The Intersection of Two Normal Subgroups of a Group

# The Intersection of Two Normal Subgroups of a Group

Recall from The Intersection and Union of Two Subgroups page that the intersection of two subgroups of a group $G$ is always a subgroup of $G$. The following proposition tells us more when we intersect normal subgroups.

Proposition 1: Let $G$ be a group and let $H_1$ and $H_2$ be normal subgroups of $G$. Then $H_1 \cap H_2$ is a normal subgroup of $G$. |

**Proof:**Let $H_1$ and $H_2$ be normal subgroups of $G$.

- Let $h \in H_1 \cap H_2$. Then $h \in H_1$ and $h \in H_2$.

- Since $H_1$ is a normal subgroup of $G$ we have that $gh_1g^{-1} \in H_1$ for all $h_1 \in H_1$. Similarly, since $H_2$ is a normal subgroup of $G$ we have that $gh_2g^{-1} \in H_2$ for all $h_2 \in H_2$.

- Since $h \in H_1 \cap H_2 \subseteq H_1, H_2$ we have that $ghg^{-1} \in H_1$ and $ghg^{-1} H_2$ for all $g \in G$. So $ghg^{-1} \in H_1 \cap H_2$ for all $g \in G$. Since this holds for all $h \in H_1 \cap H_2$ we see that for all $g \in G$:

\begin{align} \quad g(H_1 \cap H_2)g^{-1} \subseteq H_1 \cap H_2 \end{align}

- So $H_1 \cap H_2$ is a normal subgroup of $G$. $\blacksquare$