$o(G)=p^n$. Then $\exists N_i\le G$ s.t. $N_{i-1} \supset N_i$, $N_i \unlhd N_{i-1}$ and $N_{i-1}/N_i$ is abelian This is problem 12, p. 91, in Topics in Algebra by Herstein. I am not sure of my solution and I cannot find solutions online. I would be glad if someone on here could tell me if it's correct and if there are simpler solutions (without invoking Sylow if possible, as I haven't studied it yet).

If $o(G)=p^n$, $p$ a prime number, prove that there exist subgroups $N_i, i=0,1,\dots,r$ (for some $r$) such that $$G=N_0 \supset N_1 \supset N_2 \supset \cdots \supset N_r = (e)$$ where $N_i$ is a normal subgroup of $N_{i-1}$ and where $N_{i-1}/N_i$ is abelian.

My solution. The commutator subgroup of $G$, $G'$, is normal and of course the group $G/G'$ is abelian. Also, $G' \not = G$ (Claim)
(proof of Claim) For suppose there is no proper normal subgroup of $G$ such that the quotient group is abelian. It can be shown that $G$ has a normal subgroup of order each power of $p$ up to $p^n$; in particular there is $H \unlhd G$ of order $p^{n-1}$, making $G/H$ a subgroup of order $p$, which is necessarily cyclic and thus abelian, a contradiction. (end of proof)
So we can set $N_1=G'$. Since $G'$ too has order a power of $p$, we can repeat this procedure. It will eventually stop because $G$ is finite. $\square$
 A: Herstein is really asking you to prove that $p$-groups are solvable. Here's one way to do it (assuming the only thing you know about $p$-groups is that they have non-trivial centres).
Argue by induction on $k$, where $|G|=p^k$, the base case $G=C_p$ being evidently true. Now suppose you have proved the claim for all $p$-groups of order $p^{k-1}$. You know already that $Z(G)>1$, so choose an element $g$ of order $p$ in $Z(G)$
and let $N_1 = \langle g \rangle$ (since $g$ commutes with everything, $\langle g \rangle$ is normal in $G$). Consider the group $G/N_1$. The inductive hypothesis gives you subgroups $N_2/N_1$, ...,$N_k/N_1 = G/N_1$ such that $N_i/N_1$ is normal in $N_{i+1}/N_1$ and $N_{i+1}/N_1\big/N_i/N_1$ is abelian.
The series $1, N_1, \ldots, N_k =G$ has the desired properties. To see that $N_i$ is normal in $N_{i+1}$ use the Correspondence Theorem. To see that $N_{i+1}/N_i$ is abelian, use the fact that $N_{i+1}/N_1\big/N_i/N_1$ is abelian and apply the Third Isomorphism Theorem.
