# How do I prove this group is closed under multiplication?

Let G be a group and let $H \leq G$ and N $\leq G$ be subgroups of $G$ with $N$ being a normal subgroup.

I have to show that the set $HN = \{ hk$ : $h\in H$, $k\in N \}$ is a subgroup of $G$. I found the identity and the inverse elements for any $hk \in HN$, but I'm having trouble showing that $HN$ is closed under multiplication.

• $h_1k_1$ and $h_2k_2$ are elements of $HN$, then $h_1k_1 h_2k_2=h_1h_2k_1k_2$ because elements $k_i$ are elements of the normal subgroup, so they commute with $h$'s Commented Mar 13, 2017 at 10:48
• 1) The question title is a bit ill-formed. A group is always closed under multiplication by it's definition. You actually want to show that a certain set is closed under multiplication. 2) Since you've already shown closure for identity and inverses, show you work - it can help others understand your level and where are you stuck. Commented Mar 13, 2017 at 10:48
• @user160738 $k_1h_2=h_2k_1$ is not necessarily true. Commented Mar 13, 2017 at 10:52

Since $N\lhd G$, $gN=Ng$ for all $g\in G$
Let $ab,hk\in HN$. Here note that $a,h\in H$ and $b,k\in N$.
Then $(ab)(hk)=a(bh)k$
Note that $Nh=hN$, so $bh=hc$ for some $c\in N$.
Thus $a(bh)k=a(hc)k=(ah)(ck)$ where $ah\in H$ and $ck\in N$.

Let $h_i\in H$ and $k_i\in N$ for $i=1,2$.

Then:

$$h_1k_1h_2k_2=h_1k_1h_2k_2h_2^{-1}h_2k_2=h_1k_3h_2k_2\text{ for }k_3=k_1h_2k_2h_2^{-1}$$

Observe that $h_2k_2h_2^{-1}\in N$ since $N$ is normal and consequently $k_3=k_1h_2k_2h_2^{-1}\in N$.