Suppose $N \cap H = \{I \}$ and $h , h' \in H$. Show that $N \circ h = N \circ h'$ if and only if $h = h'$. I'm given that $G$ is a group, $N$ is a normal subgroup, and $H$ is any subgroup of $G$. I am also given that $NH = \{ n \circ h : n \in N, h \in H\}$.
I have already shown that NH is a subgroup of G. I now need to show that:
Suppose $N \cap H = \{I \}$ and $h , h' \in H$. I need to show that $N \circ h = N \circ h'$ if and only if $h = h'$.
So then suppose that  $N \circ h = N \circ h'$. Then $\forall n \in N$, $n \circ h = n \circ h'$. So $n^{-1} n h = n^{-1} n h'$ which implies that $h = h'$. I know I still need to do the converse of this but, I don't think what I did here is right. I think I am missing something because the book gives the fact that $N \cap H = \{I \}$.  Any help going further is appreciated.
 A: It's not necessarily true that $nh=nh^{\prime}$ for all $n\in N$. Instead, what we know is that there are $n,n^{\prime}\in N$ such that $nh=n^{\prime}h^{\prime}$.
But this implies that $h^{\prime}h^{-1}=(n^{\prime})^{-1}n$. Since the left-hand side of this equality is an element of $H$, and the right-hand side is an element of $N$, it follows that $h^{\prime}h^{-1}\in N\cap H=\{e\}$, so $h=h^{\prime}$.
A: When we write $N\cdot h=N\cdot h'$, it is meant that we have equality as sets, not some strange pointwise equality. For each $n\in N$, we have $n\cdot h\in N\cdot h'$, meaning $n\cdot h\cdot (h')^{-1}=n'$ for some $n'\in N$. Equivalently, given $n\in N$ there is some $n'
\in N$ such that $n\cdot h = n'\cdot h'$. We cannot say that $n\cdot h' = n\cdot h$ for each $n\in N$ since that just isn't what the notation means. This is why you didn't use all of your assumptions.
What we do now is fix $n\in N$ to get $n\cdot h = n'\cdot h'$ for some $n'\in N$. Then after rearranging, we have $h\cdot (h')^{-1} = n^{-1}n'\in N$. Therefore $h\cdot (h')^{-1}\in N\cap H$, so it must be that $h\cdot (h')^{-1}=e$. Therefore $h=h'$, as desired.
You will find the converse pretty easy, since $h=h'$ implies $x\in N\cdot h$ if and only if $x\in N\cdot h'$.
