G is solvable iff factors have prime order A group $G$ is said to be solvable if, and only if, there exists a subnormal series of subgroups $\{e\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G$ such that each factor $\displaystyle\frac{G_{i+1}}{G_i}$ of the series is an abelian group. Prove that a finite group $G$ is solvable if, and only if there exists a subnormal series of subgroups $\{e\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G$ such that each factor $\displaystyle\frac{G_{i+1}}{G_i}$ of the series has a prime order (i.e. $\left| \displaystyle\frac{G_{i+1}}{G_i} \right | = p$ for some $p$ prime number).
This exercise was proposed to us by our algebra teacher and I just keep thinking that there is something wrong here. Every time I looked up in the internet, other conditions were necessary for this to be valid. If it isn't, what are possible counter-examples? If it is valid, however, how can I prove it? Tried a huge amount of stuff and can't figure it out.
NOTE: I do not have learned the definitions of solvable groups with derived series and commutators yet, so please do not use them. I mean, i want to learn but I think that this exercise just needs to be solved without them.
NOTE 2: No, this is not my homework, this exercise will not be checked for points or anything like that, I just want to learn.
 A: This is false in this generality. You require $G$ to be finite for this to be true. A counterexample would be $\mathbb{Z}$ or $\mathbb{Q}$.
For finite groups it is true. In order to solve it you first need to prove it for finite abelian groups. Once you have that, then you 'insert' the series you find for each abelian subquotient (quotient of one subgroup by another) into the abelian series you started with, using the correspondence between subgroups of a quotient group and subgroups of a group containing the kernel of the quotient map.
For abelian groups $G$, the easiest way is to use induction on $|G|$. If $G$ has no subgroups at all then $G$ is cyclic of prime order, and $1\lhd G$ is a series. If $H$ is a subgroup, by induction both $H$ and $G/H$ have such a series and then you append the series for $G/H$ onto the series for $H$ using the correspondence mentioned above. Thus given
$$1=H_0\lhd H_1\lhd H_2\lhd \cdots \lhd H_n=H,$$
$$H/H=G_0/H\lhd G_1/H\lhd \cdots \lhd G_m/H=G/H,$$
we obtain
$$ 1=H_0\lhd H_1\lhd H_2\lhd \cdots \lhd H_n=H=G_0\lhd G_1\lhd G_2\lhd \cdots \lhd G_m=G.$$
Now we have a series for an abelian group, we do the general case using exactly the trick in the abelian case. Stitch the series together for each factor $G_{i+1}/G_i$.
A: If $G$ is finite then the fact is true. Indeed, it is enough to prove it for finite Abelian groups. But by the converse of Lagrange theorem for finite Abelian groups, if the order of the group is $n$ and $m \mid n$ then there is a subgroup of order $m$. This implies the statement. For infinite solvable groups the statement is not true. Example: the group of integers under addition.
