# Intersection with Normal Subgroup is Normal/Examples/Subset Product of Normal Subgroup with Intersection

## Theorem

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:

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

## Proof

Consider arbitrary $x_n \in N_1, x_h \in H_1 \cap H_2$.

Thus:

$x_n x_h \in N_1 \paren {H_1 \cap H_2}$

Note that as $x_h \in H_1 \cap H_2$ it follows that $x_h \in H_1$.

We aim to show that:

$x_n x_h N_1 \paren {H_1 \cap N_2} \paren {x_n x_h}^{-1} \subseteq N_1 \paren {H_1 \cap H_2}$

thus demonstrating $N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$ by the Normal Subgroup Test.

We have:

 $\displaystyle x_n x_h N_1 \paren {H_1 \cap N_2} \paren {x_n x_h}^{-1}$ $=$ $\displaystyle x_n x_h N_1 \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}$ Inverse of Group Product $\displaystyle$ $=$ $\displaystyle x_n x_h N_1 {x_h}^{-1} x_h \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}$ $\displaystyle$ $=$ $\displaystyle x_n N_1 x_h \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}$ as $N_1 \lhd H_1$ $\displaystyle$ $=$ $\displaystyle x_n N_1 \paren {H_1 \cap N_2} {x_n}^{-1}$ $H_1 \cap N_2$ is normal in $H_1$ from Intersection with Normal Subgroup is Normal $\displaystyle$ $=$ $\displaystyle N_1 \paren {H_1 \cap N_2} {x_n}^{-1}$ $\displaystyle$ $=$ $\displaystyle \paren {H_1 \cap N_2} N_1 {x_n}^{-1}$ Subset Product of Subgroups $\displaystyle$ $=$ $\displaystyle \paren {H_1 \cap N_2} N_1$ $\displaystyle$ $=$ $\displaystyle N_1 \paren {H_1 \cap N_2}$ Subset Product of Subgroups

$\blacksquare$