# multiplying two normal subgroups is still normal?

Let $$G_i \triangleleft G_{i+1}$$ both subgroups of $$G$$. Let $$N$$ be a normal group.

Does $$G_iN \triangleleft G_{i+1}N$$?

Does $$(G_iN/N) \triangleleft (G_{i+1}N/N)$$?

I know that $$q:G\longrightarrow G/N$$ preserves normality. Hence if $$G_iN \triangleleft G_{i+1}N$$ then their quotients would be normal.

I have tried by considering an element in $$g\in G_iN$$ and $$h\in G_{i+1}N$$ and trying to find out if $$ghg^{-1}\in G_iN$$. But I am a bit confused of those multilpication groups. And that is why I can not play with this expression.

• What do you know? What have you tried? – verret Jan 22 at 20:55
• Edited. @verret – idriskameni Jan 22 at 20:58
• Have you heard of the second/third isomorphism theorem? – DonAntonio Jan 22 at 21:00
• Yes I do. I have been playing with that expression too. $G_i/(G_i\cap H)\triangleleft G_{i+1}/(G_{i+1}\cap H)$. But still nothing. – idriskameni Jan 22 at 21:02
• I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things. – idriskameni Jan 22 at 21:02

I want to change the notation to make stuff clearer: Suppose $$H\lhd K< G$$, and let $$N\lhd G$$ (so really $$H=G_i$$ and $$K=G_{i+1}$$). Then a proof that $$HN\lhd KN$$ by considering a conjugator is:

Suppose $$k\in K$$, $$h\in H$$, and $$n, n'\in N$$. Firstly, note that $$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)\cdot (n^{-1}k^{-1}n'kn)$$ Clearly $$(n^{-1}k^{-1}n'kn)\in N$$, while $$(n^{-1}k^{-1}hkn)\in HN$$ by the following: \begin{align*} k^{-1}hk&\in H\\ \Rightarrow k^{-1}hk\cdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&\in HN\\ n^{-1}(k^{-1}hk)&\in HN\\ n^{-1}(k^{-1}hk)n&\in HN \end{align*} The result follows.

Hint:

The first one should be almost immediate, and for the second one:

$$G_iN/N\cong G_i/(G_i\cap N)\;,\;\;\; G_{i+1}/(G_{i+1}\cap N)\cong G_{i+1}N/N$$

• I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward. – Derek Holt Jan 23 at 8:47
• @DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything? – DonAntonio Jan 23 at 8:55
• I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.) – Derek Holt Jan 23 at 9:11