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I'm aware that for a $p$-group $G$ of order $p^{n}$, say, that $G$ must have a nilpotency class between 1 and $n - 1$ for $n\geq 2$. My question is why can $G$ not have nilpotency class $n$? Take for example a group of order $p^{3}$. Why is it not possible to make a lower central series with orders $p^{3} \rightarrow p^{2} \rightarrow p \rightarrow 1$? I know all the 5 possible groups in this case have nilpotency class 1 or 2. But since there's theoretically 'room' for there to be a chain of length 4, how do we know this is never the case in general?

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    $\begingroup$ Because groups of order $p^2$ are abelian. $\endgroup$ – Qiaochu Yuan Sep 5 '17 at 20:55
  • $\begingroup$ Thanks a lot, quite an oversight from me. Clearly the chain either goes (listing orders here) $\dots \rightarrow p^{3} \rightarrow p \rightarrow 1$ or $\dots \rightarrow p^{3} \rightarrow p^{2} \rightarrow 1$. $\endgroup$ – Lightful Sep 5 '17 at 21:02
  • $\begingroup$ That' not right! It's the first step, at the top of the group, that must have order at east $p^2$. All of the other steps can have order $p$. $\endgroup$ – Derek Holt Sep 5 '17 at 22:08
  • $\begingroup$ Could you elaborate on that Derek? If that's the case then for me it's not obvious why. $\endgroup$ – Lightful Sep 5 '17 at 23:01
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    $\begingroup$ No. If you have a term in the central series such that the next one is trivial, then it should be central, which is stronger than abelian. (What you said would be true if we were talking about the derived series instead.) $\endgroup$ – verret Sep 6 '17 at 22:03
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This question was answered in the comments by Derek Holt over a year ago. So that this question does not remain "unanswered", I will include his argument as an answer here but then also include a different way of seeing this corresponding to the upper central series, based on some lecture notes of Gustavo Fernández-Alcober.

Any mistakes in the following are my own.

Argument involving the Lower Central Series

Based on comments by Derek Holt

It is the case that $G/[G,G]$ must have order at least $p^{2}$. To see this, consider any normal subgroup $N$ of $G$, such that $N$ has index $p^{2}$ (as we are in a p-group with order at least $p^{2}$ we know such a subgroup must exist). Then $G/N$ is a group of order $p^{2}$ and so must be abelian, and hence $[G,G] \leq N $. It follows that
$|G/[G,G]| \geq p^{2}$. Then if $G$ has class $c$, the lower central series has $c+1$ terms and $c$ factors. Each of these factors must have order at least $p$ (or else we would have a redundant term), and the factor $G/\gamma_{2}(G)$ must have order at least $p^{2}$. Then as the order of $G$ is equal to the product of these factors, we obtain that $|G|=p^{n}\geq p^{c+1}$ and thus $c \leq n-1$.

Argument involving the Upper Central Series This is based on Theorem 1.15 in the lecture notes "An introduction to finite $p$-groups: Regular $p$-groups and groups of maximal class", by Gustavo Fernández-Alcober .

In a similar way, we can consider the Upper Central series of $G$, where $G$ has class $c$: $$G=Z_{c}(G)>Z_{c-1}(G)\dots >Z_{0}(G)=1$$ and we can show that $[G:Z_{c-1}(G)| \geq p^{2}$. Suppose for contradiction that the index was $p$. If $c=1$ then that would mean $G$ is cyclic of order $p$, contradicting the assumption that $|G|\geq p^{2}$. Thus we may assume $c \geq 2$.

We consider the quotient group $G/Z_{c-2}(G)$. Recall from the definition of the upper central series that $Z(G/Z_{c-2}(G))=Z_{c-1}(G)/Z_{c-2}(G)$. Then $$\frac{G/Z_{c-2}(G)}{Z(G/Z_{c-2}(G))}= \frac{G/Z_{c-2}(G)}{Z_{c-1}(G)/Z_{c-2}(G)} \cong G/Z_{c-1}(G)$$ by the Isomorphism theorems. But $G/Z_{c-1}(G)$ is cyclic of order $p$ and so is abelian. Hence $G=Z_{c-1}(G)$ contradicting that the class is $c$. Thus we may assume that the order of the first factor is at least $p^{2}$, and then as before multiplying the orders of the factors shows us that $p^{n} \geq p^{c+1}$ and the result follows.

Remarks

  1. For yet another proof of this, see Corollary 4.2 in the book "$p$-Automorphisms of finite $p$-groups" by Evgenii Khukhro.
  2. $p$-groups of order $p^{n}$ and class $n-1$ are known as groups of maximal class. These groups are well studied and searching this phrase should bring up plenty of information on them.
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  • $\begingroup$ $p$-groups of maximal class were classified by Aner Shalev. A complete account can be found in the book "The Structure of Groups of Prime Power Order" by C. R. Leedham-Green and S. McKay. $\endgroup$ – user1729 Jun 3 at 13:27

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