Let $Q_{2^{n+1}} = \langle x, y \mid x^{2^{n}}=y^4 = 1, x^{2^{n-1}}=y^2, y^{-1}xy = x^{-1} \rangle$ be the generalized quaternion group of order $2^{n+1}$.
The automorphism group: $${\rm Aut}(Q_{2^{n+1}}) \cong \left\{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} : a \in \mathbb{Z}^*_{2^{n}}, b\in \mathbb{Z}_{2^{n}} \right\} \cong AGL(1, \mathbb{Z}_{2^{n}})$$ such that for $\varphi \in {\rm Aut}(Q_{2^{n+1}})$ $$\varphi (x) = x^a, \varphi (y) = x^by$$
$\operatorname{Hol}(Q_{2^{n+1}})$ - Holomorph of this group.
Let $R = GR(2^n, 2) \cong \mathbb{Z}_{2^n} / (x^2 + x + 1)$ be the Galois ring of characteristic $2^n$ with $2^{2n}$ elements.
$Aff(R) = AGL(1, R)$ - affine group of $R$ (acts on $R^1$). Elements of $Aff(R)$ are of the form: $$h_{(c,k)}: y\mapsto c\cdot y + k \qquad c\in R^{*}, k\in R.$$ $Aff(R) = \{ h_{(c,k)} \mid c \in R^{*}, k \in R \}$.
It can be shown that $Q_{2^{n+1}}$ is embedded in $R$. For example, let $$c = 2^{n-1} + 1,\; k = x,\; d = 2^{n-1} x + 1,\; b = 1,$$ then $$\langle h_{(c,k)}, h_{(d,b)} \rangle \cong Q_{2^{n+1}}.$$
Naturally the question arises: is the $\operatorname{Hol}(Q_{2^{n+1}})$ embedded in $Aff(R)$?
I guess probably not. Is there any way to show this?
$\langle X\mid R\rangle$for $\langle X\mid R\rangle$. – Shaun Mar 23 '23 at 20:47