# 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 $K \le G$ | $\quad$ |

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \eta$