Intersection with Normal Subgroup is Normal

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Theorem

Let $G$ be a group.

Let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$.

Then $H \cap N$ is a normal subgroup of $H$.


Proof

By Intersection of Subgroups is Subgroup, $H \cap N$ is a subgroup of $N$.

It remains to be shown that $H \cap N$ is normal in $H$.

Let $g \in H$ and $x \in H \cap N$.

However, $H \le G$ and $ g \in H$ imply:

$g \in G$

Because $N \lhd G$:

$\forall n \in N: \forall g \in G: g n g^{-1} \in N$

Because $H \le G$ and therefore closed:

$\forall x \in H \cap N: \forall g \in H: g x g^{-1} \in H$

Therefore, from the definition of normal subgroups:

$N \lhd G \implies g x g^{-1} \in N$
$x \in N$
$g \in G$

Thus:

$g x g^{-1} \in H \cap N$

Because $H$ is a subgroup, $g x g^{-1} \in H \cap N$, $g \in H$, and the definition of normal subgroups:

$H \cap N \lhd H$

$\blacksquare$


Examples

Subset Product of Normal Subgroup with Intersection

Let $\struct G$ be a group whose identity is $e$.

Let $H_1, H_2$ be subgroups of $G$.

Let:

$N_1 \lhd H_1$
$N_2 \lhd H_2$

where $\lhd$ denotes the relation of being a normal subgroup.


Then:

$N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$


Subset Product of Intersection with Intersection

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $H_1, H_2$ be subgroups of $G$.

Let:

$N_1 \lhd H_1$
$N_2 \lhd H_2$

where $\lhd$ denotes the relation of being a normal subgroup.


Then:

$\paren {H_1 \cap N_2} \paren {H_2 \cap N_1} \lhd \paren {H_1 \cap H_2}$


Sources

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