Find the number of composition series of D24 (with 24 elements) I got this question and trying to find a way that is not fully brute force.
I found a composition series which is of length 4, so i know they are all of length 4. looking at the divisors of 24 we have $1,2,3,4,6,8,12,24$. so the corresponding sizes, by Lagrange's theorem can be:
Case 1: $1,3,6,12,24$
Case 2: $1,2,4,12,24$
Case 3: $1,2,4,8,24$
From here on I'm stuck. I can't see anything by brute force finding almost every subgroup of $D24$. Is there a better way? or maybe finding subgroups isn't too hard?
Thanks!
 A: $\newcommand{\normal}{\trianglelefteq}$
I think there's a nice recursive way of decomposing $D_n$ (I'm going to use this notation to mark the group with $2n$ elements) when $n=2^k \cdot p$ for some prime $p$ (which is your case for $n=12$). Let $D_n = \left<\rho, \varepsilon \right>$ where $\rho^n=1$ and $\varepsilon^2=1$. If $n$ is even then $N=\left<\rho^2,\varepsilon\right>$ is a normal subgroup as it has index $2$. For the same reason, this shows that the quotient group is simple. 
Next, you can show that $N_1 \cong D_{n/2}$. Now you continue in the same manner until you get $N_k \cong D_{n/2^k}=D_p$. At this point we know that $p$ is prime, and you can check that the normal subgroups are $\left<\rho^d \right>\cong \mathbb{Z}_{n/(2^k\cdot d)}$ for any $d \ | \ n/2^k=p$, thus the only proper normal subgroup is $\left<\rho\right> \cong \mathbb{Z}_p$, which finishes the composition series, since the next subgroup must be $\{1\}$.
This gives you one possible composition series, in your case it is:
$$\{e\}\underset{\mathbb{Z}_3}{ \normal} \mathbb{Z}_3  \underset{\mathbb{Z}_2}{ \normal} D_3 \underset{\mathbb{Z}_2}{ \normal} D_{6} \underset{\mathbb{Z}_2}{ \normal} D_{12}$$
and you can simply check the permutations of the quotient orders to see if you have other possibilities
