Let $q=p^e$ be a prime power and $\mathbb F_q$ the finite field with $q$ elements. Let $k$ be a field of characteristic $p$ and let $\varphi: SL_2(\mathbb F_q) \to GL_n(k)$ be a representation.

Is it true that $\varphi$ maps a diagonal matrix $D$ to a matrix $\varphi(D)$ which can be diagonalized over $k$?

(The question in the title more or less boils down to this question.)

Background: Since $\mathbb F_q$ is finite, $\varphi$ can be described by a polynomial map $f$ which is defined over $k$. Then $f$ induces a map $SL_2({\bar k}) \to GL_n({\bar k})$, where $\bar k$ is an algebraic closure of $k$. Suppose that the induced map actually is a morphism of algebraic groups. As the diagonal matrices are in particular a $k-$split torus and the map is defined over $k$, the image again will be a $k-$split torus, from which the claim would follow.

But to all appearances the induced map will not be a group homomorphism, so I might as well have asked about the smallest counterexample...

  • $\begingroup$ If the representation lifts to characteristic zero, the answer is "yes", because the group is finite and every finite order matrix in char 0 is semisimple. $\endgroup$ May 20, 2010 at 11:04
  • $\begingroup$ Thank you for your answer. I don't see, though, where you used the specific group $SL_2$. For example, a cyclic group has a representation over the reals which is not diagonalizable over the reals. $\endgroup$
    – Guntram
    May 20, 2010 at 11:29
  • $\begingroup$ Sorry, I was carried away by your algebraic group language - I meant "semisimple". Indeed, semisimple matrices aren't diagonalizable in general. Nonetheless, an element of order dividing $q-1$ will a fortiori be diagonalizable over any field containing $\mathbb{F}_q$ (and if it doesn't, it cannot possibly be true), by a root of unity argument. But it looks like Jim has already answered your question without the lifting assumption. $\endgroup$ May 20, 2010 at 11:54
  • $\begingroup$ As with Jim's answer also the comment I added there vanished, I'll repeat the simple counterexample here: take the embedding $SL_2(q) \to GL_{2e}(p)$ by simply considering a $2$-dimensional $F_q$-vector space as $2e$-dimensional $F_p$-vector space. $\endgroup$
    – Someone
    May 21, 2010 at 8:29
  • $\begingroup$ OK, but this doesn't contradict Xandi, right? A representation preserves diagonalizable iff $k$ contains $\mathbb F_q$. $\endgroup$
    – Guntram
    May 21, 2010 at 9:07

2 Answers 2


This is an edited version of an earlier hasty try, which went off the tracks and was removed. As both Xandi Tuni and Someone have made clear in their comments, the answer to the stated question is "no": even if the representation is otherwise well-behaved, it can easily take a diagonal matrix to one which is not diagonalizable over too small a field (but is semisimple and therefore does diagonalize over an extension field).

The question gets a little confusing when algebraic group language is added. But on the level of finite groups in nonzero characteristic, it's useful to keep in mind that Jordan decomposition in the algebraic group sense has intrinsic meaning: given a prime $p$, an element of a finite group decomposes uniquely as a product of two commuting elements, one of $p$-power order and the other of order prime to $p$. This decomposition is preserved under any finite group homomorphism. So for matrix groups a semisimple element is sent to a semisimple element, but a larger field may be needed to diagonalize such a matrix.

In any case, the initial question concerns only finite groups of Lie type. Just the case when the representation is nontrivial needs to be considered.

EDITED (with more details, as suggested in comments):

(1) For the given group $G =$ SL$_2(q)$, whose order is $q(q-1)(q+1)$, the matrices which are already diagonal have orders dividing $q-1$ (while other semisimple elements have orders dividing $q+1$ and can be diagonalized over a quadratic extension of $\mathbb{F}_q$). On the other hand, when $q>3$ the group $G$ is simple if $p=2$ and simple modulo a center of order 2 if $p$ is odd, while it has an element of order $q-1$. Suppose the representation takes such an element to a matrix which is diagonalizable over $k$. This matrix has order at least $(q-1)/2$, since the representation is nontrivial and $G$ is almost simple. Using the easy inequality $(p^e−1)/2 > p^{e-1}−1$, it follows that $k$ includes the $(q-1)$st roots of unity and hence includes $\mathbb{F}_q$ (with $q=p^e$).

(2) For the converse $G$ can be arbitrary. The image of an element of order dividing $q-1$ still has order dividing $q-1$. So it can be diagonalized over $k$ provided $k$ includes $\mathbb{F}_q$ (and thus contains all roots of unity of order dividing $q-1$).

All of this uses the standard characterization of subfields of a given finite field. Xandi's argument involves some of the same ideas, but fails to use the fact that the ambient group $G$ is almost simple. (The third paragraph isn't convincing.) For (1) it isn't enough to consider just a representation of an arbitrary finite linear group containing a diagonal matrix.

  • 1
    $\begingroup$ Why must $k$ have $\geq q$ elements? Any finite group has a faithful representation over any field with say $p$ elements, e.g. permutation representations (given by permuting elements of a basis). $\endgroup$
    – Xandi Tuni
    May 20, 2010 at 12:03
  • $\begingroup$ Take $k=F_p$, $q=p^e$ and define $SL_2(q)\to GL_{2n}(p)$ by interpreting $F_q^2$ as $2e$−dimensional $k$-vector space. $\endgroup$
    – Someone
    May 20, 2010 at 12:17
  • $\begingroup$ That's even simpler... $\endgroup$
    – Xandi Tuni
    May 20, 2010 at 12:22
  • $\begingroup$ @Xandi Tuni: Sorry, my version was too hasty. I'll cancel it to you can post your comment as a negative answer to the question. $\endgroup$ May 20, 2010 at 13:07
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    $\begingroup$ Just for completeness: you should exclude some small $q$ when claiming that $G/Z(G)$ is simple. $\endgroup$
    – Someone
    Jun 10, 2010 at 16:06

I think the answer is yes if and only if $k$ contains all $(q-1)$-st roots of unity, for the following reason:

If $D$ is a diagonal matrix in $SL_r(\mathbb F_q)$, then its order divides $q-1$ because all entries on the diagonal are in $\mathbb F_q^\ast$. But every endomorphism of $k^n$ of finite order $n$ prime to $p$, and in particular $\varphi(D)$, is diagonalisable if $k$ contains all $n$--th roots of unity.

On the other hand: Take $D$ of order $q-1$ and suppose $\varphi(D)$ is diagonalisable over $k$. Then all eigenvalues of $\varphi(D)$ are in $k$. These eigenvalues are roots of unity, such that the lcm of their orders is $q-1$. This implies that $k$ contains primitive $(q-1)$-st roots of unity.

As usual I'm posting in a hurry and have not checked anything.

Edit: Just to make the statement clear: Given $p, q$ and $k$ as in the question, I claim the followimg to be equivalent:

(a) the field $k$ contains all $(q-1)$-th roots of unity
(b) for every representation $\varphi: SL_2(\mathbb F_q) \to GL_n(k)$ and every diagonal matrix $D\in SL_2(\mathbb F_q)$, the matrix $\varphi(D)$ is diagonalisable over $k$.

  • $\begingroup$ Why should the lcm of the orders of the eigenvalues be $q - 1$? For example, this is false when $\varphi = 1$. Perhaps you are just claiming that one can cook up a $\varphi$ with this property? If so, I don't see it immediately. $\endgroup$
    – LSpice
    May 28, 2011 at 14:57

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