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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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    @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 $K$.2012-10-28

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