I've seen this problem on here and I'm having a hard time understanding the solution:
Let $G$ be a group of order $780 = 2^2 \cdot 3 \cdot 5 \cdot 13$. Assume that $G$ is not solvable. What are the composition factors of $G$? (Assume that the only nonabelian simple group of order $\le 60$ is the alternating group $A_5$.)
The solution states:
The Sylow $13$-subgroup $N$ is normal, since $1$ is the only divisor of $60$ that is $\equiv 1 \pmod{13}$. Using the fact that the smallest simple nonabelian group has order $60$, we see that the factor $G/N$ must be simple, since otherwise each composition factor would be abelian and $G$ would be solvable. Thus the composition factors are $\Bbb Z_{13}$ and $A_5$.
What I don't understand is why does $G/N$ not being simple imply that all composition factors are abelian?
In fact $780 = 13 \cdot 60 = 13 \cdot 2^2 \cdot 3 \cdot 5$.
NB: this explains the argument "$N$ is normal, since $1$ is the only divisor of $60$ that is $≡ 1$ (mod $13$)", using Sylow's theorem(s).
Assuming $G$ has order $780$, the order of $G/N$ is $60$.
Now, assume $G/N$ isn't simple. Then, its composition factors would haver order $<60$, hence would be abelian. Then, look at a composition series $$ \{1\} = M_0 \triangleleft M_1 \triangleleft \cdots \triangleleft M_k = G/N $$ of $G/N$. Setting $N_i = \{g\in G: gN \in M_i\}$, we would get a composition series $$ \{1\} \triangleleft N_0 \triangleleft N_1 \triangleleft \cdots \triangleleft N_k = G/N $$ of $G$, where $N_0 = N$ would be abelian (cyclic of order $13$) and each $N_{i+1}/N_i$ would be isomorphic to $M_{i+1}/M_i$, hence abelian.