Equivalence of conjugacy and commutator definitions of normality

Statement
Suppose $$H$$ is a subgroup of a group $$G$$ and $$g \in G$$ is an element. Then the following are equivalent:


 * 1) $$gHg^{-1} \subseteq H$$.
 * 2) The commutator $$[g,H] \subseteq H$$, where $$[g,H]$$ is the subgroup generated by $$[g,h] = ghg^{-1}h^{-1}, h \in H$$.

If these equivalent conditions hold for all $$g \in G$$, then $$H$$ is a normal subgroup of $$G$$. Note that the second condition on all $$g \in G$$ would translate to saying that $$[G,H] := \langle [g,h] \mid g \in G, h \in H\rangle$$ is contained in $$H$$.

Note that $$gHg^{-1} \subseteq H$$ is a weaker condition than $$gHg^{-1} = H$$ for a particular $$g \in G$$.

More on the proof techniques
The survey article manipulating equations in groups discusses similar proof techniques involving equations that deal with elements and subsets of groups.

(1) implies (2)
Given: Group $$G$$, subgroup $$H$$, $$g \in G$$ such that $$gHg^{-1} \subseteq H$$.

To prove: $$[g,H] \subseteq H$$.

Proof: Since $$H$$ is a subgroup, it suffices to show that all the generators of $$[g,H]$$ are in $$H$$, i.e., that $$[g,h] \in H$$ for every $$h \in H$$. For this, note that:

$$\! [g,h] = ghg^{-1}h^{-1} = (ghg^{-1})h^{-1}$$

By assumption, since $$h \in H$$ $$ghg^{-1} \in H$$, and so the quotient $$(ghg^{-1})h^{-1}$$ is also in $$H$$. This completes the proof.

(2) implies (1)
Given: Group $$G$$, subgroup $$H$$, $$g \in G$$ such that $$[g,H] \subseteq H$$.

To prove: $$gHg^{-1} \subseteq H$$.

Proof: For any $$h \in H$$, we have:

$$\! ghg^{-1} = (ghg^{-1}h^{-1}h = [g,h]h$$

By assumption, both $$[g,h]$$ and $$h$$ are in $$H$$, hence $$ghg^{-1} \in H$$.