Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$.
The product of $H$ and $N$ is defined to be the subset
\[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\]
Prove that the product $H\cdot N$ is a subgroup of $G$.

A subgroup $N$ of a group $G$ is called a normal subgroup if for any $g\in G$ and $n\in N$, we have
\[gng^{-1}\in N.\]

Proof.

We prove that the product $H\cdot N$ is closed under products and inverses.

Let $h_1n_1$ and $h_2n_2$ be elements in $H\cdot N$, where $h_1, h_2\in H$ and $n_1, n_2\in N$.
Let $e$ be the identity element in $G$.
We have
\begin{align*}
(h_1n_1)(h_2n_2)&=h_1en_1h_2n_2\\
&=h_1(h_2h_2^{-1})n_1h_2n_2 && \text{since $h_2h_2^{-1}=e$}\\
&=(h_1h_2)(h_2^{-1}n_1h_2n_2). \tag{*}
\end{align*}
Since $H$ is a subgroup, the element $h_1h_2$ is in $H$.
Also, since $N$ is a normal subgroup, we have $h_2^{-1}n_1h_2$ is in $N$. Hence
\[h_2^{-1}n_1h_2n_2=(h_2^{-1}n_1h_2)n_2\in N.\]
It follows from (*) that the product
\[(h_1n_1)(h_2n_2)=(h_1h_2)(h_2^{-1}n_1h_2n_2)\in H\cdot N.\]
Therefore, the product $H\cdot N$ is closed under products.

Next, let $hn$ be any element in $H\cdot N$, where $h\in H$ and $n\in N$.
Then we have
\begin{align*}
(hn)^{-1}&=n^{-1}h^{-1}\\
&=en^{-1}h^{-1}\\
&=(h^{-1}h)n^{-1}h^{-1} &&\text{since $h^{-1}h=e$}\\
&=h^{-1}(hn^{-1}h^{-1}).
\end{align*}
Since $N$ is a normal subgroup, we have $hn^{-1}h^{-1}\in N$, and hence
\[(hn)^{-1}=h^{-1}(hn^{-1}h^{-1})\in H\cdot N.\]
Thus, the product $H\cdot N$ is closed under inverses.

This completes the proof that the product $H\cdot N$ is a subgroup of $G$.

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