# Every Finite Group Has a Nilpotent Contranormal Subgroup

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 $$K$$.

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 $$K^G = G$$.