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Let $G$ be a group of order $p^n$, with $p$ prime. By Sylow's first theorem, there exists at least one subgroup of order $p^n$ (the number of subgroups of order $p^i$ is $1$ mod $p$ per $i$). The subgroups with order $p^n$ are all Sylow-$p$ groups.

Now, by Sylow's third theorem, because the group is of order $p^n$, the number $m_{p^{n}}$ of such subgroups must divide $\#G/p^n =1$, and only $1$ divides $1$, so there is only one subgroup of order $p^n$.

By Sylow's second theorem, all Sylow-$p$ groups are conjugated to each other by at least one element $g\in G$, so, for any $S$ and $S'$, we have $S=gS'g^{-1}$. In this case, there is only one Sylow-$p$ group, so it is conjugated to itself.

Of course, that one subgroup is the group itself. We now have $gG=Gg$ for some $g$ in $G$. Can we get to the entire group being abelian, from here?

I ask because my textbook on Abstract Algebra states that any group of order $p^2$ is abelian, and I'm curious whether it generalises.

Edit: As has been pointed out, everything I've proved above is quite trivial. Below it is discussed that the essential question is actually "How does one prove that groups of order $p^2$ are abelian using Sylow theory?", since my textbook explicitly mentions this property as an application of Sylow's theorems.

Edit 2: One of the authors has confirmed that they accidentally mixed some classic classification theorems into the list of applications of Sylow theory, and that this was one of them.

Mew
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    Hint: look at groups of order $8$. – lulu Jun 11 '20 at 15:41
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    Nope—consider $D_4$, the dihedral group with eight elements – boink Jun 11 '20 at 15:41
  • Ah, thanks, the both of you. Any idea which further steps are needed to prove that it does work for $n=2$? – Mew Jun 11 '20 at 15:46
  • Are the first two paragraphs necessary? Isn't the only subgroup of the same order of the (finite) group the group itself? –  Jun 11 '20 at 16:57
  • @csn: True. In that sense, the question can be boiled down to "Prove that groups of order $p^2$ are abelian from the POV of Sylow theory", see also my comment under hdighfan's answer. – Mew Jun 11 '20 at 17:07
  • @MattSamuel: I'll contact the authors, in that case. Thanks for adding. – Mew Jun 11 '20 at 17:15
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    You can't really mean that a group of order $p^n$ has at least one subgroup of order $p^n$. True, yes, but not very illuminating. – TonyK Jun 11 '20 at 17:22
  • @TonyK: On the other hand, proofs by induction would be much nastier if mathematics didn't allow for evidencies to be made explicit. :) – Mew Jun 11 '20 at 20:33

3 Answers3

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It doesn't generalise -- the dihedral group with $8$ elements is an example, and a more general one is the set of matrices $$\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix}$$ with $x,y,z \in \mathbb{Z_p}$.

For a group of order $p^2$, the most common way to prove that it is abelian is to look at its center, $Z(G)$, the set of terms which commute with every other term. The center has to be nontrivial: if you consider the conjugacy classes of $G$ on itself, each must be of size $p^k$ for some non-negative $k$. But the conjugacy class of $e$ is trivial, and thus there exist at least $p-1$ other such cases.

Suppose the center has size $p$. Then $G/Z(G)$ must be a cyclic group on $p$ elements, and thus $G$ must be abelian. (for any group $H$, if $H/Z(H)$ is cyclic then $Z(H)=H$).

hdighfan
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  • Alright. Is my proof a dead-end, as far as proving $n=2$ goes? I've seen the proof using the centre floating around, but I was never taught about the centre of a group, and since the property w.r.t. $p^2$ is mentioned in the context of Sylow theory (but without proof, sadly), I'm wondering whether I can work from that angle. – Mew Jun 11 '20 at 16:55
  • What csn said in the last comment is quite accurate. Everything you've proven is trivial -- for any element $g\in G$, $gG=Gg$, so that's saying nothing. The preceding two paragraphs are actually even more trivial. In general, Sylow's theorems applied directly to $p$-sylows are trivial and don't tell you anything. – hdighfan Jun 11 '20 at 17:00
  • I see what you both mean; it's indeed all trivial in the end. Again, though, it makes me wonder why the textbook would state the commutativity of groups of order $p^2$ directly inside an enumeration "Applications of Sylow's theorems". I was in the headspace of Sylow's theorems, and thus thought it'd only make so much sense to prove it that way. – Mew Jun 11 '20 at 17:04
  • Hmm, that's very strange. It could be that results stemming from the proof of the Sylow theorems can be used, but that depends on how they were proved, of course. – hdighfan Jun 11 '20 at 17:07
  • The class formula, which is essentially how you showed that the center is nontrivial, is an ingredient of a rather standard proof of the Sylow theorems. – Andreas Blass Jun 11 '20 at 18:16
  • I contacted the authors, and it turns out this was an error on their part: they mixed up unrelated classification theorems into the applications of Sylow theory. There's the answer. (This is the university's AbsAlg textbook, so not much scrutiny happened before publishing.) – Mew Jun 12 '20 at 07:24
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You can construct a nonabelian group of order $p^n$, for $p$ an odd prime, $n\gt2$, by selecting a nontrivial homomorphism $\varphi:\Bbb Z_p\to\rm{Aut}(\Bbb Z_{p^{n-1}})\cong\Bbb Z_{p^{n-1}-p^{n-2}}$. Let $G=\Bbb Z_{p^{n-1}}\rtimes_\varphi\Bbb Z_p$.

If, on the other hand, $p=2$, consider dihedral groups.

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A group of order $p^n$ is always nilpotent. This is a natural generalisation of abelian. The examples of $Q_8$ and $D_4$ of order $8$ are nilpotent but non-abelian. The group of upper-unitriangular matrices over $\Bbb F_p$ is the Heisenberg group, which is $2$-step nilpotent, and also non-abelian.

Reference: Prove that every finite p-group is nilpotent.

Dietrich Burde
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