# Equivalence of conjugacy and commutator 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

## 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.

## Proof

### (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 $ghg^{-1} \in H$.