Suppose that $G$ is a linear group of positive dimension, defined over some field $k$. Is that true, that $G$ admits a (closed) one-dimensional subgroup?


I'm pretty much sure this is true in characteristic 0, or at least for $k=\mathbb{C}$. It seems that the main obstacle in positive characteristic is that there may not exist elements of infinite order.


**EDIT1:** As @Daniel Loughran pointed out in the answer below, one need at least to assume that $k$ is algebraically closed
(or it may not even be necessary, see the comments of @Marguax abaut $k$ being separably closed, or real closed in characteristic 0).


**EDIT2:** Since my first statement of the problem seems to be genereting a lot of confusion (after getting a comment from @Jim Humphreys I'm even a little ashamed), here is the final version of the question:


*Suppose that $G$ is a linear algebraic group, of positive dimension, defined over an algebraically closed field $k$ (but of arbitrary characteristic). Is it true, that $G$ posses a one dimensional (closed) subgroup? (so in fact, either $\mathbb{G}\_a$ or $\mathbb{G}\_m$)*



Does every linear group admit a subgroup of dimension 1?
Every positive dimensional linear algebraic group $G$ over an algebraically closed field has a one dimensional subgroup. 


**Case 1** $G$ is reductive. In that case, $G$ contains a torus $T$. Since we are over an algebraically closed field, $T \cong \mathbb{G}\_m^r$ for some $r>0$. In particular, $\mathbb{G}\_m \subseteq T \subseteq G$.


**Case 2** $G$ has a nontrivial unipotent radical $U$. In turn, $U$ has a nontrivial center $Z$. We have $Z \cong \mathbb{G}\_a^s$ for some $s>0$ (not quite right, see Peter McNamara's correction below). In particular, $\mathbb{G}\_a \subseteq Z \subseteq U \subseteq G$.


