# Equivalence of conjugacy and coset definitions of normality

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