Let $G$ be a diagonalizable algebraic group, and $G_{n}$ denotes the subgroup of elements of order dividing $n$. Assuming $k$ is algebraically closed and has characteristic $p$ and $p$ does not divide $n$. Here $H^{\perp}$ denotes elements in $X(G)$ such that $X(h)=1,\forall h\in H$. I want to show:
1) $G_{n}^{\perp}=nX$
2) The subgroup of elements of finite order is dense in $G$.
I have some ideas but both questions remained unsolved. If I use the structure theorem on $G$, this reduces to the case $G=G_{m}$ or $G=\bf Z_{n_{j}}$ with $n_{j}\in \bf Z,gcd(n_{j},p)=1$. It is trivial that for any $\chi\in nX$, $\chi(G_{n})=1$. But it is not clear to me how to show if $\chi(g)=1,\forall g\in G_{n}$, then $\chi=nX$.
Similarly I do not really know how to solve the second problem, where the subgroup of finite order has infinite cardinality, but this alone does not help me to assert it is dense in $G$.
