Composition Factors cyclic IFF finite group soluble

[TaliskerBA]

Hey, just trying to get my head around the logic of this. I can see that if composition factors are cyclic then clearly the group is soluble, since there exists a subnormal series with abelian factors, but I am struggling to see how the converse holds. If a group is soluble, then it has a subnormal series with abelian factors, and if this is already the composition series then given that the factors are both simple and abelian, they are cyclic. But when refining a series and adding in new "terms", why should the new factors be abelian? I know they are simple if they are in the composition series, but just trying to see why it must be that they are abelian.

 

[fresh_42]

Given a solvable group $G$ and a subnormal series $\{\,0\,\} =G_0 \triangleleft G_1 \triangleleft \ldots \triangleleft G_n=G$ with Abelian factors. Now let's look at a refinement, say $G_i \triangleleft H \triangleleft G_{i+1}$. Then $H/G_i$ is Abelian as a subgroup of the Abelian group $G_{i+1}/G_i$. But then by the isomorphism theorem $G_{i+1}/H \cong \left( G_{i+1}/G_i \right)/\left( H/G_i \right)$ is also Abelian. By the main theorem of subnormal series, our given series can be refined to a composition series, which must have simple Abelian factors. But simple and Abelian means cyclic of prime order.