The commutator of Holomorph of generalized quaternion is abelian?

Question

Let $Q_{2^{n}} = \langle x, y | x^{2^{n-1}}=y^4 = 1, x^{2^{n-2}}=y^2, y^{-1}xy = x^{-1} \rangle$ - generalized quaternion group of order $2^{n}$.

$\operatorname{Hol}(Q_{2^{n}})$ - Holomorph of this group

Let $G = [ \operatorname{Hol}(Q_{2^{n+1}}), \operatorname{Hol}(Q_{2^{n}})]$ - commutator of $\operatorname{Hol}(Q_{2^{n+1}})$

I'm wondering whether $G$ is abelian.

Using the SAGE, I checked that this is true for $n = 3 ... 9$.

Some auxiliary facts:

1. $${\rm Aut}(Q_{2^n}) \cong \left\{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} : a \in \mathbb{Z}^*_{2^{n-1}}, b\in \mathbb{Z}_{2^{n-1}} \right\} \cong AGL(\mathbb{Z}_{2^{n-1}}, 1)$$ such that for $\varphi \in {\rm Aut}(Q_{2^n})$ $$\varphi (x) = x^a, \varphi (y) = x^by$$
2. $[Q_{2^n}, Q_{2^n}] = \langle x^2 \rangle$
3. $[{\rm Aut}(Q_{2^n}), {\rm Aut}(Q_{2^n})] = \left\{ \varphi : \varphi(x) = x, \varphi (y) = x^by,\space b\in 2\mathbb{Z}_{2^{n-1}}\right\}$

Answer 1

Note that

• $\forall \varphi \in {\rm Aut}(Q_{2^n}), \forall g \in Q_{2^n} \;\; \varphi (g) g^{-1} \in \langle x \rangle$.
• $\forall \tau \in [{\rm Aut}(Q_{2^n}), {\rm Aut}(Q_{2^n})] \;\; x^{\tau} = x$. ($x^{\tau} = \tau (x)$)
• $[{\rm Aut}(Q_{2^n}), {\rm Aut}(Q_{2^n})]$ and $[Q_{2^n},Q_{2^n}]$ are abelian.

Consider two elements from the Holomorph $\varphi g, \phi h\in {\rm Hol}(Q_{2^n})$. Note that $\varphi g \cdot \phi h = \varphi \phi g^{\phi}h$ and $(\varphi g)^{-1} = \varphi^{-1} (g^{-1})^{\varphi^{-1}}$.

As anyone can see, $[\varphi g, \phi h] = [\varphi, \phi] \cdot (g^{\phi} h g^{-1} (h^{-1})^{\varphi})^{\varphi^{-1} \phi^{-1}} = [\varphi, \phi] \cdot \alpha$. $$\alpha = (g^{\phi} h g^{-1} (h^{-1})^{\varphi})^{\varphi^{-1} \phi^{-1}} = (g^{\phi}g^{-1} [g,h] h (h^{-1})^{\varphi})^{\varphi^{-1} \phi^{-1}}$$ As above, $g^{\phi}g^{-1}$, $h (h^{-1})^{\varphi}$ $\in \langle x \rangle$ and $[g,h]\in \langle x^2 \rangle \subset \langle x \rangle$, then $\alpha \in \langle x \rangle$, because $\varphi^{-1} \phi^{-1}$ - automorphism.

In that way, for some elements $\tau \alpha, \sigma \beta \in [{\rm Hol}(Q_{2^n}),{\rm Hol}(Q_{2^n})]$ we have that $\beta^{\tau} = \beta$ and $\alpha^{\sigma} = \alpha$.

Then $$\tau \alpha \cdot \sigma \beta = \tau \sigma \alpha^{\sigma} \beta = \tau \sigma \alpha \beta = \sigma \tau \beta \alpha = \sigma \tau \beta^{\tau} \alpha = \sigma \beta \cdot \tau \alpha,$$ e.g. $[{\rm Hol}(Q_{2^n}),{\rm Hol}(Q_{2^n})]$ is abelian.