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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. $\square$

The Footprint
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  • No, this does not show that the centre of $G$ is non-trivial. Something stronger than the exercise asks you to prove is actually true: there are subgroups $N_i$ with these properties which are all normal in $G$. See here: https://math.stackexchange.com/questions/3276373/let-g-be-a-p-group-g-pr-prove-that-g-contains-a-normal-subgroup-o/3276421?noredirect=1#comment6737992_3276421 – the_fox Jun 29 '19 at 20:30
  • @the_fox but why does it not follow from my construction? – The Footprint Jun 29 '19 at 20:36
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    The group $S_3$ has a normal abelian subgroup of order $3$ (which is its commutator subgroup), yet $Z(S_3)=1$. – the_fox Jun 29 '19 at 20:41
  • @the_fox Do you think my solution is correct then? – The Footprint Jun 29 '19 at 22:58
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    In your proof, you are using a theorem that is stronger than what you are trying to prove. (Also, you are basically using induction, so spell it out.) That's not a good idea. Here's a better one: $N_G(H)>H$ whenever $H$ is a proper subgroup of a $p$-group $G$. Thus every maximal subgroup is normal in $G$ and thus has index $p$, so the factor group is certainly abelian. – the_fox Jun 29 '19 at 23:08
  • @the_fox yes, I use the fact that every subgroup of a p-group is normal, but I also have to show that $N_{i-1}/N_i$ is commutative so it's not a stronger theorem altogether? (is it?) And I understand what you are hinting, but I don't know how to show that $N_G(H) > H$. What elements $x \not \in H$ commute with $H$? – The Footprint Jun 29 '19 at 23:18
  • What basic theorems about $p$-groups has Herstein covered up to that point? – the_fox Jun 29 '19 at 23:20
  • @the_fox Basically, only the fact that p-groups have non trivial centers. The rest is in the exercises, without solutions. (it's quite frustrating) For me it was easier to prove that every subgroup of a p-group is normal than your latest claim about $N(H)$, so that's why I used that in my solution. – The Footprint Jun 29 '19 at 23:23
  • It is not true that every subgroup of a $p$-group is normal. – the_fox Jun 29 '19 at 23:24
  • @the_fox you're right. I proved that a p-group has a normal subgroup of order $p^k$ for each $k$. I hope that's true at least! Can you give me a hint for proving that $N(H)>H$? – The Footprint Jun 29 '19 at 23:28
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    Herstein's book should cover that property at some point because it is quite fundamental. Otherwise see here: https://math.stackexchange.com/questions/234426/if-h-is-a-proper-subgroup-of-a-p-group-g-then-h-is-proper-in-n-gh – the_fox Jun 29 '19 at 23:44

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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.

the_fox
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    A very similar proof shows that a p-group has a normal subgroup of order $p^k$ for each $k$. I had proved that in a previous exercise, so that's why I used it, instead of repeating your argument. – The Footprint Jun 29 '19 at 23:49