Strongly paranormal implies paranormal

Statement
Any strongly paranormal subgroup of a group is paranormal.

Strongly paranormal subgroup
A subgroup $$H \le G$$ is termed strongly paranormal if, for any $$g \in G$$, $$[[g,H], H] = [g,H]$$. Here, $$[g,K]$$ is the subgroup generated by all commutators between $$g$$ and elements of $$K$$.

Paranormal subgroup
A subgroup $$H \le G$$ is termed paranormal if, for any $$g \in G$$, $$H$$ is a contranormal subgroup of $$\langle H, H^g \rangle$$: in other words, the normal closure of $$H$$ in $$\langle H, H^g \rangle$$ is the whole group $$\langle H, H^g \rangle$$.

Facts used

 * 1) Product with commutator equals join with conjugate:

$$H[g,H] = \langle H, [g,H] \rangle = \langle H, H^g \rangle$$

Proof idea
The key idea is to show, using the identity, that any normal subgroup of $$\langle H, H^g \rangle$$ that contains $$H$$ must also contain $$[g,H]$$, and hence must be the whole of $$\langle H, H^g$$.

Proof details
Given: A group $$G$$, a subgroup $$H$$ such that $$[[g,H],H] = [g,H]$$ for any $$g \in G$$.

To prove': If $$K$$ is a normal subgroup of $$\langle H, H^g \rangle$$ that contains $$H$$, $$K = \langle H, H^g \rangle$$.

Proof:


 * 1) $$[g,H] \le \langle H, H^g \rangle$$: This follows from fact (1).
 * 2) $$[[g,H],K] \le K$$: Since $$[g,H]$$ is in $$\langle H, H^g \rangle$$, we have $$[[g,H],K] \le [\langle H, H^g \rangle, K]$$. The latter is contained in $$K$$, since $$K$$ is normal in $$\langle H, H^g \rangle$$.
 * 3) $$[[g,H],H] \le K$$: This follows from the previous step, and the fact that $$H \le K$$.
 * 4) $$[g,H] \le K$$: This follows from the previous step, and the fact that $$[[g,H],H] = [g,H]$$.
 * 5) $$K = \langle H, H^g \rangle$$: We already have $$H \le K$$, and the previous fact yields $$[g,H] \le K$$. Thus, we get $$H[g,H] \le K$$. By fact (1), this yields $$\langle H, H^g \rangle \le K$$. Since $$K \le \langle H, H^g \rangle$$, we get equality.