Groups of order 12 with a normal 3-subgroup contain an element of order 6

Question

Let $G$ be a group of order $12$ with a normal $3$-subgroup (which is unique by Sylow's theorems). Does it contain an element of order $6$? I just need a hint to prove it without classifying all the groups of order $12$.

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If $G$ contained just one $2$-subgroup, there would necessarily be an element of order $6$. So, we can assume there are three $2$-subgroups.

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This is a similar question, but I cannot understand j.p.'s comment...

Answer 1

Let $S=G\setminus\langle a\rangle$ where $a$ is of order $3$. As $\langle a\rangle\lhd G$, we have a homomorphism $$\phi\colon G\to\operatorname{Aut}(\langle a\rangle)=\{a\mapsto a^{-1},a\mapsto a\}\cong \Bbb Z/2\Bbb Z$$

Pick $b\in S$ and $c\in S\cap b^{-1}S$ (which is possible because $ S\cap b^{-1}S$ has at least six elements). Then also $bc\in S$. It follows that at least one of $b,c,bc$ is in $\ker\phi$.

But if $x\in S\cap \ker\phi$ then $$\langle a,x\rangle \cong\langle a\rangle\oplus\langle x\rangle\cong\Bbb Z/3\Bbb Z\oplus \Bbb Z/k\Bbb Z$$ with $k\in\{2,4,6,12\}$, and this contains an element of order $6$.