Let $G$ be a linear algebraic group of dimension one.
The proof that I am looking at, in t.a springer's book (thm 3.4.9) proceeds by showing that $G$ must be either equal to its semisimple part $G_s=\{g \in G | g \text{ is diagonalizable in some( and hence every ) embedding into GL} \}:=\{g \in G| g \text{ is semisimple } \}$
or its unipotent part =$\{g \in G | g \text{ is unipotent} \}$.
In the case where $G=G_s$ she mentions that $G_s$ can be embedded into the $D_n$ the group of $n-$ dimensional diagonal matrices.
I don't understand this because $G_s$ is not by hypothesis $\textbf{simultaneously}$ diagonalizable. One can't choose one $P$ that will diagonalize all of the elements in $G$. I would be grateful if someone could point out why this not a problem.
It seems that the missing link is that commuting diagonalizable elements are simultaneously diagonalizable, since then we may conclude by the fact that $G$ is commutative.
We may invoke the general Lie theorem about solvable Lie algebras, but the proof is quite elementary for commuting elements. So let $(x_i)_{i\in I}$ be a family of pairwise commuting elements of $End_K(V)$ for some finite-dimensional $K$-vector space $V$, such that each $x_i$ is diagonalizable. We want to show that the $x_i$ are simultaneously diagonalizable.
Then up to extracting a basis of $Vect(x_i)$, we may assume that $I$ is finite and prove the result by induction on $|I|$. If $|I|=1$, the result is trivial. If $|I|>1$, write $I=\{0\}\cup J$. then the $(x_j)_{j\in J}$ are simultaneously diagonalizable : let $V = \bigoplus_{k\in X} V_k$ be a decomposition such that all $x_j$ with $j\in J$ act as scalars on each $V_k$. Then since $x_0$ commutes with the $x_j$ it stabilizes the $V_k$ and its restriction to each $V_k$ is still diagonalizable, and we can write $V_k = \bigoplus_{l\in Y_k} V_{k,l}$ such that $x_0$ acts on each $V_{k,l}$ as a scalar. We conclude by the fact that $V = \bigoplus_{k\in X}\bigoplus_{l\in Y_k} V_{k,l}$ and all $x_i$ with $i\in I$ act as scalar on each $V_{k,l}$.