Let $G$ be a group, $H \leq G$ and $N \unlhd G$. Let $HN=\{hn │h \in H, n \in N\}$.
I finished
(a) $H \cap N \unlhd H$,
(b) $HN \leq G$,
(c) $N \subset HN$ and $N \unlhd HN$.
I have a question in
(d) $\frac {HN} N \simeq \frac H {H \cap N}$.
I think it will be proved by using the Fundamental Homomorhism Theorem (FHT):
Let $f: HN \to \frac H {H \cap N}, f(hn)=(H \cap N)h$.
Then $f(h_1 n_1 h_2 n_2) = f(h_1 h_2 n_1 n_2) = H \cap N(h_1 h_2)=(H \cap N h_1)(H \cap N h_2) = f(h_1 n_1) f(h_2 n_2)$ for $h_1, h_2 \in H, n_1, n_2 \in N$.
And $\ker f = N$.
Thus, by FHT, $\frac {HN} N \simeq \frac H {H \cap N}$.
My questions are:
(1) May I write $f(h_1 n_1 h_2 n_2) = f(h_1 h_2 n_1 n_2)$?
(2) If it is wrong, how can I fix $f$? Or what other methods can I use?
Hint:
It's much simpler the other way* \begin{align*} H&\rightarrow HN/N\\h&\mapsto hN \end{align*} It is indeed a homomorphism: $$(hh')N=(hh')NN=h(h'N)N=h(Nh')N=(hN)(h'N).$$ It is surjective: $$(hn)N=h(nN)=hN$$ The kernel is clealy made up of the $h\in H$ such that $\,hN=N h\in N\iff h\in H\cap N$.
Hence the assertion by the first isomorphism theorem.