Invertible functions of connected reductive linear algebraic groups Let $G$ be a connected reductive linear algebraic group over a field $k$, where $k$ is algebraically closed and characteristic $0$. The coordinate ring of $G$ is denoted $k[G]$.
Clearly $k^*\subseteq k[G]^*$ (unit groups), but my main question is:

For what $G$ do we have $k[G]^*\neq k^*$?

I know that when $G$ is a torus $\mathbb{G}_m^n$ we have lots of nontrivial units in $k[\mathbb{G}_m^n]=k[x_1^{\pm},\ldots,x_n^{\pm}]$.
For a general $G$ we have a finite cover $G'\to G$ where $G'=G^{ss}\times T$ with $G^{ss}$ a semisimple simply connected linear algebraic group and $T$ a torus (I'm getting this from a paper that I'm reading: The Cox ring of a spherical embedding by Giuliano Gagliardi). So when this torus is nontrivial, $G'$ will have lots of nontrivial units. But is this the only time this can happen? That is,

If $G$ is semisimple simply connected, do we have $k[G]^*=k^*$?

 A: This is sometimes referred to as the Rosenlicht unit theorem.

Proposition (Corollary to Rosenlicht's unit theorem, [1, Corollary 1.2]): Let $k$ be a field, $G$ a smooth connected group scheme over $k$, and $T$
a torus over $k$. If $f\colon G\to T$ is a morphism of $k$-schemes such
that $f(e)=1$, then $f$ is a group homomorphism.

From this you can easily deduce the following. For a reductive group $G$ we denote by $G^\mathrm{ab}$ its abelianization $G/G^\mathrm{der}$.
NB: Most of what I say below applies in the general case, but the reductive case is that of most interest.

Corollary 1: Let $G$ be a reductive (connected) group over a field $k$. Then, the natural map $G\to G^\mathrm{ab}$ induces an isomorphism
$\mathcal{O}(G^\mathrm{ab})^\times\to \mathcal{O}(G)^\times$ is an
isomorphism. In particular, if $G$ is semi-simple then
$\mathcal{O}(G)^\times\cong k^\times$.

Proof: Since $G\to G^\mathrm{ab}$ is faithfully flat the map $\mathcal{O}(G^\mathrm{ab})\to \mathcal{O}(G)$ is injective, and thus so is $\mathcal{O}(G^\mathrm{ab})^\times\to\mathcal{O}(G)^\times$. To show this map is surjective, let $f\in\mathcal{O}(G)^\times$. We may interpret $f$ as a morphism of $k$-schemes $f\colon G\to \mathbb{G}_{m,k}$. Set $c=f(e)\in k^\times=\mathbb{G}_{m,k}$. Then, composing $f$ with the left multiplication by $c^{-1}$ map $\ell_{c^{-1}}\colon \mathbb{G}_{m,k}\to \mathbb{G}_{m,k}$ we see that $\ell_{c^{-1}}\circ f\colon G\to \mathbb{G}_{m,k}$ is a morphism of $k$-schemes sending $e$ to $1$. By the above proposition we must have that $\ell_{c^{-1}}\circ f$ is a group homomorphism and thus admits a factorization $\ell_{c^{-1}}\circ f=g\circ p$ where $p\colon G\to G^\mathrm{ab}$ is the quotient map and $g\colon G^\mathrm{ab}\to \mathbb{G}_{m,k}$ is some homomorphism of group $k$-schemes. Thus, $f=(\ell_c\circ g)\circ p$, and as $\ell_c\circ g\colon G^\mathrm{ab}\to \mathbb{G}_{m,k}$ is a $k$-scheme homomorphism we see see that $f$ lies in the image of $p\colon \mathcal{O}(G^\mathrm{ab})^\times\to\mathcal{O}(G)$ as desired. $\blacksquare$
In fact, if you look at the proof a slightly stronger thing is proven. For a $k$-scheme $X$ let us set $RU(X):=\mathcal{O}(X)^\times/k^\times$ (the 'reduced units').

Corollary 2: Let $G$ be a reductive group over a field $k$. Then, the group $RU(G)$ is isomorphic to the $k$-character
group $X^\ast_k(G^\mathrm{ab}):=\mathrm{Hom}(G^\mathrm{ab},\mathbb{G}_{m,k})$ of $G^\mathrm{ab}$.

These results are not entirely obvious in practice.

Example: The units of $k[x,y,z,w]/(xy-zw-1)$ is $k^\times$.

(By the way, what's true is that for a general reductive group one may write $G$ as the quotient of a product of a torus and semi-simple simply connected group by a finite group).
References:
[1] http://math.stanford.edu/~conrad/papers/unitthm.pdf
