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Let $G$ be a finite group. Suppose that every element of order $2$ of $G$ has a complement in $G$, then $G$ has no element of order $4$.

Proof. Let $x$ be an element of $G$ of order $4$. By hypothesis, $G=\langle x^{2} \rangle K$ and $\langle x^{2} \rangle \cap K=1$ for some subgroup $K$ of $G$. Clearly, $G=\langle x \rangle K$ and $\langle x\rangle \cap K=1$, but $|G|=|\langle x^{2} \rangle||K|<|\langle x \rangle ||K|=|G|$, a contradiction. Therefore $G$ has no element of order $4$.

Is above true? Thanks in advance.

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How did you conclude $\langle x\rangle\cap K=\{1\}$? (By the way, the code for that is \langle x\rangle\cap K=\{1\}. Note that your use of binary comparison operators instead of angled brackets messes up the spacing not only around $x$ but also around $K$ because the intersection operator isn't being recognized as a binary operator when it directly follows the comparison operator.) – joriki Oct 28 '12 at 1:14
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$\langle x \rangle \cap K=1$ because if not, then $\langle x \rangle \cap K=\langle x \rangle$ or $\langle x^{2} \rangle$ and in both cases $\langle x^{2} \rangle \leq K$, a contradiction. – user28083 Oct 28 '12 at 1:20
Sorry, I will edit right a way. – user28083 Oct 28 '12 at 1:24
@user28083 Yes, the argument seems correct. You can shorten it to a direct proof using a more general $x$, i.e. let $2$ divide $|x|$ and show that $x\not\in K$ for a slightly different $K$. – peoplepower Oct 28 '12 at 1:24
Thank you very much – user28083 Oct 28 '12 at 15:58

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