# Subgroup is Normal iff Contains Conjugate Elements

## Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $N$ be a subgroup of $G$.

Then $N$ is normal in $G$ if and only if:

 $\text {(1)}: \quad$ $\ds \forall g \in G: \,$ $\ds n \in N$ $\iff$ $\ds g \circ n \circ g^{-1} \in N$ $\text {(2)}: \quad$ $\ds \forall g \in G: \,$ $\ds n \in N$ $\iff$ $\ds g^{-1} \circ n \circ g \in N$

## Proof

By definition, a subgroup is normal in $G$ if and only if:

$\forall g \in G: g \circ N = N \circ g$

### Necessary Condition

Suppose that $g \circ N = N \circ g$, by definition 1 of normality in $G$.

Let $n \in N$.

Then:

 $\ds g \circ n$ $\in$ $\ds N \circ g$ Definition of Coset $\ds \leadstoandfrom \ \$ $\ds \exists n_1 \in N: \,$ $\ds g \circ n$ $=$ $\ds n_1 \circ g$ Definition of Coset $\ds \leadstoandfrom \ \$ $\ds g \circ n \circ g^{-1}$ $=$ $\ds n_1 \circ g \circ g^{-1}$ $\ds$ $=$ $\ds n_1 \circ e$ Definition of Inverse Element $\ds$ $=$ $\ds n_1$ Definition of Identity Element $\ds \leadstoandfrom \ \$ $\ds g \circ n \circ g^{-1}$ $\in$ $\ds N$ Definition of $n_1$

$\Box$

### Sufficient Condition

Suppose that:

$\forall g \in G: \paren {n \in N \iff g \circ n \circ g^{-1} \in N}$

Let $g \circ n \circ g^{-1} \in N$.

 $\ds \exists n_1 \in N: \,$ $\ds g \circ n \circ g^{-1}$ $=$ $\ds n_1$ $\ds \leadsto \ \$ $\ds g \circ n$ $=$ $\ds n_1 \circ g$ Group Axioms $\ds \leadsto \ \$ $\ds g \circ n$ $\in$ $\ds N \circ g$ Definition of Coset $\ds \leadsto \ \$ $\ds g \circ N$ $\subseteq$ $\ds N \circ g$ Definition of Subset

Similarly:

$n \circ g \in N \circ g \implies N \circ G = g \circ N$
 $\ds \exists n_1 \in N: \,$ $\ds g \circ n \circ g^{-1}$ $=$ $\ds n_2$ $\ds \leadsto \ \$ $\ds n \circ g^{-1}$ $=$ $\ds g^{-1} \circ n_2$ Group Axioms $\ds \leadsto \ \$ $\ds n \circ g^{-1}$ $\in$ $\ds g^{-1} \circ N$ Definition of Coset $\ds \leadsto \ \$ $\ds N \circ g^{-1}$ $\subseteq$ $\ds g^{-1} \circ N$ Definition of Subset

As $g$ is arbitrary, then so is $g^{-1}$.

Thus:

$N \circ g \subseteq g \circ N$

By definition of set equality:

$g \circ N = N \circ g$