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$\newcommand{\sc}{\mathrm{sc}}$All the groups below are algebraic groups over an algebraically closed field,

From Page $163$ of Malle and Testerman's book "Linear algebraic groups and finite groups of Lie type", we see that $\operatorname{SO}_n$ is a maximal subgroup of $\operatorname{SL}_n$ as algebraic groups.

On Page $165$, it states that "the maximal closed positive-dimensional subgroups of any homomorphic image $\phi(G_{\sc})$ of $G_{\sc}$ are images of the corresponding subgroups of $G_{\sc}$." (sc stands for simply connected)

Let $\rho$ be the natural homomorphism from $\operatorname{SL}_n$ to $\operatorname{PGL}_n$ by quotienting out the center of $\operatorname{SL}_n$.

When $n$ is odd, $\rho(\operatorname{SO}_n) =\operatorname{SO}_n\cong\operatorname{PSO}_n=\operatorname{PO}_n$ and $\operatorname{PO}_n$ is a maximal subgroup of $\operatorname{PGL}_n$ indeed.

When $n$ is even, $\rho(\operatorname{SO}_n) =\operatorname{PSO}_n.$ But $\operatorname{PSO}_n$ is an index $2$ subgroup of $\operatorname{PO}_n$ which is in $\operatorname{PGL}_n$. So $\operatorname{PSO}_n$ is not a maximal subgroup of $\operatorname{PGL}_n$ but it should be based on the info from P$165$... Is there any tricky subtlety (like the difference between algebraic groups and abstract groups) I've missed?

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  • $\begingroup$ The subgroups in class $\mathcal{C}_6$ are the normalizers of classical groups in $SL_n$. For $n$ even, the normalizer of $SO_n$ in $SL_n$ is $SO_n . 2$. So $SO_n$ is not maximal in $SL_n$, but $SO_n . 2$ is. And the image of $SO_n . 2$ is $PO_n$ in $PGL_n$. $\endgroup$ Commented Oct 5, 2023 at 2:39
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    $\begingroup$ For OP: The group ${\rm PGL}_n$ is a simple algebraic group, and so it has no index 2 subgroups. (You should consider points over your algebraically closed field of characteristic different from 2.) $\endgroup$ Commented Oct 5, 2023 at 15:42
  • $\begingroup$ @testaccount Thanks for the input. May I ask what the outside $2$ in the normalizer is? $\endgroup$ Commented Oct 6, 2023 at 1:38
  • $\begingroup$ @user488802: For example, a suitable scalar multiple of a reflection from $O_n = SO_n:2$. $\endgroup$ Commented Oct 6, 2023 at 3:56
  • $\begingroup$ But it has to be in $SL_n$ where $n$ is even... $\endgroup$ Commented Oct 6, 2023 at 6:51

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Let me try an answer, after the caveat that im totally out of my depth: i have never done any group theory and did not imagine how subtle and tedious it can get. I got my hands on Malle-Testerman's book [MT] and also looked up some articles. Some of the authors who worked on the structure of classical groups that i found are EB Dynkyn, RH Dye, O King, M Liebeck, G Seitz, D Testerman. And some articles that i looked at: Liebeck-Seitz which comprises a large part of chapter 18 of the book [MT], whence the OP quote comes, King 86, King 81.

I will assume $K$ is an algebraically closed field, and usually denote the $K$ points of a group variety $G$ by $G$. If you look at the literature much of the work on linear groups deals with finite groups, in particular non algebraically closed fields, so it's not always easy to find the result for the field you like. Regarding your example the presentation of table 18.2 in [MT] is ambiguous, but as testaccount points out in comment for $n$ even $SO_n(K)$ is never maximal in $SL_n(K)$: on page 162 the authors write: « The subgroup $H$ of $G = Cl(V)$ is obtained as the intersection with $G$ of the normalizer in $GL(V)$ of the group in the second column. » And so one must take the normalizer of $SO_n(K)$ in $GL_n(K)$ and intersect it with $SL_n$: this is what King calls the "special general orthonal group", $SGO_n=GO_n\cap SL_n$, with $GO_n$ the group of matrices $A$ such that $^tAA=\lambda I$ for some $\lambda\in K^*$ -writing groups in terms of matrices as usual. King calls this group "general orthogonal group" while [MT] call it conformal orthogonal group. I recommend being careful with terminology and notation, because for instance some authors will have conflicting choices -King uses "singular" for "isotropic" or "totally isotropic". Coming back to the maximal subgroup containing $SO_n$ in $SL_n$, for example over the complex numbers $\text{diag}(i,-i)\in GO_2\cap SL_2(\mathbb C)=SGO_2$ but $(\text{diag}(i,-i))^2=-I$ so $\text{diag}(i,-i)\not\in O_2\supset SO_2$, which shows that $SO_2$ is not maximal in $SL_2(\mathbb C)$. This extends to all even special orthogonal groups over $\mathbb C$ and over all algebraically closed fields $K$ of characteristic not 2, replacing $i\in\mathbb C$ by $\sqrt{-1}\in K$. Note that you would have recovered $SGO_n$ from $PO_n$, simply taking the inverse image: $SGO_n=\rho^{-1}PO_n$, so $PO_n=\rho(SGO_n)=PSGO_n$ which is also equal to $PGO_n$ in our situation where $K$ is algebraically closed.

A few more remarks though, the OP quotes only part of the relevant text, on page 164 the authors write: « The classification of the maximal positive-dimensional closed subgroups of the simple algebraic groups of types $A_n$, $B_n$, $C_n$ and $D_n$ follows from the above considerations. Let $G_{sc}$ be a simply connected simple algebraic group with the given root system. » In particular the quoted correspondence does not apply a priori to $\rho:SL_n\rightarrow PSL_n$, as $SL_n$ is not simple. And then to be honest the formulation « the maximal closed positive-dimensional subgroups of any homomorphic image $𝜙(𝐺_{sc})$ of $𝐺_{sc}$ are images of the corresponding subgroups of $G_{sc}$ » is not totally clear: what is "the corresponding subgroup" for a given maximal subgroup of $𝜙(𝐺_{sc})$. In fact $\phi$ induces a bijection between maximal subgroups if we further assume (unlike in [MT]) that it is not trivial, because $G_{sc}$ is simple and simply connected. For a general map of groups or algebraic groups $\phi:X\twoheadrightarrow \phi(X)=Y$ we can only prove that the image of a maximal subgroup $Z<X$ is not contained in any proper subgroup of $Y$, but it may be equal to $Y$, thus not maximal proper -and "proper" is always assumed in the definition of "maximal subgroup". For example $X=SL_3(3)<GL_3(3)$ is maximal closed -as the only maximal subgroup of $\mathbb F_3^*$ is trivial- but, with $\pi:GL_3\rightarrow PGL_3$ the projection, $\pi(X)=\pi(SL_3)=PSL_3=PGL_3$ -as $\mathbb (F_3^*)^3=\mathbb F_3^*$- and thus is not maximal in $PGL_n$ -$K$ is not algebraically closed here, and i think there is no maximal closed subgroup of $GL_n$ containing $SL_n$ if $K$ is ac. We can also take an abelian (trivial) example: $\pi:\mathbb Z/2\rightarrow 0$, then $(0)\subset\mathbb Z/2$ is maximal but its image is not. In the other direction: if we take inverse images, still with $\phi:X\twoheadrightarrow Y$, it always holds that $\phi^{-1}(Z)<X$ is maximal for $Z<Y$ maximal; and $\phi^{-1}(Z)$ is of course algebraic, closed, and positive dimensional if $Z$ is -and $\phi$ is a morphism of algebraic groups of course.

I think i'll stop here, hopefully without having dropped in too many mistakes -surely there are some. I began hinting that the topic is awfully complicated, but it is sort of hard to stop looking for further details after catching a first glimpse. I cannot imagine the suffering of people working on the classification of finite simple groups and related topics. Actually i think that the $\mathcal C_i$ notation used in [MT] was introduced by Michael Aschbacher in his study of finite groups of Lie type -i believe he goes up to $\mathcal C_8$ instead of stopping at $\mathcal C_6$ as in [MT].

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  • $\begingroup$ Appreciate very much your writing! question: If $G_{sc}$ is simple, and $SL_n$ is simply connected, why is it not simple? $\endgroup$ Commented Oct 6, 2023 at 19:40
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    $\begingroup$ [1/4] Hi @user488802. To begin, here's from wiki « Authors differ on whether a simple Lie group has to be connected, or on whether it is allowed to have a non-trivial center, or on whether $\mathbb{R}$ is a simple Lie group. » Next we can mention Lie's third theorem: equivalence between the category of real Lie algebras and that of simply connected Lie groups -i keep it real for simplicity. So a sc group can have arbitrary Lie algebra, thus arbitrarily many normal subgroups. Now we can look at the classification of simple real Lie algebras and take the corresponding simply connected Lie group. $\endgroup$ Commented Oct 6, 2023 at 22:32
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    $\begingroup$ [2/4] Sometimes, as above, we are asking that our Lie group be simple as a group. But it seems that the most common definition is looser: that its closed connected normal subgroups be trivial -this allows a disconnected center. With this definition the resulting simply connected Lie group has discrete center -cf encyclopediaofmath.org/wiki/Simply-connected_group - which is the only normal subgroup so we are done -we can quotient by the center, then we have an "almost" simply connected group. We can also start with simple Lie groups with trivial center and take their universal cover, $\endgroup$ Commented Oct 6, 2023 at 22:32
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    $\begingroup$ [3/4] as described well here: en.wikipedia.org/wiki/Simple_Lie_group , its center will be the fundamental group of the Lie group we started with. It turns out that for simple compact real form Lie algebras, and thus simple compact Lie groups, the only trivial fundamental groups are for the exceptional groups $E_8$, $F_4$, and $G_2$, but as you see the fundamental groups/centter of universal cover are never very big. I also think the fundamental group can be read off the diagrams, and Humphreys defines a related notion of fundamental group of a Dynkin diagram, $\endgroup$ Commented Oct 6, 2023 at 22:32
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    $\begingroup$ [4/4] but i have to read about it. In conclusion we see that simple groups have rather small fundamental groups. So there is a loose implication in this direction, but no converse in general -passing to complex numbers as base does not change this general observation, though it changes fundamental groups. Finally the wiki pages on Lie and algebraic groups are pretty well done, with many examples, explanations, and references -this cannot be said of all science topics. Again, i hope there aren't too many mistakes, and that i did not miss some fundamental fact. $\endgroup$ Commented Oct 6, 2023 at 22:38

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