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In a group $G$, a subgroup $K$ is said to be contranormal if its normal closure $K^G$ is the whole group $G$. I have been asked to show that every finite group $G$ contains a nilpotent contranormal subgroup $K$.

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If $G$ is nilpotent, simply choose $K = G$.

Otherwise, we proceed by induction on the order of the group. Consider a maximal subgroup $M$ of $G$ which is not normal in $G$ (remember that if every maximal subgroup of $G$ were normal, then $G$ would be nilpotent). By induction, there is a subgroup $K$ of $M$ such that $K^M = M$. Then $M \subseteq K^G$ and since $M$ is maximal, we have $K^G = M$ or $K^G = G$. However, $K^G$ is normal, and $M$ is not. Therefore, we conclude that $K^G = G$.

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