Let $k$ be an arbitrary field with $\operatorname{char}(k) \neq 2$. Let $G$ be a linear algebraic group over $k$. Let $X$ be the conjugacy class of a semisimple element $s \in G(k)$ of order 2 (or a union of such sets).

Can we say something about the subgroup of $G(k)$ generated by $X$? Is it closed under the Zariski topology? Does it make sense to speak of the algebraic subgroup generated by X?

A few facts that might be relevant:

Theorem 1.45 of Milne, Algebraic Groups, tells us that if $\langle X \rangle$ is a closed subgroup of $G(k)$, there exists a unique reduced algebraic subgroup $H$ of $G$ such that $\langle X \rangle = H(k)$ and $H$ is geometrically reduced.

If $k$ is algebraically closed, $X$ itself is closed. See 18.2 in Humphreys, Linear Algebraic Groups.

In the case where I am particular interested in, $G$ is the automorphism group of a (not necessarily associative) algebra and $X$ corresponds to idempotents of the algebra.

  • $\begingroup$ You can certainly speak of the algebraic subgroup generated by $X$. It is the intersection of all closed subgroups containing $X$. $\endgroup$
    – S. Carnahan
    Commented Jan 31, 2018 at 14:05
  • $\begingroup$ If $s$ belongs to the subgroup of $G(k)$ generated by unipotents and the latter is not equal to $G(k)$ then the answer is no. For instance, $PGL_2$ over any field in which some element is not a square yields counterexamples. $\endgroup$
    – YCor
    Commented Jan 31, 2018 at 17:28
  • $\begingroup$ @S.Carnahan Is there nothing more to it? Is an algebraic subgroup really the same as a closed subgroup? Or do you use something like the Theorem 1.45 that I mentioned in my question? $\endgroup$ Commented Feb 1, 2018 at 9:45
  • $\begingroup$ @YCor Do you mean that in that case it will never be true of that there exist counterexamples? Can you be more precise with your counterexample in $PGL_2$? I find it hard to see whether a subgroup is closed but that is probably because I am not aware of some relevant Theorems... Probably you want $X$ to generate $PSL_2$ or so which is not closed? $\endgroup$ Commented Feb 1, 2018 at 9:49
  • $\begingroup$ Per @YCor's answer, you should probably specify that you're speaking of the $G(k)$- (i.e., rational) conjugacy class of $s$, not its algebraic conjugacy class. $\endgroup$
    – LSpice
    Commented Aug 3, 2018 at 12:45

1 Answer 1


We can assume that the $G(k)$-conjugacy class of $s$ generates a Zariski-dense subgroup $Q$ of $G$.

Let me assume that $k$ has characteristic zero and provide consequences. So $G=U\rtimes S$ for the unipotent radical $U$, which is defined over $k$, and some $k$-defined reductive subgroup $S$, and $G(k)=U(k)\rtimes S(k)$.

I claim that $Q$ contains $U(k)$.

If $U$ is abelian, $Q\cap U(k)$ contains all commutators $quq^{-1}u^{-1}$ ($q\in Q$, $u\in U(k)$), which can be rewritten as additively $q\cdot u-u$ and such element form a $k$-linear subspace, which is the whole of $U$, because $G$ is generated by unipotent elements. In general, this applies to $G/[U,U]$ and one deduces that $Q\cap U(k)$ has a surjective image in $[U(k),U(k)]=[U,U](k)$ and hence a standard lemma about nilpotent groups implies $Q\cap U(k)=U(k)$.

This completely reduces the question (in char 0) to reductive groups.

In the abelian case there is not much to say: a conjugacy class is a singleton. So mostly things amount to understand the semisimple case.

Then in the isotropic case we can get somewhat fine results. Let $Z$ be the largest finite normal subgroup of $G$, that is, the centralizer of the unit component $G^0$. Let $G(k)^+$ be the subgroup of $G(k)$ generated by unipotent elements.

Suppose that $G$ is connected, $k$-isotropic and $k$-simple. It is known (I think it's in Margulis' book) that every normal subgroup of $G(k)$ not contained in $Z$ contains $G(k)^+$. [This extends immediately to the case when $G$ is semisimple, $k$-isotropic, $G(k)$ is Zariski-dense (i.e., meets all connected components), and $G/G^0$ acts transitively on minimal nontrivial $k$-defined semisimple subgroups (these latter assumptions implies that they are all $k$-isotropic). Typical example: the wreath product $S\wr\mathrm{Sym}_n=S^n\rtimes\mathrm{Sym}_n$, with $S$ $k$-simple $k$-isotropic.]

Example: $G=\mathrm{PGL}_2(k)$ (here $\mathrm{PGL}_2$ is connected, simple, trivial center). Then the subgroup generated by the $G(k)$-conjugacy class of $s\in G(k)\smallsetminus\{1\}$ is equal to the subgroup of those $t\in\mathrm{PGL}_2(k)$ such that $\overline{\det}(t)\in\{1,\overline{\det}(s)\}$, where $\overline{\det}$ is the determinant viewed as function into $k^\ast/(k^\ast)^2$, which is well-defined as homomorphism $\mathrm{PGL}_2(k):k^\ast/(k^\ast)^2$.

The anisotropic case is far worse-understood. Still in the real case it behaves well and even better (in the above setting with "isotropic" replaced by "anisotropic", we automatically have $Q$ Zariski-closed.). In the $p$-adic case, for $G$ semisimple anisotropic, $G(k)$ is profinite and thus has no "minimal" normal subgroup; anyway some easy $p$-adic analysis probably ensures that $Q$ is an open subgroup, which is a reasonable conclusion.


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