Subnormal-to-normal is normalizer-closed

Statement
Suppose $$H$$ is a subnormal-to-normal subgroup of a group $$G$$: either $$H$$ is normal in $$G$$ or $$H$$ is not subnormal in $$G$$. Then, the normalizer of $$H$$ in $$G$$ is also a subnormal-to-normal subgroup of $$G$$.

Related facts

 * Intermediately subnormal-to-normal is normalizer-closed
 * Intermediately normal-to-characteristic is normalizer-closed
 * Automorph-conjugacy is normalizer-closed
 * Intermediate automorph-conjugacy is normalizer-closed

Proof
Given: A subnormal-to-normal subgroup $$H$$ of a group $$G$$, with normalizer $$N_G(H)$$.

To prove: $$N_G(H)$$ is also a subnormal-to-normal subgroup of $$G$$.

Proof: If $$N_G(H)$$ is not subnormal in $$G$$ we are done. So, we need to prove that if $$N_G(H)$$ is a subnormal subgroup of $$G$$, $$N_G(H)$$ is normal in $$G$$.

If $$N_G(H)$$ is subnormal in $$G$$, then, since $$H$$ is normal in $$N_G(H)$$, $$H$$ is subnormal in $$G$$. Since $$H$$ is subnormal-to-normal, $$H$$ is in fact normal in $$G$$, so $$N_G(H) = G$$. Since every group is normal as a subgroup of itself, $$N_G(H)$$ is a normal subgroup of $$G$$.