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Recall the definition of the finite Suzuki 2-groups: These are finite non-abelian 2-groups that contain more than one involution such that a solvable group of automorphisms permutes the involutions transitively. The Suzuki 2-groups were classified by Higman in [Hig].

If $G$ is a Suzuki 2-group, then G has nilpotency class equal to 2, and $G' = Z(G) = \Phi(G) = \Omega( Z(G) ) = \Omega(G)$ and $|G| = |Z(G)|^2$ or $|G| = |Z(G)|^3$. The Suzuki 2-groups $G$ with $|G| = |Z(G)|^2$ are also known as the Suzuki 2-groups of type $A$ and a concrete description of them has been given in [Hig]. For any integer $n \geq 3$ which is not a power of 2 and any odd order automorphism $\theta$ of $\mathbb F_{q}$ where $q = 2^n$ there is such a group of type $A$ denoted by $A(n, \theta)$. And one knows that $A(n,\theta) \cong A(n, \theta')$ if and only if $\theta^{\pm 1} = \theta'$.

I would like to understand the finite 2-groups $P$ whose center $Z(P)$ is of order 2 such that $P/Z(P)$ is a Suzuki 2-group of type $A$. If such a $P$ exists, is it unique?

Moreover, I am happy to impose the following further restrictions on $P$ (if this helps):

  1. $P$ should have exponent 4
  2. $\Omega(P)$ is elementary abelian of rank $n+1$ and $P/\Omega(P)$ is elementary abelian of rank $n$.
  3. $\Omega(P) = \Phi(P) = P' = \mho(P)$

I tried to check some examples with GAP: There is a unique Suzuki 2-group of type $A$ of order 64 and for this group I do know that a group $P$ of order 128 as above exists (and it is unique up to isomorphism). At least the next larger Suzuki 2-groups of type $A$ of order 1024 are capable (i.e. they can be written as $H/Z(H)$ for some group $H$). However, for these one cannot choose $H$ to be a 2-group with $Z(H)$ of order 2.

I would also be interested to know which of the Suzuki 2-groups (not necessarily of type $A$) are capable.

[Hig] Higman, G. "Suzuki 2-groups.", Ill. J. Math. 7, 79-96 (1963).

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I can give a partial answer now, at least regarding the capability of the groups $A(n, \theta)$. For the following we do not assume that $\theta$ is of odd order. The group $A = A(n, \theta)$ is in general defined by the set $\mathbb F_q \times \mathbb F_q$ (here $q =2^n$) with mulitplication defined via $$ (\zeta, \alpha) \cdot ( \eta, \gamma) = (\zeta + \eta + \alpha \cdot \gamma^{\theta}, \alpha + \gamma) $$ ($A(n, \theta)$ is only a Suzuki 2-group if $\theta$ is of odd order, but here we consider this more general setup). Now we can write down a cocycle $f \colon A \times A \to \mathbb F_q$ by the following formula: $$ f((\zeta, \alpha), (\eta, \gamma)) = \zeta \cdot \gamma^{\theta^2} + \alpha \cdot \eta^{\theta} $$ Then in corresponding group $\mathbb F_q \times_{f} A$ the multiplication is defined by $$ (a, g) \cdot (b,h) = (a+b+f(g,h), g \cdot h) $$ One checks that this group has center $\mathbb F_q$ provided that $\theta^{2} \neq 1$. Hence

The group $A(n, \theta)$ is capable if $\theta$ is not of order two.

(If $\theta = \operatorname{id}$, then $A(n, \theta) \cong (C_4)^n$ and this group is capable)

So the only case left to consider is if $\theta$ is of order 2, so let us assume this from now on. It is not hard to check that in this case $$ A' = \{ (\zeta, 0) \mid \zeta \in \operatorname{Fix}(\theta) \} = \mho(A) $$ Then any element of $A'$ is a basic commutator. Moreover $$ \Omega(A) = Z(A) = \{ (\zeta, 0) \mid \zeta \in \mathbb F_q \} $$ A calculation with GAP now suggests that the epicenter $Z^*(A) = A'$ (hence $A$ would not be capable in this case) and it also suggests that the abelianization map $A \to A/A'$ induces an isomorphism of the non-abelian tensor squares $$ A \otimes A \to A/A' \otimes A/A' $$ Recall that the non-abelian tensor square $G \otimes G$ is generated by the symbols $g \otimes h$ subject to the relations: \begin{align*} (gg') \otimes h &= (^g{g}' \otimes ^g{h}) (g \otimes h) \\ g \otimes (hh') &= (g \otimes h) (^h{g} \otimes ^h{h}') \end{align*} (here $^h g = hgh^{-1}$)

In the case I am interested in $G = A$ is nilpotent of class 2. And one knows that $A \otimes A$ is abelian in this case. Moreover, the following identity holds (regardless of the nilpotency class) $$ [h,g] \otimes h' = (g \otimes h) ^{h'}{(g \otimes h)}^{-1} $$ This implies, for example, that $x \otimes z = 1_{\otimes}$ whenever $x \in A'$ and $z \in Z(A)$.

Does $[h,g] \otimes h' = 1_{\otimes}$ for all $h,h',g \in A$ hold in $A \otimes A$?

Maybe someone with more knowledge of non-abelian tensor squares can help with this question.

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