# If $\frac{G_1}{G_2}$ is minimal normal in $\frac{G}{G_2}$ then is $\frac{G_1N}{G_2N}$ minimal normal in $\frac{G}{G_2N}$?

Question:

Let $$G, G_1, G_2$$ be three groups such that $$G_1 \lhd G$$, $$G_2 \lhd G$$ and $$G_2 \lhd G_1$$. These are strict containments. Moreover:

1. $$\frac{G_1}{G_2}$$ is a minimal normal subgroup of $$\frac{G}{G_2}$$.

2. $$N$$ is a minimal normal subgroup of $$G$$.

Then is $$\frac{G_1N}{G_2N}$$ minimal normal in $$\frac{G}{G_2N}$$?

My Attempt:

Using the 2nd isomorphism theorem, I have that $$\frac{G_1N}{G_2N} \cong \frac{G_1}{G_1 \cap G_2N}$$ and then we know $$G_2 \leq G_1 \cap G_2N \leq G_1$$. I'm not sure how exactly to proceed after this though. Any thoughts?

I was intuitively thinking of this as: if there can't be a normal subgroup of $$G$$ properly contained in between $$G_1$$ and $$G_2$$, then there can't be a normal subgroup of $$G$$ properly contained in between $$G_1N$$ and $$G_2N$$ either. Is this true? Why or why not?

I thought of approaching it as a proof by contradiction as well. Suppose there exists a normal subgroup of $$G$$, named $$A$$, properly contained in between $$G_1N$$ and $$G_2N$$. That is, $$G_1N \lhd A \lhd G_2N$$. Then quotienting this out by $$N$$, we'd have $$\frac{G_1N}{N} \lhd \frac{A}{N} \lhd \frac{G_2N}{N} \implies \frac{G_1}{G_1 \cap N} \lhd \frac{A}{N} \lhd \frac{G_2}{G_2 \cap N}$$. I'm stuck after this. Can we show from here that there exists some normal subgroup of $$G$$ properly contained in between $$G_1N$$ and $$G_2N$$?

• I understand minimal normal to mean normal, nontrivial, and does not properly contain a nontrivial normal subgroup of $G$. Consider $G=C_2\times C_2\times C_2$, $G_2=C_2\times \{1\}\times\{1\}$, $G_1=C_2\times C_2\times\{1\}$, and $N=\{1\}\times C_2\times\{1\}$. Then $G_1/G_2$ is minimal normal of $G/G_2$, $N$ is minimal normal of $G$, but $G_1N=G_2N$, so that $G_1N/G_2N$ is the trivial subgroup, which is not minimal normal in $G/G_2N$. Oct 11, 2021 at 4:31
• I thinks this will be true if $G_1N\neq G_2N$. Note that $G_1\cap G_2N$ contains at least $G_2$, and since $G_1/G_2$ is simple, that means that $G_1\cap G_2N$ is either $G_2$ or $G_1$. Oct 11, 2021 at 4:35
• @ArturoMagidin I don't think $\frac{G_1}{G_2}$ is simple though? It's only true that there is no normal subgroup of $G$ properly contained in between $G_1$ and $G_2$, since $\frac{G_1}{G_2}$ is minimal normal in $\frac{G}{G_2}$.
– S.D.
Oct 11, 2021 at 4:49
• Right; misspoke. The argument still holds, because $G_1\cap G_2N$ is normal in $G$, because $G_1$ and $G_2N$ are normal in $G$. This is a normal subgroup of $G$ that contains $G_2$ and is contained in $G_1$, so it is either $G_2$ or $G_1$. Oct 11, 2021 at 4:56
• @ArturoMagidin Thanks; I agree with this. Let's say $G_1 \cap G_2N = G_2$. Then $\frac{G_1N}{G_2N} \cong \frac{G_1}{G_1 \cap G_2 N} = \frac{G_1}{G_2}$. However, we only know that $\frac{G_1}{G_2}$ is minimal normal in $\frac{G}{G_2}$. From there can we directly say that $\frac{G_1N}{G_2N}$ is minimal normal in $\frac{G}{G_2N}$? Note that there's $G_2N$ in the denominator and not just $G_2$.
– S.D.
Oct 11, 2021 at 5:02