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**moont14263** Let $\displaystyle G$ be a finite group. Suppose that every element of order $\displaystyle 2$ of $\displaystyle G$ has a complement in $\displaystyle G$, then $\displaystyle G$ has no element of order $\displaystyle 4$.

Proof. Let $\displaystyle x$ be an element of $\displaystyle G$ of order $\displaystyle 4$. By hypothesis, $\displaystyle G=\langle x^{2} \rangle K$ and $\displaystyle \langle x^{2} \rangle \cap K=1$ for some subgroup $\displaystyle K$ of $\displaystyle G$. Since $\displaystyle [G:K]=2$, then $\displaystyle K$ is normal in $\displaystyle G$. Clearly, $\displaystyle G=\langle x \rangle K$ and $\displaystyle \langle x\rangle \cap K=1$