I'd like to understand why Hom$(G,G_a)$ for $G$ of multiplicative type is trivial. Recall that $G$ is of multiplicative type if $G_{k_s}$ is diagonalizable, where $k_s$ is the splitting field of the field $k$ we are working in.
Hints? How do I use the information that $G$ is of multiplicative type here?
First notice that there is an isomorphism $$ \textrm{Hom}_k(G,G_a) \otimes_k k_s \cong \textrm{Hom}_{k_s}(G_{k_s},G_{{a}_{k_s}}) $$ so that if we can show that $\textrm_{k_s}(G_{k_s},G_{{a}_{k_s}})$ is trivial, then we will be done. Assume therefore that $k=k_s$. Then this question comes down to finding elements $a \in k[G]=:A$ such that $\Delta(a)=a \otimes 1+1 \otimes a$ and $\epsilon(a)=0$. Assume that such an $a\neq 0$ exists, $a=\sum_i k_i a_i$, $a_i$ group-like elements. Then $$ \sum_i k_i^2 a_i \otimes a_i=\sum_i k_i (a_i\otimes 1+1 \otimes a_i) $$ so that $a_i$ can only be $1$. Therefore $a=k_0$ for some $k_0 \in k$. But $\epsilon(a)=0$ and this forces $a=0$. It follows that there are no nontrivial homomorphism from $G$ to $G_a$ if $G$ is of multiplicative type.