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I was reading an article where it is claimed that

If $G$ is a finite $p$-group with $|G|=p^n$ and nilpotency class of $n-2$ where $n\ge 7$ then $p\le|Z(G)|\le p^2$ and $p^{n-3}\le |G'|\le p^{n-2}$.

I am not sure how these follow, I tried but couldn't succeed. I know from this post that nilpotency class of $G'$ is at most $n-3$ but I can not use it to get the required bound. I am really sorry if I am missing something easy.

If anyone can help me I will be really grateful.

Thanks

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You have a finite $p$-group $G$ of order $p^{n}$ and nilpotency class $c=n-2$, with $n\geq 7$. You are seeking to show four things, labelled 1 to 4 below: $$p \leq^{(1)} |Z(G)| \leq^{(2)} p^{2} $$ $$ p^{n-3} \leq^{(3)} |G^{\prime}| \leq^{(4)} p^{n-2} $$

(1) For a finite $p$-group we must have $p \leq |Z(G)|$. One of the most important facts about finite $p$-groups is that their center is non trivial. As this is a standard result in basic courses I will not prove it here. For a reference see here.

(2) We wish to show that $|Z(G)| \leq p^{2}$. Consider the upper central series of $G$: $$1=Z_{0}(G)<Z_{1}(G)<\dots<Z_{c-1}(G)<Z_{c}(G)=G$$ The series consists of $c+1$ terms and $c$ factors. It is clear that each factor has order at least $p$, or else we would have a redundant term. I claim that the last factor, $\frac{G}{Z_{c-1}(G)}$, must have order at least $p^{2}$. Suppose for contradiction that this last factor has order $p$. We work modulo $Z_{c-2}(G)$, and observe that $\frac{Z_{c-1}(G)}{Z_{c-2}(G)}$ is the center of $\frac{G}{Z_{c-2}(G)}$. Notice that $\frac{G/Z_{c-2}(G)}{ Z_{c-1}(G)/Z_{c-2}(G)} \cong \frac{G}{Z_{c-1}(G)} $ by the isomorphism theorems, and that by our assumption this is of order $p$, hence the quotient is cyclic. Thus we are taking a quotient by the center and end up with something cyclic, and so by a well known result $G/Z_{c-2}(G)$ must be abelian. Hence $[G,G]\leq Z_{c-2}(G)$. However this now means if we consider the upper central series, the term $Z_{c-1}(G)$ is redundant, and so we can obtain a central chain of shorter length, contradicting the assumption that the class is $c$. Thus we see the order of the last factor must be at least $p^{2}$. Now if $|G|=p^{n}$ the product of all the factors must equal $n$. Explicitly $p^{n}=|G|=|\frac{Z_{1}(G)}{1}||\frac{Z_{2}(G)}{Z_{1}(G)}|\dots|\frac{G}{Z_{c-1}(G)}|$. There are $c=n-2$ factors and each factor has order at least $p$ with the final factor $\frac{G}{Z_{c-1}(G)}$ at least $p^{2}$. Now by considering the order of the first factor, it is clear that $|Z_{1}(G)| \leq p^{2}$ or else the order of the group would be too large.

(3) Now we will show $p^{n-3} \leq |G^{\prime}|$. Consider the lower central series of $G$. $$G>\gamma_{2}(G)> \dots > \gamma_{c}(G)>1$$ where $\gamma_{2}(G)=[G,G]$. Each of the $c-1$ factors in the product below must have order at least $p$ and so $|[G,G]|=|\gamma_{2}(G)|=|\frac{\gamma_{2}(G)}{\gamma_{3}(G)}|\dots |\frac{\gamma_{c}(G)}{\gamma_{c+1}(G)}| \geq p^{c-1}=p^{n-3} $.

(4) Finally we show $|G^{\prime}|\leq p^{n-2}$. Consider the upper central series as in (2). We have that $[G,G]\leq Z_{c-1}(G)$ (this is clear from the definition of the upper central series). Given what we have shown in (2), that the factor $\frac{G}{Z_{c-1}(G)}$ must have order at least $p^{2}$, it follows that the order of $Z_{c-1}(G)$ is at most $p^{n-2}$. Hence $|G^{\prime}| \leq |Z_{c-1}(G)| \leq p^{n-2}$.

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  • $\begingroup$ Thanks for your answer. Such a detailed answer is so helpful. The claim was not clear to me form the article, I understand now. Thanks. It is also nice to see (from your profile, sorry!) that you are interested in GAP and p-groups :) I also like these topics. Thanks a lot again. $\endgroup$ – usermath Jun 5 at 1:32

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