Normalizer condition implies every maximal subgroup is normal

Statement
Suppose $$G$$ is a group satisfying normalizer condition: it has no proper self-normalizing subgroup. In other words, every subgroup $$H$$ of $$G$$ satisfies the property that $$H$$ is properly contained in its normalizer $$N_G(H)$$.

Then, any maximal subgroup of $$G$$ is a normal subgroup.

Proof
Given: A group $$G$$ satisfying the normalizer condition, a maximal subgroup $$H$$ of $$$$.

To prove: $$H$$ is normal in $$G$$.

Proof: $$N_G(H)$$ is a subgroup of $$G$$ containing $$H$$. Since $$H$$ is maximal in $$G$$, $$N_G(H) = G$$ or $$N_G(H) = H$$. Since $$G$$ satisfies the normalizer condition, we cannot have $$N_G(H) = H$$. Thus, $$N_G(H) = G$$, and we obtain that $$H$$ is normal in $$G$$.