A theorem says a finite group G is solvable if and only if for every divisor n of $\vert G\vert$ such that (n,$\frac{\vert G\vert}{n})=1$,G has a subgroup of order n.
- Does this imply that G must containt prime factors upto power 1.(i.e. $\vert G\vert$=$p_{1}p_{2}...p_{n}$ where all $p_{i}\neq p_{j}$ for $i\neq j$? I thought by moding out operation say a composition series for solvable group $1=G_{0},G_{1},...,G_{n-1},G=G_{n}$ such that $G_{i}/G_{i-1}$ is abelian, G/N is abelian by definition of G solvable. Then every prime factor has an element of order of that prime number. Hence, $\vert G\vert$=$p_{1}p_{2}...p_{n}$ and they are all different.
- Can one also conclude that any subgroup of G such that N normal to G, there will be a composition series $1=G_{0},...,G_{m-1},N=G_{m},G_{m+1},...,G=G_{n}$ such that $G_{i}/G_{i-1}$ is abelian? I knew this is true for composition series but I do not know whether this is true for solvable groups.
No, $|G|$ need not be square-free. For example, any abelian group satisfies that condition.
Yes. Since $G$ is solvable, $N$ and $G/N$ are both solvable. A composition series of $G$ can be constructed from that of $N$ and $G/N$:
$1,G_0,\dots, G_{m-1},G_m=N$ is a composition series of $N$,
$1,G_{m+1}/N,\dots, G/N$ is a composition series of $G/N$.