# Why is the semisimple quotient of a reductive group with semisimple rank 1 equal to PGL2?

Lets work over $\mathbb{C}$. Let $G$ be a reductive group with semisimple rank 1 (this means that a maximal torus in $G / R(G)$ has dimension 1). In section 25 of Humphreys book "linear algebraic groups", it is claimed that $G / Z(G)$ is isomorphic to ${\rm PGL}_2$, but I don't understand the proof. Let me be a little more precise. Write $I$ for the intersection of all borel subgroups of $G$. I understand why $G / I \cong {\rm PGL}_2$ and why $I / Z(G)$ is finite. I just don't understand how Humphreys concludes that $I = Z(G)$.

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I don't know what Humphreys's proof is, but this is certainly a consequence of the classification of adjoint semsimple groups (over algebraically closed fields) by Dynkin diagrams: There is only one Dynkin diagram with a single vertex. –  Keerthi Madapusi Pera Feb 25 '14 at 23:45
@KeerthiMadapusiPera: I think your suggestion is circular, since the classification with Dynkin diagrams (which, oddly enough, is not discussed in Borel's textbook) requires a huge amount of structure theory, the entire foundation of which rests on knowing the classification in semisimple-rank 1 to get a root system... –  user76758 Feb 26 '14 at 3:11
That's certainly possible! I'm no expert on the classification. –  Keerthi Madapusi Pera Feb 26 '14 at 3:15
@OP: Now that I've read your question more carefully, I see that what you're really asking is: Why is the identity component of the intersection of the Borels the radical? This is because all Borels are conjugate, which means that their intersection is a normal solvable sub-group of $G$. But then its connected component must be contained in the radical! –  Keerthi Madapusi Pera Feb 26 '14 at 3:30
@KeerthiMadapusiPera: Is $I$ defined as the identity component of the intersection? I think one has to control the entire intersection, and its centrality in the entirety of $G$ has to be proved. The "problem" is that the good behavior of centrality under quotients, a familiar feature of connected reductive groups, is not true for connected linear algebraic groups more generally, so at this early stage in the theory some care is needed for handling centrality claims. –  user76758 Feb 26 '14 at 3:37

Your question comes down to knowing that if $G$ (your $G/Z(G)$) is a smooth connected affine extension of ${\rm{PGL}}_2$ by a finite group then the quotient map $q:G \rightarrow {\rm{PGL}}_2$ is either an isomorphism or the central extension ${\rm{SL}}_2$ by $\mu_2$. (In the latter case, the full kernel from $G$ onto the ${\rm{PGL}}_2$ quotient would then be a subgroup $H$ that is an extension of the cyclic $\mu_2$ by the central $Z(G)$, so $H$ would be visibly commutative and hence of multiplicative type, thus central in the connected $G$, a contradiction.)

Observe that $\ker q$ is finite and normal in the connected $G$, so it is central (as you're in char. 0). Such a $G$ must be reductive, even semisimple (definitions via vanishing of unipotent radical and radical, not via representation theory), since ${\rm{PGL}}_2$ is visibly semisimple. Thus, every maximal torus in $G$ is its own centralizer (!) and hence contains the central $\ker q$. These maximal tori are 1-dimensional, so $\ker q$ is a finite subgroup of a 1-dimensional torus and hence is $\mu_n$ ($n$th roots of unity) for some $n \ge 1$. Hence, for any maximal torus $T$ in $G$, $\ker q$ is exactly its unique subgroup $T[n]$ of order $n$ (namely, the group $\mu_n$ of $n$th roots of unity).

The image of $T$ in the isogenous quotient ${\rm{PGL}}_2$ is a maximal torus $D$ in the latter. By conjugacy of maximal tori in ${\rm{PGL}}_2$ and inspection for the diagonal torus, there is $w \in {\rm{PGL}}_2$ that normalizes $D$ with $w$-conjugation acting on $D$ via inversion. Since $T$ is the preimage of its image $D$ (as the central $\ker q$ is contained in all maximal tori of $G$), any $g \in G$ lifting $w$ must normalize the 1-dimensional $T$ with $g$-conjugation acting on $T$ via its unique nontrivial automorphism, namely via inversion. Ah, but the subgroup $\ker q = T[n]$ is central in $G$, so $g$-conjugation is trivial on this. In other words, inversion (on $T$) has trivial effect on $T[n]$, forcing $n \le 2$. Voila, so either $q$ is an isomorphism or it is a central extension of degree 2.

Now we can go a bit further and show that if $n = 2$ then $G = {\rm{SL}}_2$ (as a central extension of ${\rm{PGL}}_2$). Namely, consider the pullback $$\widetilde{G} = G \times_{{\rm{PGL}}_2} {\rm{SL}}_2,$$ or more specifically its identity component. This is an isogenous smooth connected affine cover of ${\rm{PGL}}_2$ that has degree at least 2 and yet dominates the degree-2 cover ${\rm{SL}}_2$. The preceding shows that its degree over ${\rm{PGL}}_2$ is at most 2, whence exactly 2, so its covering map onto ${\rm{SL}}_2$ is degree 1 (i.e., an isomorphism).

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This is such a great answer! I just worked through all the details and everything works out. Thanks! –  Daniel Barter Feb 26 '14 at 19:54