# Subgroup is Superset of Conjugate iff Normal

## Theorem

Let $\struct {G, \circ}$ be a group.

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

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

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

## Proof

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:

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

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

### Necessary Condition

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

Then:

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

Similarly:

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

$\Box$

### Sufficient Condition

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

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

and so from $(1)$ above:

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

Then:

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

Similarly:

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

Thus we have:

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

By definition of set equality:

$g \circ N = N \circ g$

Hence the result.

$\blacksquare$

## Also known as

Because of its convenience as a technique for determining whether a subgroup is normal or not, this result is often seen referred to as the Normal Subgroup Test.