# Normal Subgroup Test

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It has been suggested that this page or section be merged into Definition:Normal Subgroup/Definition 3.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Mergeto}}` from the code. |

## Theorem

Let $G$ be a group and $H \le G$.

Then $H$ is a normal subgroup (by definition 1) of $G$ if and only if:

- $\forall x \in G: x H x^{-1} \subseteq H$.

## Proof

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

Suppose $H$ is normal in $G$.

Then $\forall x \in G, a \in H: \exists b \in H: x a = b x$.

Thus, $x a x^{-1} = b \in H$ implying $x H x^{-1} \subseteq H$.

Conversely, suppose $\forall x \in G: x H x^{-1} \subseteq H$.

Then for $g \in G$, we have $g H g^{-1} \subseteq H$, which implies $g H \subseteq H g$.

Also, for $g^{-1} \in G$, we have $g^{-1} H (g^{-1})^{-1} = g^{-1} H g \subseteq H$ which implies $H g \subseteq g H$.

Therefore, $g H = H g$ meaning $H \lhd G$.

$\blacksquare$