Equivalence of conjugacy and coset definitions of normality

Statement
The following are equivalent for a subgroup $$H$$ of a group $$G$$ and an element $$g \in G$$:


 * 1) $$gHg^{-1} = H$$. Here, $$gHg^{-1} = \{ ghg^{-1} \mid h \in H \}$$ is the conjugate subgroup of $$H$$ by $$g$$.
 * 2) $$gH = Hg$$. Here, $$gH = \{ gh \mid h \in H\}$$ is the fact about::left coset of $$g$$ for the subgroup $$H$$ and $$Hg = \{ hg \mid h \in H \}$$ is the right coset of $$g$$ for the subgroup $$H$$.

The set of $$g \in G$$ such that the equivalent conditions (1) and (2) hold is termed the fact about::normalizer of $$H$$ in $$G$$. $$H$$ is a fact about::normal subgroup of $$G$$ if the above two conditions hold for all $$g \in G$$.

The techniques used in this proof
For a survey article describing these techniques in more detail, see manipulating equations in groups.

(1) implies (2)
Given: $$H$$ a subgroup of $$G$$, $$g \in G$$, and $$gHg^{-1} = H$$.

To prove: $$gH = Hg$$.

Proof: Informally, we start with:

$$gHg^{-1} = H$$

and multiply both on the right by $$g$$. We get:

$$gHg^{-1}g = Hg$$

which simplifies to:

$$gH = Hg$$.

A clearer justification of the manipulation done on the left side can be obtained by looking at things elementwise. We have $$gHg^{-1} = \{ ghg^{-1}\mid h \in H \}$$. Thus, $$(gHg^{-1})g = \{ (ghg^{-1})g \mid h \in H \} = \{ hg : h \in H\}$$.

(2) implies (1)
Given: $$H$$ a subgroup of $$G$$, $$g \in G$$, and $$gH = Hg$$.

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

Prof: We start with:

$$gH = Hg$$

We multiply both sides on the right by $$g^{-1}$$, and obtain:

$$gHg^{-1} = Hgg^{-1}$$

which simplifies to:

$$gHg^{-1} = H$$.

A clearer justification of the manipulation done on the right side can be obtained by looking at things elementwise. We have $$Hg = \{ hg \mid h \in H \}$$ and $$(Hg)g^{-1} = \{ (hg)g^{-1} \mid h \in H \} = \{ h \mid h \in H \} = H$$.