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Let $k$ be a $p$-adic field and $G$ be a connected non-abelian reductive algebraic group over $k$. I am asking for a proof of the following lemma:

Lemma. Assuming that $p$ is "good" for $G$, there exist two elements of finite orders $x_1,x_2\in G(k)$ such that $x_1x_2\neq x_2 x_1$.

Motivation. For $i=1,2$, write $n_i={\rm ord}(x_i)$ and consider the homomorphisms $$ c_i\colon {\rm Gal}(\bar k/k)\to G(k) $$ that factor as follows: $$ {\rm Gal}(\bar k/k) \twoheadrightarrow \widehat {\Bbb Z} \twoheadrightarrow {\Bbb Z}/n_i {\Bbb Z}\overset\sim\longrightarrow\langle x_i \rangle\hookrightarrow G(k).$$ Then $c_1,c_2\in Z^1(k,G)$: they are 1-cocycles, but $c:=c_1c_2$ is not a 1-cocycle because $x_1x_2\neq x_2 x_1$. Based on this fact, one can conjecture that the set $H^1(k,G)$ has no natural group structure. However, this conjecture is false. An abelian group structure on $H^1(k,G)$ is given by Theorem 1.2 of Robert E. Kottwitz, Stable trace formula: elliptic singular terms, Math. Ann. 275(1986), no.3, 365–399. Moreover, this abelian group structure is functorial in $G$ and $k$ (Section 6 of Borovoi and Kaletha, Galois cohomology of reductive groups over global fields, arXiv:2303.04120 - sorry for self-promotion!)

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    $\begingroup$ Is the group split? In that case, you can use representatives of non-commuting elements of the Weyl group. $\endgroup$
    – Jason Starr
    Commented Oct 15 at 12:45
  • $\begingroup$ @JasonStarr: Thank you, this helps. What about a non-split group? $\endgroup$
    – Mikhail Borovoi
    Commented Oct 15 at 13:10
  • $\begingroup$ If $G$ is an isotropic semisimple group, then the question can be easily reduced to the cases $G={\rm SL}_2$ and $G={\rm PGL}_2$. These cases can be done "by hand". Namely, let $T\subset G$ be a split torus. Then $T(k)\simeq k^\times$. Write $\kappa$ for the residue field of $k$. Then $\kappa\simeq {\Bbb F}_q$ for some $q=p^l$, and $T(k)\simeq k^\times$ has an element $x_1$ of order $q-1$. Assume that $q\ge 7$; then the centralizer of $x_1$ in $G$ is $T$. Choose $g\in G(k)\smallsetminus N_G(T)(k)$ and set $x_2=gx_1 g^{-1}$. Then $x_2\notin T(k)$, whence $x_2$ does not commute with $x_1$. $\endgroup$
    – Mikhail Borovoi
    Commented Oct 15 at 13:30
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    $\begingroup$ An anisotropic group "essentially" comes from a division algebra. If it's $D^\times/k^\times$, then a suitable depth-$0$ element and the image of a uniformiser will suffice for your elements. I'm not sure what to take if your group is the group of norm-$1$ elements $D^1$, and wonder vaguely if there might not exist such elements in that case. $\endgroup$
    – LSpice
    Commented Oct 15 at 13:47
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    $\begingroup$ In fact, I think that, if $p \ne 2$ and $D/k$ is a quaternionic division algebra, then the only finite-order elements of $D^1$ are the Teichmüller lifts of elements of $\kappa_2^1$, where $\kappa$ is the residue field of $k$ and $\kappa_2^1$ is the group of norm-$1$ elements in its quadratic extension. $\endgroup$
    – LSpice
    Commented Oct 15 at 14:04

2 Answers 2

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Let $G$ be a group where any two elements of finite order commute with each other. It follows that for $x$ an element of finite order, any conjugate of $x$ commutes with any other conjugate of $x$. Hence the subgroup generated by conjugates of $x$ is abelian and normal. It therefore is contained in the center of $G$. (This is probably well-known, but can be checked like this: Its Zariski closure is also abelian and normal. Being a normal Zariski closed subgroup of a reductive group, its identity component is reductive, and being abelian, its identity component is a torus. The automorphism group of an algebraic group whose identity component is a torus is discrete, so the action of a connected reductive group on this group by conjugation is trivial, hence it is central, so the original group is central as well.)

So an equivalent formulation of the question is that there exists an element of finite order not contained in the center of $G$.

Based on this I think LSpice's example in the comments has opposite valence: These Teichmuller lifts are finite-order and not contained in the center, hence they have conjugates which don't commute with them. Based on the reduction he sketches it seems the answer to your question is positive but I lack the expertise to completely follow the logic of the reduction.

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    $\begingroup$ This requires just the additional note that a subgroup that's normalised by $G(k)$ is normal in $G$, since $G(k)$ is Zariski dense in $G$. I otherwise agree with your argument, and yet struggle to reconcile this with a fact that I thought I knew, that every element of a central division algebra $D/k$ can be written as a sum $\sum_{i = v}^\infty \varepsilon_i\varpi^i$, where each $\varepsilon_i$ comes from $\mu_{p'}(L)$, $L/k$ a maximal unramified extension inside $D$, and $\varpi$ belongs to $\operatorname N_D(L) \setminus L$. $\endgroup$
    – LSpice
    Commented Oct 15 at 17:32
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    $\begingroup$ @LSpice Ah, good point about the Zariski density being key. Isn't the point that such a series can be conjugate to the Teichmuller lift $\varepsilon_0$ without being equal to $\varepsilon_0$? $\endgroup$
    – Will Sawin
    Commented Oct 15 at 17:35
  • $\begingroup$ Re, yes: I was arguing to myself that $x = \sum_{i = v} \varepsilon_i\varpi^i$ having finite order obviously implies $v = 0$, and then it having prime-to-$p$ order $N$ implies that $1 = x^N \equiv \varepsilon_0^N + N\varepsilon_{i_1}\varpi^{i_1} \pmod{P_D^{i_1 + 1}}$, where $i_1$ is the first positive index with $\varepsilon_{i_1} \ne 0$; but of course that would only work if $\varepsilon_{i_1}$ and $\varpi^{i_1}$ commuted, which they need not. $\endgroup$
    – LSpice
    Commented Oct 15 at 17:38
  • $\begingroup$ @WillSawin: Thank you, this is very helpful! $\endgroup$
    – Mikhail Borovoi
    Commented Oct 15 at 19:17
  • $\begingroup$ Comically, when browsing my old answers looking for a different argument, I found that I had already explicitly described something that can easily be promoted to a counterexample to my implicit belief that the maximal unramified extension of $k$ inside $D$ sits there uniquely. $\endgroup$
    – LSpice
    Commented Nov 26 at 20:01
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Here I give details of the reduction in LSpice's comment. I write it as an answer rather than a string of comments in order to have an editable text.

The reduction goes as follows. According to Will's answer, we must prove that for our non-abelian connected reductive $k$-group $G$, there is a non-central element of finite order in $G(k)$. We may and shall assume that $G$ is $k$-simple. It suffices to consider the case where $G$ is simply connected. Thus we may and shall assume that $G$ is absolutely simple.

If $G$ is isotropic, then the problem reduces to the case ${\rm SL}_2$, which is easy. If $G$ is an anisotropic simply connected absolutely simple group over a $p$-adic field $k$, then by a theorem of Kneser and of Bruhat and Tits, $G$ is isomorphic to ${\rm SL}(1,D)$ for some central division algebra $D$ over $k$. We conclude by Will's argument This gives a positive answer to my question.

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  • $\begingroup$ @WillSawin: Please edit to your taste! $\endgroup$
    – Mikhail Borovoi
    Commented Oct 15 at 19:14
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    $\begingroup$ Do you care about small-(residual-)characteristic issues, such as were avoided by assuming $q \ge 7$ in your comment? $\endgroup$
    – LSpice
    Commented Oct 15 at 19:59
  • $\begingroup$ @LSpice: Not really. However, with Will's answer, it seems that all works over a $p$-adic field $k$ with residual characteristic $p>2$. $\endgroup$
    – Mikhail Borovoi
    Commented Oct 16 at 5:19

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