Subset Product with Normal Subgroup is Subgroup/Proof 1

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 $G$.

It is clear that $e \in N H$, so $N H \ne \O$.

Suppose $n_1, n_2 \in N$ and $h_1, h_2 \in H$.

Then:

$\ds \paren {n_1 h_1} \paren {n_2 h_2}$

$=$

$\ds n_1 \paren {h_1 n_2 h_1^{-1} h_1} h_2$

$\ds $

$=$

$\ds n_1 \paren {h_1 n_2 h_1^{-1} } \paren {h_1 h_2}$

Since $N$ is normal in $G$:

$\exists n \in N: n = h_1 n_2 h_1^{-1}$

Thus:

$\ds \paren {n_1 h_1} \paren {n_2 h_2}$

$=$

$\ds \paren {n_1 n} \paren {h_1 h_2}$

$\ds $

$\in$

$\ds N H$

Also:

$\ds \paren {n_1 h_1}^{-1}$

$=$

$\ds h_1^{-1} n_1^{-1}$

$\ds $

$=$

$\ds \paren {h_1^{-1} n_1^{-1} h_1} h_1^{-1}$

$\ds $

$\in$

$\ds N H$

so from the Two-Step Subgroup Test, $N H$ is a subgroup of $G$.

The fact that $N H = H N$ follows from Subset Product of Subgroups.

$\blacksquare$

1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.10$