Does every linear group admit a subgroup of dimension 1? 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 $\mathbb{G}_m$)
 A: Over an algebraically closed field, the answer is yes. I assume that by "linear group" you mean a smooth affine algebraic group scheme. Such a group has a filtration whose quotients (modulo finite group schemes) are successively a unipotent group, a torus, and a semisimple group, so it is only a question of checking each of these cases.
A: Here is a counter-example for $k$ not algebraically closed (for simplicitly I assume that $k$ is perfect). Recall that there is an anti-equivalence of categories  between the category of algebraic tori over $k$ and the category of free $\mathbb{Z}$-modules with a continuous action of $Gal(\overline{k}/k)$ (given by the character group $\widehat{T}$).
Choose an algebraic torus $T$ of dimension larger than one, whose character group $\widehat{T}$ is simple as a Galois module (these exist if e.g. $k=\mathbb{Q}$). Suppose that $T$ contains a one dimensional irreducible algebraic subgroup $C$. Then it is well known that such a group is isomorphic to either $\mathbb{G}_a,\mathbb{G}_m$ or an elliptic curve over $\overline{k}$. As $T$ is affine and every element is semi-simple, we see that $C$ is itself an algebraic torus. The inclusion $C \subset T$ therefore induces a non-trivial homorphism $\widehat{T} \to \widehat{C}$. The kernel of this homorphism is a Galois sub-module of $\widehat{T}$. However by assumption $\widehat{T}$ was simple, thus we deduce that $\widehat{C} = 0$ or $\widehat{T}$, which is a contradiction.
Conclusion: Such a torus $T$ cannot contain a one dimensional algebraic subgroup.
Edit: As Marguax points out I do in fact need $\widehat{T} \otimes_{\mathbb{Z}} \mathbb{Q}$ to be simple as a Galois module. I believe that such examples may also be given be given by the norm one torus
$$R^1_{K/k} \mathbb{G}_m : N_{K/k}(x_1,\ldots,x_n) = 1.$$
Here $k \subset K$ is a cyclic Galois field extension of prime degree $n > 2$ and $N_{K/k}$ denotes the norm form for $K/k$. This may be defined by choosing an isomorphism $K \cong k^n$ and considering the usual field norm as a homogeneous polynomial of degree $n$ on $k^n$.
A: 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, $\mathbb{G}_a \subseteq Z \subseteq U \subseteq G$.
