A question in section Normal and Subnormal Series of Hungerford Algebra.
If $N$ is a simple normal subgroup of a group $G$ and $G/N$ has a composition series, then prove that $G$ has a composition series.
$G/N$ has a composition series means that there exists a series $G = G_0 > G_1 > · · · > G_n = (e)$ is a normal series such that each factor $G_i/G_{i+1}$ is simple.
But I am unable to think how to deduce that G has a composition series using it.
Can you please help?
Hint:
By definition, there exists a series $G/N = G_0 > G_1 > · · · > G_n = (e_{G/N})$ is a normal series such that each factor $G_i/G_{i+1}$ is simple.
(1) By Correspondence Theorem, a subgroup $A$ of $G/N$ must be in the form $H/N$ where $H$ is a subgroup of $G$ containing $N$ and $A$ is a normal subgroup of $G/N$ if and only if $H$ is a normal subgroup of $G$. Use this theorem to get a normal series $H_0>H_1>\dots>H_n=N>(e_G)$.
(2) Use Third Isomorphism Theorem to deduce that $H_i/H_{i+1}$ is simple.