Subset Product with Normal Subgroup is Subgroup

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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

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:

\(\displaystyle \paren {n_1 h_1} \paren {n_2 h_2}\) \(=\) \(\displaystyle n_1 \paren {h_1 n_2 h_1^{-1} h_1} h_2\)
\(\displaystyle \) \(=\) \(\displaystyle n_1 \paren {h_1 n_2 h_1^{-1} } \paren {h_1 h_2}\)


Since $N$ is normal in $G$:

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

Thus:

\(\displaystyle \paren {n_1 h_1} \paren {n_2 h_2}\) \(=\) \(\displaystyle \paren {n_1 n} \paren {h_1 h_2}\)
\(\displaystyle \) \(\in\) \(\displaystyle N H\)


Also:

\(\displaystyle \paren {n_1 h_1}^{-1}\) \(=\) \(\displaystyle h_1^{-1} n_1^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {h_1^{-1} n_1^{-1} h_1} h_1^{-1}\)
\(\displaystyle \) \(\in\) \(\displaystyle N H\)

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.

$\blacksquare$


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