I am trying to find a group $G$ which has a composition series of length $1$, but has a subgroup with a composition series of length $m$ for all $m\geq 2$. I know that length $G=1 \implies G$ is simple since the composition series of $G$ would be $G=G_0 \triangleleft G_1=\{1\}$. Thus for $G$ to have a subgroup $H$ with length $m$, I know $H$ must not be normal in $G$. I am hoping to find a group $G$ with a non-normal subgroup $H \simeq C_n$ where $n=p_1^{n_1}p_2^{n_2}\cdots p_k^{n_k}$ where $\sum_{i=1}^k n_i = m$ for distinct primes $p_i$ since this is the length of $C_n$. Note $C_n$ is the cyclic subgroup of order $n$ (think $\mathbb{Z}_n$).
I was looking at different alternating groups in forms $A_{2^m+1}$ and $A_{2m+1}$. They are guaranteed to be simple for $m\geq 2$, but I am wondering if they always have such a cyclic subgroup. If they do, what is the subgroup?
If I am wrong about the alternating groups, what other groups do satisfy this?
What with the "standard" embedding of $\;A_4\;$ in $\;A_5\;$ ?
$$1\le V\lhd A_4\le A_5\;,\;\;V:=\{\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\,\}$$