Why does the commutator subgroup of a unipotent algebraic group have smaller dimension? Suppose $U$ is a unipotent linear algebraic group. Is there an explanation why the commutator subgroup $[U,U]$ has strictly smaller dimension, or at least why it is a proper subgroup? 
This fact is used in some induction arguments on the dimension of the abelianization, so I'm curious why it is true.
 A: If $U$ is a linear algebraic group, then you can embed it at a subgroup of $\operatorname{GL}_n$. Furthermore, if $U$ is unipotent, then $U$ can be embedded as a subgroup of $U_n\subseteq\operatorname{GL}_n$, where $U_n$ denotes the group of all upper triangular unipotent matrices, i.e. upper triangular matrices with $1$ on the diagonal.
Since $U_n$ is solvable, so is any subgroup of it. In particular, any unipotent linear algebraic group is solvable, and therefore $[U,U]$ must be a proper subgroup of $U$. 
From now on, we are in characteristic zero. I do not know if the following holds otherwise.
Since $U$ is connected (see below), $U/[U,U]$ is also connected. Therefore, $U/[U,U]$ can not be a finite group, otherwise $[U,U]$ would not be a proper subgroup of $U$. This implies that $[U,U]$ actually has strictly smaller dimension than $U$. 
Connectedness: Let $u\in U$, then there is a nilpotent matirx $x$ with $u=\exp(x)$.
For $t\in\Bbbk$, you have $\exp(tx)=\exp(x)^t=u^t\in U$. Since $\Bbb Z$ is dense in $\Bbbk$, this implies that $\exp(tx)\in U$ holds for all $t\in\Bbbk$. This gives you a path connecting $u$ to the identity matrix.
