If $G$ is a group with $N$ normal and $H_1$, $H_2$ subgroups of $G$ with $H_1$ a normal subgroup of $H_2$, show $NH_1$ is a normal subgroup of $NH_2$.

Question

I was unsure how to prove this property which seems necessary for the solution to a problem I did recently. Since $N$ is normal, it seems pretty clear that $NH_1 = H_1N$ and $NH_2 = H_2N$ so they are indeed subgroups of $G$. I tried to prove normality directly by taking $nh \in NH_2$ and was able to show that $$nh NH_1 = N nh H_1 = Nn H_1 h$$ using the fact that $H_1$ is normal in $H_2$ and so $h H_1 = H_1 h$. Then, it seems like this problem reduces to showing that $nH_1 = H_1n$ which isn't immediately obvious. In fact, I'm not sure if that's even true.

My latest attempt uses the fact that $NH_1 = H_1N$ and $Nn = N$ so that we can find some $n' \in N$ so that $n'H_1 = H_1n$ which would allow us to conclude $$nh NH_1 = N nh H_1 = n H_1 h = Nn' H_1 h = NH_1nh$$ which is what we wanted but I had some trouble showing that this idea actually worked.

Thank you in advance for your help!

Answer 1

Since $NH_1$ is a subgroup, we have (refer here) $$NH_1=H_1N.$$ So, $$nhNH_1=nNhH_1=nNH_1h=NH_1h=H_1Nh=H_1Nnh=NH_1nh.$$

Answer 2

Hp: $N\trianglelefteq G$, $H_1,H_2\leq G$, $H_1 \trianglelefteq H_2$
Ts: $NH_1 \trianglelefteq NH_2$
Note that, since $N\trianglelefteq G$ and $H_1,H_2 \leq G$, we have $NH_1,NH_2 \leq G$.

Take $x$ in $NH_1 \implies \exists n_1 \in N,h_1\in H_1 :x=n_1h_1$ and consider $y$ in $NH_2$, so $\exists n_2 \in N,h_2\in H_2: y=n_2h_2$. We have to show that $NH_1$ is normal in $NH_2$ (which means that $y^{-1}xy\in NH_1$), so
$$(n_2h_2)^{-1}(n_1h_1)(n_2h_2)=h_2^{-1}(n_2^{-1}n_1)h_1n_2h_2= h_2^{-1}n_3 h_1n_2h_2 = \underset {N\trianglelefteq G,H_1\trianglelefteq H_2}{n_4(h_2^{-1}h_1h_2)n_5}=n_4\bar h_1n_5=\underset{N\trianglelefteq G}{n_6\bar h_1} \in NH_1 \implies NH_1\trianglelefteq NH_2$$