Equivalence of conjugacy and coset definitions of normality

This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup
View a complete list of pages giving proofs of equivalence of definitions

This article gives a proof/explanation of the equivalence of multiple definitions for the term normalizer
View a complete list of pages giving proofs of equivalence of definitions

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 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 Normalizer (?) of $H$ in $G$. $H$ is a 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.

Proof

(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 $(Hg)g^{-1}=\{(hg)g^{-1}\mid h\in H\}=\{h\mid h\in H\}=H$.