Subset Product with Normal Subgroup is Subgroup/Proof 2

Theorem

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

Let:

$(1): \quad H$ be a subgroup of $G$

$(2): \quad N$ be a normal subgroup of $G$.

Let $H N$ denote subset product.

Then $H N$ and $N H$ are both subgroups of $G$.

Proof

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Consider the subset product $N H$.

We have that:

\(\ds N H\)

\(=\)

\(\ds \set {n h : n \in N , h \in H}\)

Definition of Subset Product

\(\ds \)

\(=\)

\(\ds \bigcup_{h \mathop \in H} \set {n h : n \in N}\)

\(\ds \)

\(=\)

\(\ds \bigcup_{h \mathop \in H} N h\)

Definition of Left Coset

\(\ds \)

\(=\)

\(\ds \bigcup_{h \mathop \in H} h N\)

$N$ is normal

\(\ds \)

\(=\)

\(\ds \bigcup_{h \mathop \in H} \set {h n : n \in N}\)

Definition of Right Coset

\(\ds \)

\(=\)

\(\ds \set {h n : n \in N , h \in H}\)

\(\ds \)

\(=\)

\(\ds H N\)

Definition of Subset Product

Therefore, $N$ and $H$ are permutable.

The result follows from Subset Product of Subgroups.

$\blacksquare$