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

Table of Contents

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 $n \in N$ . $(\star)$

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

(1)

\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:

(2)

\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., $H \cap N \trianglelefteq H$ . $\blacksquare$

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