Assume $k$ alg. closed field, with characteristic $0$. Let $G$ a linear algebraic group, whose elements have finite order, in symbols $$ \forall g \in G, \exists n \in \mathbb{N} \ \ \text{s.t. } g^n=e $$ Prove that it is finite.
My attempt At first I wanted to make use of the variety-version of this "well known result". Algebraically closed field and characteristic $0$ doesn't imply that $k$ is uncountable (Thanks to Matt S for the observation) So I need to prove this (if possible) to use the result, which was given as an hint of the exercise.
The countable family of closed would be $\{Z_i\}_{i \in \mathbb{N}}$ where $$ Z_i := \{ g \in G \mid g^i = e \}$$ obviously $G=\bigcup_i Z_i$ but I'm not able to prove (I hope it is true with these hypothesis) that $$ \forall i \in \mathbb{N} \ Z_i \neq G $$ using the fact that $G$ is an infinite group. Suppose we have proved that, we would have that $$ \bigcup_i Z_i \neq G$$ so exists an element in $G$ of infinite order, so an absurd and so $G$ cannot be infinite.
So can someone help me filling the details of this proof (I've highlighted the two problematic steps) ? In particular, the second step is motivated by the fact that I have to find an absurd using infiniteness of the group, otherwise would be true. Maybe I've chosen the wrong family of closed, but anything else came to my mind
Thanks in advance
Let $G^0$ be the connected component of the identity. Since $G^0$ is a closed group of finite index in $G$, and the same hypothesis holds for $G^0$, we can assume without loss of generality that $G$ is connected.
In a linear algebraic group over an algebraically closed field of characteristic $0$, every element with finite order is semisimple. This can be argued using representation theory.
So by hypothesis, $G$ is semisimple. Now, every semisimple element of a connected linear algebraic group $G$ is contained in a maximal torus of $G$ (Springer, 6.45). A torus is a finite product of $k^{\ast}$s, the empty product being interpreted as the trivial group. It is impossible for a nontrivial torus to have all its elements have finite order.
We have to conclude that $G$ is the trivial group.