Is the set of fixed points of hyperbolic elements from $\mathrm{PSL}_{2}(A)$ a group?

Given a subring $$A$$ of $$\mathbb{R}$$, we can consider the set $$\mathrm{PSL}_{2}(A)$$ of elements in $$\mathrm{PSL}_{2}(\mathbb{R})$$ with entries in $$A$$ and the determinant of associated matrix is equal to one.

We define a hyperbolic elements from $$\mathrm{PSL}_{2}(A)$$ as elements which the trace of the associated matrix is greater than two, in absolute value.

Let us denote the set of all fixed points of all hyperbolic elements from $$\mathrm{PSL}_{2}(A)$$ by $$\mathcal{P}_{A}$$. Notice that $$\mathcal{P}_{A}\subseteq \mathbb{R}\cup\{\infty\}$$. If $$A$$ is a proper subring of $$\mathbb{R}$$, does it exist some result about $$\mathcal{P}_{A}$$ being a group?

• When you construct $\mathcal P_A$ as a set of fixed points, on what is $\mathrm{PSL}_2(A)$ acting? You say that $\mathcal P_A \subseteq \mathbb R$, but I don't know how to make $\mathrm{PSL}_2(A)$ act on $\mathbb R$. – LSpice Dec 13 '18 at 2:42
• @LSpice I think the action is defined by the isomorphism of $PSL_2(R)$ to fractional linear transformations acting on $R \cup \{\infty\}$. – Neil Hoffman Dec 13 '18 at 5:44
• If $A$ has infinite unit group (i.e. $A$ is not $\mathbb Z$) then $\infty$ is a fixed point of an hyperbolic element of $\mathrm{PSL}_2(A)$ so $\mathcal P_A$ is not contained in $\mathbb R$. On the other hand for $A =\mathbb Z$ zero is not in $\mathcal P_A$ (as the group is discrete and zero is a parabolic fixed point). – Jean Raimbault Dec 13 '18 at 8:54
• How could $\mathcal{P}_A$ be a group if it contains $\infty$? – abx Dec 13 '18 at 10:26
• As noted above this is the case only if $A = \mathbb Z$, and then $\mathcal P_A$ is not a subring (as noted in the deleted comment, $\sqrt 2, \sqrt 3 \in \mathcal P_A$ but not $\sqrt 2 + \sqrt 3$). So the answer to your question as asked is : never. It would help if you'd provide some context to explain what you are looking for exactly. – Jean Raimbault Dec 13 '18 at 11:03