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Let $G$ be a group and let $H \leq G$ and $N\leq G$, $N$ is a normal subgroup of N.

I'm to show that the set $\{ hk, h\in H, k \in N\} = HN$ is a subgroup of $G$.

Identity: since both $H\leq G$ and $N \leq G$ we have $e_g=e\in H$ and $e\in N$.

Try $e*e = e$:

$e*hk = hk = hk*e$ for all $hk$, with $h\in H, k\in N$.

Inverse: $(hk)^{-1} = k^{-1}h^{-1} = h^{-1}hk^{-1}h^{-1} = h^{-1}k_1 \in HN$, for some $k_1 \in N$ and $h^{-1} \in H$.

Closure: let $h_1k_1 \in HN$ and $h_2k_2 \in HN$:

$h_1k_1 * h_2k_2 = h_1h_2k_3k_2 \in HN$ ($k_1h_2 = h_2k_3$ for some $k_3 \in N$, since $N$ is normal).

And finally the question, how do I prove that HN is associative? $h_1k_1*(h_2 k_2*h_3k_3) = (h_1 k_1*h_2 k_2)*h_3 k_3$

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  • $\begingroup$ $HN\subset G$, right? $\endgroup$ – Juniven Mar 17 '17 at 9:50
  • $\begingroup$ yes, that is correct $\endgroup$ – Mathaniel Mar 17 '17 at 9:52
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Let $x,y,z\in HN$.
Since $HN\subseteq G$, $x,y,z\in G$.
Thus $(xy)z=x(yz)$.
Since the operation in $HN$ and $G$ are the same, $HN$ is associative.

For clearer explanation,
Let $*$ be the operation of $G$.
We want to show that $(HN,*)$ is a subgroup $(G,*)$.
Let $x,y,z\in HN$.
Note that $x,y,z\in G$ also.
By using the associativity of $G$, we have $$(x*y)*z=x*(y*z)$$ But the operation of $HN$ is same with the operation of $G$.
This means that $HN$ is associative.

In other words, we can say that associativity is inherited in any subset of $G$ with respect to the operation $*$.

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  • $\begingroup$ but i don't know that $HN \leq G$, that's what I'm supposed to show. I do know that both $H\leq G$ and $N \leq G$ though. $\endgroup$ – Mathaniel Mar 17 '17 at 9:56
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    $\begingroup$ it's enough to know that HN is a subset (as opposed to subgroup) of G to say $(xy)z = x(yz)$ for $x,y,z \in HN$ ? $\endgroup$ – Mathaniel Mar 17 '17 at 9:58
  • $\begingroup$ @Mathaniel Exactly. $\endgroup$ – Alan Wang Mar 17 '17 at 9:58
  • $\begingroup$ @Mathaniel The clearer answer is updated. Can you get the idea? $\endgroup$ – Alan Wang Mar 17 '17 at 10:04
  • $\begingroup$ yeah, I do. Thanks :) $\endgroup$ – Mathaniel Mar 17 '17 at 10:05

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