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I am thinking on the following problem:

If $|G|=12$ and there is no element of order $2$ in its center then $3$-Sylow subgroup of $G$ cannot be normal in $G$.

I was told that to assume the $3$-Sylow subgroup of $G$, say $P$, is normal in the group and go to reach a contradiction.

My attept: $|P|=3$ so it is cyclic, $P=\langle x\rangle=\{1,x,x^2\}$. As above hint, I would have for all $g\in G$ two possibilities: $gxg^{-1}=x$ or $gxg^{-1}=x^2$. Cannot to go any further. I am in doubt if the hint leads me to a contradiction properly and if so, it does with a long proof. Any help or any other way are welcome to me. Thanks for your time.

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    This is a followup to http://math.stackexchange.com/questions/158339/g-12-and-it-is-isomorphic-to-a-42012-06-24

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Hint #1: Show that all Sylow 2-subgroups are abelian. So if an element of a Sylow 2-subgroup commutes with the elements of $P$, then it belongs to the center.

Hint #2: At this point the questions you should ask from yourself are: 1) What possibilities (up to isomorphism) are there for the Sylow 2-subgroup H? 2) Given that $P$ was assumed to be normal, what possibilities are there for the conjugation action for the non-trivial elements of $H$. Remember that conjugation action is a homomorphism from $H$ to $Aut(P)$.

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    I am on the way you paved. Yes, with that element say $y$, I can see that $G$ is being generated by $x$ and $y$. Thanks. Thanks.2012-06-24
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To follow your attempt:

For every $g \in G$, we have $gxg^{-1} = x$ or $gxg^{-1} = x^2$. Thus $x$ has $1$ or $2$ conjugates, in other words, $[G : C_G(x)] = 1$ or $[G : C_G(x)] = 2$. In both cases, $2$ divides the order of $C_G(x)$, and thus $C_G(x)$ contains an element $y$ of order $2$. This gives us the desired contradiction. Can you see why $y$ must belong to the center of $G$?

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    Thanks for your answer. It solves my problem in detail. :)2012-06-25