# Subset Products of Normal Subgroup with Normal Subgroup of Subgroup

## Theorem

Let $G$ be a group.

Let:

$(1): \quad H$ be a subgroup of $G$
$(2): \quad K$ be a normal subgroup of $H$
$(3): \quad N$ be a normal subgroup of $G$

Then:

$N K \lhd N H$

where:

$N K$ and $N H$ denote subset product
$\lhd$ denotes the relation of being a normal subgroup.

## Proof

Consider arbitrary $x_n \in N, x_h \in H$.

Thus:

$x_n x_h \in N H$

We aim to show that:

$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$

thus demonstrating $N K \lhd N H$ by the Normal Subgroup Test.

We have:

 $\displaystyle x_n x_h N K \paren {x_n x_h}^{-1}$ $=$ $\displaystyle x_n x_h N K {x_h}^{-1} {x_n}^{-1}$ $\quad$ Inverse of Group Product $\quad$ $\displaystyle$ $=$ $\displaystyle x_n N x_h K {x_h}^{-1} {x_n}^{-1}$ $\quad$ as $N \lhd G$ and $H \le G$ $\quad$ $\displaystyle$ $=$ $\displaystyle x_n N K {x_n}^{-1}$ $\quad$ as $K \lhd H$ $\quad$ $\displaystyle$ $=$ $\displaystyle N K {x_n}^{-1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle K N {x_n}^{-1}$ $\quad$ as $N \lhd G$ and $K \le G$ $\quad$ $\displaystyle$ $=$ $\displaystyle K N$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle N K$ $\quad$ as $N \lhd G$ and $K \le G$ $\quad$

$\blacksquare$