Let $S$ and $T$ be linear operators on a finite-dimensional vector space $V$ over $\mathbb{C}$. Prove or disprove the following:
If $ST$ is diagonalizable, then both $S$ and $T$ are diagonalizable.
If $S^3 = S$, then $S^2$ is diagonalizable.
I have what I think is a valid proof of number 2.
Proof: If $S^3 = S$ then $S^3-S = 0$. Let $q(x) = x^3-x$. Since $q(S) = 0$, $q(x)$ is a multiple of the minimal polynomial of $S$. We see, $$\begin{align*} q(x) &= x^3-x\\ &= x(x^2-1)\\ &= x(x-1)(x+1)\end{align*}. $$ This polynomial has roots $0,1,$and $-1$, all with multiplicity 1, so any polynomial which divides it will also have no repeated roots. Thus $S$ is diagonalizable. Let $S = PDP^{-1}$ via diagonalization. Then $$\begin{align*}S^2 &= PDP^{-1}PDP^{-1}\\ &= PD^2P^{-1}.\end{align*}$$ Hence, $S^2$ is diagonalizable. $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad{\blacksquare}$
I am not quite sure how to handle part 1. It seems to require some kind of argument using eigenspaces. I would appreciate any help/hints with part 1. as well as any critiques/corrections for part 2.