Let G be a group. Show that $G$ has a composition series iff $G$ satisfies the following two conditions.
(i) If $G=G_1\triangleright G_2 \dots$ then there is an $n$ such that $G_n=G_{n+1}=\dots$.
(ii) If $H$ is a term of a normal series of $G$ and $H_1 \subset H_2 \subset \dots $ is an ascending sequence of normal subgroups of $H$, then there exists an $m$ such that $H_m=H_{m+1}=\dots.$
To show that (i) and (ii) imply G has a composition series I want to start with a normal chain of $G$, $G=G=G_1\triangleright G_2 \dots$, where we require that $G_i\neq G_{i+1}$. This has to terminate at say $n$. Then without loss of generality we can say that $G_n=\{e\}$.
Look at $G_k\triangleright G_{k+1}$ for $k=1,2,\dots,n-1$. If $G_k/G_{k+1}$ is simple we are fine. If not we somehow want to build a normal series between $G_k$ and $G_{k+1}$ where each factor is simple.
I am sort of stuck at this step. So I could use help on this.
I also have no idea how to show that if $G$ has a composition series that $G$ satisfies (i) and (ii).
Any help would be great!