-12-and-no-elements-of-order-2-in-zg2012-06-24