# Group is Normal in Itself

Jump to navigation Jump to search

## Theorem

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

Then $\struct {G, \circ}$ is a normal subgroup of itself.

## Proof

First we note that $\struct {G, \circ}$ is a subgroup of itself.

To show $\struct {G, \circ}$ is normal in $G$:

$\forall a, g \in G: a \circ g \circ a^{-1} \in G$

as $G$ is closed by definition.

Hence the result.

$\blacksquare$