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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 $HN$ ?

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    $\begingroup$ If $N$ is a normal subgroup of $G$ then $N$ is also normal in any subgroup of $G$ that contains $N$. $\endgroup$ – Jim Sep 28 '17 at 0:53
  • $\begingroup$ Can you give me a hint how to prove it ? $\endgroup$ – user249018 Sep 28 '17 at 1:19
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    $\begingroup$ 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 $G$. $\endgroup$ – Jim Sep 28 '17 at 9:31
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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!

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    $\begingroup$ Thanks for the answer. What do you mean by $H \lor N$ ? Can you give me an advice how to prove that property ? $\endgroup$ – user249018 Sep 28 '17 at 1:35
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    $\begingroup$ $HN$ 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 $HN$ $\endgroup$ – Juan Pablo Díaz Sidaras Sep 28 '17 at 1:40
  • $\begingroup$ 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 $\endgroup$ – Juan Pablo Díaz Sidaras Sep 28 '17 at 2:11
  • $\begingroup$ 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 ? $\endgroup$ – user249018 Sep 28 '17 at 2:41
  • $\begingroup$ 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 $hN=Nn'h=Nh$ $\endgroup$ – Juan Pablo Díaz Sidaras Sep 28 '17 at 2:52
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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.

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

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