In these lecture notes about classical groups, on page 4, 2nd paragraph, the finite subgroups of the multiplicative group of some division ring/skew field are considered:
Let $G$ be a finite subgroup the multiplicative group of the division ring $F$. We claim that there is an abelian [finite] group $A$ such that $G$ is a group of automorphisms of $A$ acting semiregularly on the non-zero elements. Let $B$ be the subgroup of $(F,+)$ generated by $G$. Then $B$ is a finitely generated abelian group admitting $G$ acting semiregularly. If $F$ has non-zero characteristic, then $B$ is elementary abelian; take $A = B$. Otherwise, choose a prime $p$ such that, for all $x,g \in G$, the element $(xg - x)p^{-1}$ is not in $B$, and set $A = B / pB$.
The prime is choosen such that $G$ acts semiregularly on the elements $x + pB$ with $x \in G$ [by the way, $g \ne 1$ should be written in the last sentence after "for all $x,g\in G$"]. But I do not see that this implies that $G$ acts semiregulary on the entire group $B / pB$. The elements of $B$ are sums of elements from $G$ and their additive inverses, and the claim is that for any non-identity element $g \in G$, the difference of the image of such an element under $g$ and the sum of these elements is not in $pB$, i.e., $G \setminus \{1\}$ has no fixed points modulo $p$.
I do not see that this is the case, that $G$ acts semiregularly on $A$? What am I missing, why is this the case?
If $B \cong {\mathbb Z}^n$, then the action of $g-1$ on $B$ for $g \in G \setminus \{1\}$ defines an $n \times n$ matrix $A_g$ with entries in ${\mathbb Z}$ and, since $g$ acts semiregularly on $B \setminus \{0\}$, $A_g$ has nonzero determinant.
If you choose your prime $p$ such that it does not divide $\det(A_g)$ for any $g \in G \setminus \{1\}$, then the reduction of $A_g$ mod $p$ will be invertible and so will have zero nullspace. So each $g \in G \setminus \{1\}$ acts semiregularly on the nonzero elements of $B/pB$.