Picard groups and fundamental groups of connected algebraic groups Recently, I'm reading V. L. Popov's paper "Picard groups of homogeneous spaces of linear algebraic groups and one-dimensional homogeneous vector bundles" V. L. Popov, 1974, and I got confused about "Theorem 6" in that paper which says that
"Let $G$ be a connected linear algebraic group with radical $R$. Then $\mathrm{Pic}(G)$ is isomorphic to the fundamental group of the semisimple group $G/R$."
This theorem follows from "Theorem 3" and "Theorem 4" in that paper.
However, this seems to contradict the following example. Consider $G = \mathrm{GL}(n,\mathbb{C})$, whose radical $R$ is isomorphic to $\mathbb{G}_{m}$. Then the homogeneous space is $G/R = \mathrm{PGL}(n,\mathbb{C})$. We know that $\mathrm{Pic}(\mathrm{GL}(n,\mathbb{C}))$ is $0$, but $\pi_{1}(\mathrm{PGL}(n,\mathbb{C}))=\mathbb{Z}/n\mathbb{Z}$. Can anyone explain what I've understood incorrectly? 
 A: This is really just a long comment since I can't access the paper.
There definitely seems to be something wrong with the formula, but formulas that I do know are correct basically perfectly account for the missing factor.
Namely, here are two facts that I definitely know. Let me denote by $X^\ast(G)$ the character group $\text{Hom}(G,\mathbf{G}_m)$.

Fact 1: Let $\varphi:G'\to G$ be a map of connected algebraic groups with $\ker\varphi$ of multiplicative type (i.e. $\ker\varphi$ is geometrically a product of tori and roots of unity). Then, there is the following exact sequence:
  $$0\to X^\ast(G)\to X^\ast(G')\to X^\ast(\ker\varphi)\to \text{Pic}(G)\to\text{Pic}(G')\to 0$$

and

Fact 2: If $X^\ast(G)=0$ (and $G$ is a connected algebraic group) then $G$ has a universal cover, and $\text{Pic}(G)=X^\ast(\pi_1(G))$. 

So, now, suppose, for example, that $G$ is a reductive group. Then, of course, we know that $R(G)$ is a torus and so $\varphi:G\to G/R(G)$ has multiplicative kernel. So, applying Fact 1 we get
$$0\to X^\ast(G/R(G))\to X^\ast(G)\to X^\ast(R(G))\to \text{Pic}(G/R(G))\to \text{Pic}(G)\to 0$$
Now, since $G/R(G)$ is semi-simple, we have that $X^\ast(G/R(G))=0$ (indeed, the image of $G/R(G)$ in $\mathbf{G}_m$ would be connected, semisimple, and abelian--so trivial). Thus, the above really reduces to
$$0\to X^\ast(G)\to X^\ast(R(G))\to \text{Pic}(G/R(G))\to\text{Pic}(G)\to 0$$
Then, using Fact 2, and the just mentioned fact that $X^\ast(G/R(G))=0$, we have that $\text{Pic}(G/R(G))=X^\ast(\pi_1(G/R(G)))$. So, finally, our sequence looks like 
$$0\to X^\ast(G)\to X^\ast(R(G))\to X^\ast(\pi_1(G))\to \text{Pic}(G)\to 0$$
Now, let's run this for $G=\text{PGL}_n$ so that $R(G)=\mathbf{G}_m$, and $G/R(G)=\text{PGL}_n$. Then, evidently:
$$X^\ast(\text{GL}_n)=X^\ast(\text{GL}_n/D(\text{GL}_n))=X^\ast(\text{GL}_n/\text{SL}_n)=X^\ast(\mathbf{G}_m)=\mathbb{Z}$$
and
$$X^\ast(R(G))=X^\ast(\mathbf{G}_m)=\mathbb{Z}$$
Thus, it remains to find what $X^\ast(\pi_1(\text{PGL}_n))$ is. But, $\text{PGL}_n=\text{PSL}_n$ and $\text{SL}_n\to\text{PGL}_n$ is a central isogeny with kernel $\mu_n$. Since $\text{SL}_n$ is simply connected, this implies that $\pi_1(\text{PGL}_n)=\mu_n$. Thus, 
$$X^\ast(\pi_1(\text{PGL}_n))=X^\ast(\mu_n)=\mathbb{Z}/n\mathbb{Z}$$
Thus, we have, finally, the sequence
$$0\to \mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}\to \text{Pic}(\text{GL}_n)\to 0$$
which implies the desired result (that $\text{Pic}(\text{GL}_n)=0$) since the map $X^\ast(\text{GL}_n)\to X^\ast(\mathbf{G}_m)$ is multiplication by $n$ (since any $\varphi\in X^\ast(\text{GL}_n)$ factors through the determinant which, essentially, is multiplication by $n$ on $\mathbf{G}_m$: the composition $\mathbf{G}_m\xrightarrow{\approx}R(\text{GL}_n)\xrightarrow{\det}\mathbf{G}_m$ is multiplication by $n$).
In general, since $\text{Pic}(G/R(G))=X^\ast(\pi_1(G/R(G))$ and $\pi_1(G/R(G))$ is some finite abelian group, we have, non-canonically that $X^\ast(\pi_1(G/R(G))=\pi_1(G/R(G))$. Thus, the above analysis shows that we have a short exact sequence
$$0\to X^\ast(G)\to X^\ast(R(G))\to \pi_1(G/R(G)))\to \text{Pic}(G)\to 0$$
Thus, since $\pi_1(G/R(G))$ and $\text{Pic}(G)$ are finite groups, if they are isomorphic then the map $\pi_1(G/R(G))\to \text{Pic}(G)$ is an isomorphism (since they're of the same order and finite) and thus $X^\ast(G)\to X^\ast(R(G))$ is an isomorphism. Conversely, if $X^\ast(G)\to X^\ast(R(G))$ is an isomorphism, then evidently $\pi_1(G/R(G))\to \text{Pic}(G)$ is an ismorphism.
So, now note that the morphism $X^\ast(G/D(G))\to X^\ast(G)$ is evidently an isomorphism. Thus, we see that $X^\ast(G)\to X^\ast(R(G))$ is an isomorphism if and only if $X^\ast(G/D(G))\to X^\ast(R(G))$ is an isomorphism. But, note that since $G/D(G)$ and $R(G)$ are tori, this is equivalent to the statement that $R(G)\to G/D(G)$ is an isomorphism. But, $R(G)=Z(G)^\circ$ (we can ignore reduced subschemes because we're in characteristic $0$) and it's well known that $G=D(G)R(G)$ and thus $R(G)\to G/D(G)$ is surjective (which implies the claim above that $X^\ast(G)\to X^\ast(R(G))$ is injective) and it's an isomorphism if and only if $R(G)\cap D(G)$ is trivial. 
So, the upshot of all of this is the following:

Conclusion: There is a canonical surjection $\pi_1(G/R(G))\to \text{Pic}(G)$ which is an isomorphism if and only if $R(G)\cap D(G)$ is trivial. In fact, the kernel of this map has size $|R(G)\cap D(G)|$. In fact, since $\pi_1(G/R(G))$ and $\text{Pic}(G)$ are finite abelian groups, they are isomorphic if and only if $\pi_1(G/R(G))\to \text{Pic}(G)$ is an isomorphism. Thus, $\pi_1(G/R(G))$ is isomorphic to $\text{Pic}(G)$ if and only if $D(G)\cap R(G)$ is trivial.

I haven't thought too deeply about when $R(G)\cap D(G)$ is non-trivial--do you know an example in which $G$ is not already semisimple? Anyways, it seems that the stated theorem is wrong precisely because $R(\text{GL}_n)\cap D(\text{GL}_n)$ is non-trivial, and exactly order $n$ accounting for your $\mathbb{Z}/n\mathbb{Z}$ discrepancy. 
