Theorem. Let $G$ be a connected affine algebraic group, let ${\mathcal R}_u(G)$ be its unipotent radical, let $\varrho\colon G\to G/{\mathcal R}_u(G)$ be the canonical homomorphism, let $S$ be the derived subgroup of the connected reductive group $G/{\mathcal R}_u(G)$, and let $\iota\colon S\hookrightarrow G/{\mathscr R}_u(G)$ be the identity embedding. Then the following canonical homomorphisms are isomorphisms:

Corollary. The group $\operatorname{Pic}(G)$ is canonically isomorphic to the group $\operatorname{Hom}_{\rm alg}(\pi(S),\mathbb G_m)$ and is non-canonically isomorphic to the group $\pi(S)$.

Example. Let $G=\operatorname{GL}_n$. Then the group ${\mathcal R}_u(G)$ is trivial, and the derived group of the group $G$ is the semisimple group $\operatorname{SL}_n$. The latter is simply connected, so the group $\pi(\operatorname{SL}_n)$ is trivial. Therefore, by the theorem, the group $\operatorname{Pic}(G)$ is trivial. This agrees with the fact that the group variety of the group $\operatorname{GL}_n$ is isomorphic to an open subset of $\mathbb A^{n^2}$.

Lemma. Let $X$ be an irreducible smooth algebraic variety, let $U$ be a nonempty open subset of $\mathbb A^{d}$, and let $u_0$ be a point of $U$. Then for the morphisms

(a) the group variety of ${\mathcal R}_u(G)$ is isomorphic to an affine space;

(b) the following commutative diagram exists:

$(4)$

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In view of the lemma it follows from (a) and (4) that $\varrho^*$ is an isomorphism.

According to [6], Theorem 1, the group $G/{\mathcal R}_u(G)$ contains a torus $Z$ such that the mapping

$(5)$

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In view of the semisimplicity of the group $\widehat S$ the group $\operatorname{Hom}_{\rm alg}(\widehat S,\mathbb G_m)$ is trivial, and since the group $\widehat S$ is simply connected, the group $\operatorname{Pic}(\widehat S)$ is trivial (see [2], Proposition 1). According to [2], Theorem 4, it follows from this that $\varepsilon_{\widehat{S}, \pi(S)}$ is an isomorphism. This completes the proof of the theorem.

The first part of the corollary follows directly from the theorem, while the second part follows from the fact that $\pi (S)$ is a finite abelian group. $\Box$

Remark. The above theorem corrects Theorem 6 in [2]. The latter asserts that the group $\operatorname{Pic}(G)$ is isomorphic to $\pi\bigl(G/{\mathcal R}(G)\bigr)$, where ${\mathcal R}(G)$ is the solvable radical of $G$. If the group extension $1\to {\mathcal R}(G)\to G\to G/{\mathcal R}(G)\to 1$ splits, then the group $G/{\mathcal R}(G)$ is isomorphic to the derived group of $G/{\mathcal R}_u(G)$, and so the formulated assertion is true in view of the above theorem. But in general this is not the case, as the example above shows: in it the group $G/{\mathcal R}(G)$ is isomorphic to $\operatorname{PGL}_n$, and $\pi(\operatorname{PGL}_n)$ is isomorphic to the group of all $n$th roots of $1$ in the field $k$. This latter group is non-trivial if $n$ is not a power of the characteristic of the field $k$ (however, the group $\operatorname{Pic}(G)$ is trivial for every $n$).