# Subgroup equals Conjugate iff Normal

## Theorem

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

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

Then $N$ is normal in $G$ (by definition 1) if and only if:

$\forall g \in G: g \circ N \circ g^{-1} = N$
$\forall g \in G: g^{-1} \circ N \circ g = N$

## Proof 1

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

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

First note that:

$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} = N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g = N}$

which is shown by, for example, setting $h := g^{-1}$ and substituting.

### Necessary Condition

Suppose that $N$ is normal in $G$.

Then:

 $\ds \forall g \in G: \,$ $\ds g \circ N$ $=$ $\ds N \circ g$ Definition of Normal Subgroup of $G$ $\ds \leadsto \ \$ $\ds \paren {g \circ N} \circ g^{-1}$ $=$ $\ds \paren {N \circ g} \circ g^{-1}$ Definition of Subset Product $\ds \leadsto \ \$ $\ds g \circ N \circ g^{-1}$ $=$ $\ds N \circ \paren {g \circ g^{-1} }$ Subset Product within Semigroup is Associative: Corollary $\ds \leadsto \ \$ $\ds g \circ N \circ g^{-1}$ $=$ $\ds N \circ e$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds g \circ N \circ g^{-1}$ $=$ $\ds N$ Coset by Identity

Similarly:

 $\ds \forall g \in G: \,$ $\ds N \circ g$ $=$ $\ds g \circ N$ Definition of Normal Subgroup $\ds \leadsto \ \$ $\ds g^{-1} \circ \paren {g \circ N}$ $=$ $\ds g^{-1} \circ \paren {N \circ g}$ Definition of Subset Product $\ds \leadsto \ \$ $\ds g^{-1} \circ N \circ g$ $=$ $\ds \paren {g^{-1} \circ g} \circ N$ Subset Product within Semigroup is Associative: Corollary $\ds \leadsto \ \$ $\ds g^{-1} \circ N \circ g$ $=$ $\ds e \circ N$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds g^{-1} \circ N \circ g$ $=$ $\ds N$ Coset by Identity

$\Box$

### Sufficient Condition

Let $N$ be a subgroup of $G$ such that:

$\forall g \in G: g \circ N \circ g^{-1} = N$

and so from $(1)$ above:

$\forall g \in G: g^{-1} \circ N \circ g = N$

Then:

 $\ds \forall g \in G: \,$ $\ds g \circ N \circ g^{-1}$ $=$ $\ds N$ $\ds \leadsto \ \$ $\ds \paren {g \circ N \circ g^{-1} } \circ g$ $=$ $\ds N \circ g$ Definition of Subset Product $\ds \leadsto \ \$ $\ds \paren {g \circ N} \circ \paren {g^{-1} \circ g}$ $=$ $\ds N \circ g$ Subset Product within Semigroup is Associative: Corollary $\ds \leadsto \ \$ $\ds \paren {g \circ N} \circ e$ $=$ $\ds N \circ g$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds g \circ N$ $=$ $\ds N \circ g$ Coset by Identity

$\blacksquare$

## Proof 2

$\forall g \in G: N \supseteq g \circ N \circ g^{-1}$
$\forall g \in G: N \supseteq g^{-1} \circ N \circ g$
$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$

The result follows by definition of set equality.

$\blacksquare$