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I am trying to look for a finite $p$-group of maximal class of order at least $p^{2p+1}$ exponent at least $p^3$ which possibly has the following property :

If s is an element in $G-G_1$ ($G_1$ is the unique two step centralizer) so that order of s is $p^2$, then for any element $s_1 \in G_1-G_2$ ($G_2 = [G,G]$) we have order of the product $ss_1$ is $p$.

So to speak, the only conjugacy classes in $G-G_1$ represented by order $p^2$ elements are given by $s^j G_2$ ($1 \leq j \leq p-1$).

Following work of Miech (70's) I checked that this cannot be possible for metabelian $p$-groups of maximal class.

However, I have a feeling that these groups may not exist. I am trying to go through the general cases, yet unsuccessful. If there is any obvious answer, please let me know.

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  • $\begingroup$ Here is what I tried : starting with assuming $G_3$ abelian (so that $G$ is not metabelian) and relations $s^p = s^y_{n-1}, (ss_1)^p = 1, [s_1, s_2] = s_{n-p+1}, [s_2, s_3] = s_{n-p+3}$ and the counting commutator identities by Miech (Thm. 4, Counting commutators, 1974) yields a strange relation $(ss^{\zeta}_1)^p = s^{4\zeta(\zeta-1)}_{n-2} s^{y(1-\zeta) + {\zeta \choose 2} + 6\zeta(\zeta-1)}_{n-1}$. This is obviously not true as $(ss^{\zeta}_1)^p \in Z(G)$. $\endgroup$
    – Siddhartha
    Commented Aug 21, 2018 at 15:14

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