Center of $p$-group of maximal class

Asked Sep 12 '19 at 09:37

Active Sep 17 '19 at 19:47

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If $G$ is a finite $p$-group of maximal class, then I was trying to prove that the upper and lower central series coincide.

I am able to prove the result if center of $G$ is of order $p$. Then I found it as an exercise in the book "Groups of prime power order Vol.1" by Yakov Berkovich that for a finite $p$-group of maximal class, center of $G$ is always of order $p$.

Can anyone help me to prove why the other possibility of the size of center i.e. $p^2$ is not there?

edited Sep 17 '19 at 19:47

Shaun

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asked Sep 12 '19 at 09:37

math seeker

See https://math.stackexchange.com/questions/1467774/on-p-groups-having-maximal-class – Nicky Hekster Sep 12 '19 at 10:32
1

Since $K/Z(K)$ cannot be cyclic of order $p$, the topmost proper term of the upper central series is of index $p^2$; in order to have the correct number of terms, all the remaining ones must have index $p$ in the succeeding term, hence $Z_0={e}$ must have index $p$ in $Z_1(G)=Z(G)$. – Arturo Magidin Sep 12 '19 at 11:28
I was unable to get why topmost proper term of the upper central series can only have index p^2 – math seeker Sep 12 '19 at 11:33
@SurinderKaur: Because otherwise, the center of $G/Z_{k-1}(G)$ (which is $Z_k(G)/Z_{k-1}(G)$ would have prime index in the group, which is impossible. The center cannot have prime index in the group, because the central quotient of a group cannot be nontrivial cyclic. So the index must be at least $p^2$, and if it were any larger, there wouldn’t be enough room for an upper central series with the right number of terms for maximal class. – Arturo Magidin Sep 12 '19 at 20:38
Now I get it. Thank you very much @Arturo – math seeker Sep 13 '19 at 04:11

Center of $p$-group of maximal class

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