Transitivity of normality in $H_0 \triangleleft H_1 \triangleleft H_2 \triangleleft H_3$.

Question

Supposing that $H_0 \triangleleft H_1 \triangleleft H_2 \triangleleft H_3$ , $H_0 \triangleleft H_3$ and $H_3 / H_0 $ is abelian. Prove that $H_0 \triangleleft H_i$ and $H_i \triangleleft H_3$ for $i=1,2$.

My attempt

I thought that if $H_0=\{e\}$, then $H_3$ is abelian, and then the conclusion is trivial.

But I got stuck on the general cases. Could you please give me some hints? Thanks in advance!

Answer 1

The claim follows from the correspondence principle. Actually, given that $H_3/H_0$ is abelian, it follows that $K\unlhd H_0$ for any intermediate group $K$, $H_3\le K\le H_0$. All because $K\unlhd H_0$ if and only if $K/H_0\unlhd H_3/H_0$, and the latter is obcious.