Let $G$ be a group and let $N,H$ be subgroups of G. Suppose that $N\cap H=$ {$I$}. Let $h,h'\in H.$ Prove $Nh=Nh'$ iff $h=h'.$
I am trying to teach myself a little abstract algebra and came across this problem. I know this probably seems trivial, but I would just like some verification/advice on the simple proof I came up with for this. I am just trying to make sure I am not making any invalid arguments. Thank you.
$(\Rightarrow)$ Suppose $Nh=Nh'$. Then $N=N(h'h^{-1})$ $\Rightarrow$ $h'h^{-1}\in N$, but $N\cap H$= {$I$} $\Rightarrow$ $h'h^{-1}=I$ $\Rightarrow$ $h'=h.$
$(\Leftarrow)$ Suppose $h'=h$. Then $h'h^{-1}=I$ $\Rightarrow$ $N(h'h^{-1})=N(I)=N$ $\Rightarrow$ $Nh=Nh'.$