# Is $N$ normal to $HN$ if $H$ subgroup and $N$ normal subgroup of the group $G$

If $H$ is a subgroup of $G$ and $N$ is a normal subgroup of $G$, then whats the relation between $N$ and the subgroup $HN$ in respect to normality, i.e. must $N$ be normal to $HN$ ?

• If $N$ is a normal subgroup of $G$ then $N$ is also normal in any subgroup of $G$ that contains $N$. – Jim Sep 28 '17 at 0:53
• Can you give me a hint how to prove it ? – user249018 Sep 28 '17 at 1:19
• Say $N \leq Q \leq G$. Write down the condition for $N$ being normal in $Q$ and for $N$ being normal in $G$. It is essentially immediate that the conditions for being normal in $Q$ are a subset of the conditions for being normal in $G$. – Jim Sep 28 '17 at 9:31

## 2 Answers

There is a property that says: $N \le G$ Normal and $H \le G$ $\Rightarrow H \lor N = HN = NH$.

And if $H$ and $N$ are both normal, then $HN \le G$ is also normal.

If you need a double click in the Demo, let me know!

Regards!

• Thanks for the answer. What do you mean by $H \lor N$ ? Can you give me an advice how to prove that property ? – user249018 Sep 28 '17 at 1:35
• $HN$ is not by definition a subgroup, so $H \lor N$ is the interseccion of every subgroup of $G$ that contains $HN$, So the first thing that you have if $N$ is normal is that $HN$ is a subgroup, in the other hand, $N$ is normal so the normality works fine with every g in $G$, in particular $hn \in HN \Rightarrow hn \in G$ So the only missing part is that $N$ is a subgroup of $HN$ is that happen then is normal with $HN$ – Juan Pablo Díaz Sidaras Sep 28 '17 at 1:40
• I'm new to "MathExchange" so I just notice the change in the question :), let review the following, If $H \lor N$ is the interseccion of all the groups that contains $HN$, we just need to see that $HN$ is subgroup when $N$ is normal, So, i) is $HN$ closed?, let's grab $h_1n_1$ and $h_2n_2$ and operate them $(h_1n_1)(h_2n_2)=h_1(n_1h_2)n_2$ so, because $N$ is normal is valid to say that $n_1h_2=h_2n'$ , so now we have $h_1h_2n'n_2 \in HN$, $e \in HK$ is trivial and the inverse can be done similarly – Juan Pablo Díaz Sidaras Sep 28 '17 at 2:11
• To prove that $N$ is a subgroup of $HN$ is trivial. If I write $hnN=Nhn$ for $hn \in HN$, this is equivalent to $hN=Nhn$. But now i have another difficulty: since $h \in G$ and $N$ is normal in $G$, i need to obtain $hN=Nh$, which i dont see how. Can you elaborate on that ? – user249018 Sep 28 '17 at 2:41
• By definition of normal $\forall g \in G/ g^{-1}Ng=N$ this mean that $\forall g \in G$, $\forall n \in N$, $\exists n' \in N / g^{-1}ng=n'$ in particular each $h \in H$ belongs to $G$, then $h^{-1}nh=n' \Rightarrow nh=n'h \Rightarrow Nh =hN$, following the thing that you have in your question, $hN = Nhn$ but $hn = n'h$ so $hN=Nn'h=Nh$ – Juan Pablo Díaz Sidaras Sep 28 '17 at 2:52

Given:

1. $$G$$ is a group.
2. $$H\lt G$$.
3. $$N$$ normal to $$G$$.

To Show: $$N$$ normal to $$H\circ N$$.

Possible Proof: Now,
$$H\circ N$$ = {$$h\circ n | h\in H, n\in N$$}$$= H\cup N$$.

Since, $$H\lt G$$ and $$N$$ normal to $$G$$
$$\implies H, N \lt G \implies (H\cup N)\subseteq G\implies H\circ N\subseteq G$$.

Now,
$$N$$ normal to $$G$$
$$\iff g\circ N\circ g$$-1 $$= N, \forall g\in G$$.
[This comes from the basic properties of a normal subgroup.]
$$\implies g\circ N\circ g$$-1 $$= N, \forall g\in H\circ N$$. [Since, $$H\circ N\subseteq G$$, already proved.]
$$\iff N$$ normal to $$H\circ N$$.
QED

PS: Please let me know if there are any errors in the proof. I myself wanted a proof for this theorem.

• I think I still have a bit of a problem somewhere towards the end. – Aakash Singh Bais Jun 28 at 14:14