Let $S$ be the group of invertible $n \times n$ matrices over $\mathbb{F}_p.$ Alice and Bob take turns choosing elements of $S$ such that no element is repeated and every element must commute with all previous matrices. Whoever cannot move loses. Who wins?
My solution: Let $A = \{ X \in S : X^2 = I\}, B = \{ X \in S : X^2 = -I\}, C = S \setminus (A \cup B).$ If $p \ne 2,$ then $X \ne -X$ for all $X \in S.$ Thus, Bob may win by playing $f(X)$ if Alice plays $X,$ where $$f(X) = \begin{cases} -X, \, X \in A \text{ or } X \in B \text{ and n odd}\\ X^{-1}, \text{ else} \end{cases}.$$ Suppose Alice just played $X$ and previously played $Y \ne X.$ To show Bob never repeats a move, it suffices to show that $f(X) \notin \{Y, f(Y)\}.$ Since $f$ is an involution, $f(X) \in \{Y,f(Y)\} \Rightarrow X \in \{f(Y), Y\},$ which means Alice repeated a move and already lost. Thus, Bob wins if $p \ne 2.$
Suppose $p = 2$ and $X \in A.$ The minimal polynomial of $X$ divides $t^2 - 1 = (t-1)^2 \mod 2,$ so the only eigenvalue of $X$ is $1.$ Let $X = PJP^{-1}$ and let $J_i = I + N$ where $N$ is the matrix with ones on the diagonal right above the main one be a Jordan block of $J.$ We must have $I = J_i^2 = I + N^2 \Rightarrow N^2 = 0 \Rightarrow J_i$ has size $1$ or $2.$
I suspect that in the case $p = 2,$ Alice wins by playing $I,$ then playing $X^{-1}$ whenever Bob plays $X \notin A,$ but I have no idea what she should do in the case $X \in A$ is played. For $n=1,2,$ we can see that Alice wins easily, but trying to analyze all $n$ choices for $J$ is unfeasible for larger $n$ since we have tons of choices for $P$ as well. I believe the catch is that this final case is much harder than the rest of the problem, and I've only scratched the surface of the tip of the iceberg when it comes to solving it. How should I proceed? Any hints, approaches, or ideas?
If we look for a strategy involving just polynomials, we will observe that $\mathbb{F}_2[X]/\langle X^2 - 1 \rangle = \{ O, I, X, X+I\}.$ However, $(X+I)^2 = X^2 + I = 2I = O$ and the other choices are not viable either. Thus, our strategy (which ostensibly must rely on performing some sensible operations on the set of previously played matrices) has to rely on the products of matrices. But which matrices shall we multiply?