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.