I came across this question while doing some research work. In the following let $\mathbb Q_2,\mathbb Z_2$ denote the field of $2$-adic numbers and the ring of $2$-adic integers respectively. That is $\mathbb Z\le \mathbb Z_2$ and $\mathbb Q \le \mathbb Q_2$.
Let $G\le \rm{GL}(2^d,\mathbb Z_2)$ be a finite $2$-group acting fixed point free and faithfully on $\mathbb Z_2^{2^d}$ where $d$ is a fixed positive integer. Then does there exist any $u\in U$ such that $u$ has no fixed point in $\mathbb Q_2^{2^d}$?
My understanding is that $U$ has no fixed point in $\mathbb Z_2^{2^d}$ which should imply it has no fixed point in $\mathbb Z^{2^d}$ and possibly this means $U$ has no fixed point in $\mathbb Q^{2^d}$. Am I correct here?
But I am stuck after this.
If in general not true any $U$ then does the above hold if $U$ acts (in addition to the avove) uniserially on $\mathbb Z_2^{2^d}$ that is there exists a unique series of submodule namely $\mathbb Z_2^{2^d}=T_0> T_1> T_2> \ldots$ such that $[T : T_i]$ = $2^{i}$ for all $i \ge 0 $ and $T_i= [T_{i-1}, P]$ is the subgroup of $T_{i-1}$ generated by the elements $-t+t^g$ where the action of $g\in U$ on $t\in T_{i-1}$ is denoted by $t^g$.
Any help will be greatly appreciated.
I don't understand your question.
With $g\in GL_n(\Bbb{Q}_2)$ then $g(v)=v$ for some $v\in \Bbb{Q}_2^n$ iff $g(2^m v)=2^mv$ for some $2^mv\in \Bbb{Z}_2^n$. It happens iff $\det(g-xI)\in \Bbb{Q}_2[x]$ has $1$ in its root.
Moreover if $g\in GL_n(\Bbb{Z}_2)$ then $\| g(v)\|\le\|v\|$ and $\|v\|=\|g^{-1}(g(v))\|\le \|g(v)\|$ thus $\|g(v)\|=\|v\|$ where $\|v\|=\sup_j |v_j|_2$.
What is $U$. What does the last paragraph mean.