Showing $NH$ is a subgroup of $G$ 
Question :
  If $N$ is a normal subgroup of $G$ and $H$ is a subgroup of $G$, prove that $NH$ is a subgroup of $G$.

Thread is constructed on a mobile so I will attempt to be as succinct as possible.
Attempt:
$NH=gNg^{-1}h$ for all $h \in H$, $g\in G$
By the one-step subgroup test,  $H$ is a subgroup of a group $G$ iff for all $h_{1}, h_{2} \in H$ we have $h_{1} h_{2}^{-1}\in H $
However, I am unable to simplify my current results to the form showing that the elements are indeed in $NH$.
$g_{1}Ng_{1}^{-1}h_{1} \cdot (g_{2}Ng_{2}^{-1}h_{2})^{-1}$
Any help is appreciated.
 A: $NH$ is just the set of products $\{nh \in G \mid n \in N, h \in H\}$. The trick is to use normality of $N$ to move elements around (recall that $N$ is normal iff it is stable under conjugation):
$$(n_1h_1)(n_2h_2) = n_1(h_1n_2h_1^{-1})h_1h_2 = n_1n_2'h_1h_2 \in NH.$$
Similarly, for inverses:
$$(nh)^{-1} = h^{-1}n^{-1} = (h^{-1}n^{-1}h)h^{-1} = n'h^{-1} \in NH.$$

This is a key result in the construction of semidirect products, where we turn this procedure around and build many groups out of pairs of smaller groups. This is also a reason why it is particularly valuable to keep the two steps above separate in this case instead of applying the "one-step test": the two formulas above suggest that we could define a "twisted" group operation on pairs $(n,h) \in N \times H$ via the operations
$$(n_1,h_1)(n_2,h_2) = (n_1\tau_{h_1}(n_2),h_1h_2) \\ (n,h)^{-1} = (\tau_{h_1^{-1}}(n^{-1}),h^{-1})$$
where the twisting map $\tau:H \to \operatorname{Aut}N$ acts like conjugation inside the group created this way.
A: I think you need to be a bit clearer about the difference between, on one hand, what you know, and on the other, what you need to do. We know that


*

*$N$ is a normal subgroup, so $\forall g \in G, gNg^{-1} = N$

*$H$ is a subgroup


We want to show that $NH$ is a subgroup, and we do that by picking two elements in that set (they are on the form $nh$ for $n \in N, h \in H$, and not of the form $gNg^{-1}h$) and doing the one-step test on it. So what we need to do is to take $n_1h_1, n_2h_2 \in NH$, and calculate the following
$$
n_1h_1(n_2h_2)^{-1} = n_1h_1(h_2n_2')^{-1} = n_1h_1n_2'^{-1}h_2^{-1} = n_1n_2''^{-1}h_1h_2 \in NH
$$
wher we used twice that for any $n \in N$ and $h \in H$ (actually any $h\in G$) there is an $n'\in N$ such that $nh = hn'$ (this is because $N$ is normal), and also that $n_1n_2''^{-1} \in N$ and $h_1h_2^{-1} \in H$ because they are both subgroups of $G$.
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 3 on p.53 in Herstein's book.
I solved this problem as follows:

We show that $NH=HN$.
Let $x\in NH$.
Then, we can write $x=nh$ for some $n\in N$ and $h\in H$.
Since $N$ is a normal subgroup of $G$, $h^{-1}x=h^{-1}n(h^{-1})^{-1}\in N$.
So, we can write $h^{-1}x=n^{'}$ for some $n^{'}\in N$.
So, $x=hn^{'}\in HN$.
Let $x\in HN$.
Then, we can write $x=hn$ for some $h\in H$ and $n\in N$.
Since $N$ is a normal subgroup of $G$, $xh^{-1}=hnh^{-1}\in N$.
So, we can write $xh^{-1}=n^{'}$ for some $n^{'}\in N$.
So, $x=n^{'}h\in NH$.
Therefor $NH=HN$.
By LEMMA 2.5.1 on p.44, $NH$ is a subgroup of $G$.

