The Int. of a Normal Subgroup with a Subgroup is a Normal Subgroup

# The Intersection of a Normal Subgroup with a Subgroup is a Normal Subgroup

Theorem 1: Let $G$ be a group and let $H$ and $N$ be subgroups of $G$ such that $N \trianglelefteq G$. Then $H \cap N \trianglelefteq H$. |

**Proof:**Let $H$ and $N$ be subgroups of $G$ and let $N \trianglelefteq G$. From The Intersection and Union of Two Subgroups page, we see that $H \cap N \subseteq H$ is a subgroup of $H$. Thus, all that remains to show is that $H \cap N$ is a normal subgroup of $H$.

- Since $N$ is a normal subgroup of $G$ we have that $gng^{-1} \in N$ for all $g \in G$ and for all $n \in N$. $(\star)$

- Let $x \in H \cap N$ and let $h \in H$.

- Firstly, since $x \in H \cap N \subseteq H$ and since $h \in H$, we have that since $H$ is a group that:

\begin{align} \quad hxh^{-1} \in H \quad (*) \end{align}

- Secondly, since $x \in H \cap N \subseteq N$ and since $h \in H \subseteq G$ we have from $(\star)$ that:

\begin{align} \quad hxh^{-1} \in N \quad (**) \end{align}

- So from $(*)$ and $(**)$ we see that $hxh^{-1} \in H \cap N$. Since this holds true for all $x \in H \cap N$ and for all $h \in H$, we conclude that $h(H \cap N)h^{-1} \subseteq H \cap N$. So $H \cap N$ is a normal subgroup of $H$, i.e., $H \cap N \trianglelefteq H$. $\blacksquare$