Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recently I have been working with a certain subgroup of $GL_{10}(\mathbb{F}_2)$ and for various reasons was fairly sure it contained a normal subgroup isomorphic to $A_5$. Today I was able to affirmatively show that this presumed copy of $A_5$ does indeed exist in the bigger group. Now the matrices composing this copy of $A_5$ are themselves $2\times 2$ matrices over a certain commutative 5-dimensional algebra $\mathcal{S}$ (necessarily with zero divisors) over $\mathbb{F}_2$; this gives rise to a 2-dimensional representation of $A_5$ over $\mathcal{S}$ which I am fairly sure is irreducible (as opposed to just indecomposeable).

Now it is a fairly standard exercise early in learning representation theory to show that a simple group cannot have an irreducible representation of dimension 2. As the proof of this simple fact relies only on considerations involving the characters of elements of order 2, does the existence of the above representation rely only on the fact that I am working in characteristic 2, or is it related to the fact that I am working with rings with zero divisors as opposed to an algebraically closed field, or something else more subtle? I am primarily curious as to whether it is solely related to the characteristic, as I am also working with some related groups defined over commutative rings in other characteristics which I suspect also contain subgroups isomorphic to $A_5$. Hopefully someone with more background in modular representation theory than I can shed some helpful light on this situation.

share|improve this question

3 Answers 3

up vote 11 down vote accepted

To reinforce Geoff's answer, I'd emphasize that the "fairly standard exercise" mentioned in the question is only standard when you consider characteristic 0 irreducible representations as is usually done in a first course (then it's also usual to start with a big field like $\mathbb{C}$ to avoid extra complications). When dealing with modular representations, there are lots of further issues.

Qiaochu attempts an outline of the "standard exercise" which shows clearly where problems arise. In his first step, you need characteristic 0 (and a splitting field) to be confident that the degree of an irreducible representation divides the group order. This is definitely false in general. And then of course characteristic 2 is especially dangerous for dealing with the negative of the $2 \times 2$ identity matrix.

The bottom line is that you have to avoid this "standard exercise" and focus just on the special situation you are studying, in order to avoid confusion.

[ADDED] 1) Concerning the "standard exercise" (say over $\mathbb{C}$), I recall that decades ago it appeared as a Math Monthly problem (which I wrote an answer for) and later became an exercise in at least one advanced algebra text. Given an absolutely irreducible representation of degree 2 for a (necessarily nonabelian) finite simple group, classical character theory implies that the group order is even, so you can apply Cauchy etc. Of course, Feit-Thompson gets you there as well, but that's overkill. However, in odd characteristic that does help to get through the other steps of the proof, but in characteristic 2 you'd still face $I = -I$.

2) While the degrees of irreducible representations of simple groups of Lie type are far from being known explicitly in general, over a splitting field of the defining characteristic you can see easy examples where degrees fail to divide the group order. For instance, take $\mathrm{PSL}_2(\mathbb{F}_7)$ in characteristic 7. Here you have an irreducible of degree 5 whereas 5 doesn't divide the group order 168.

share|improve this answer
How to describe 5dim irrep of pal(2 7) ? I an playing with this group now. Actually over complex but would be happy to know more. –  Alexander Chervov Aug 18 '12 at 18:27
I mean psl (2,7)... –  Alexander Chervov Aug 18 '12 at 18:28
@Alexander: For this group in characteristic 7 (but not over the complex field), the irreducibles have dimensions 1, 3, 5, 7, and arise from classical Lie theory combined with reduction mod 7. This starts with the usual Lie algebra matrix representations for the special linear case but then involves an integral (say Chevalley) lattice and reduction mod 7. Since 7 divides the group order, the ordinary and 7-modular theories diverge to some extent but are then related by Brauer theory. –  Jim Humphreys Aug 19 '12 at 23:58
P.S. I should add that the most direct construction for $SL_2$ (say over any field of char 7) just uses the spaces of homogeneous polynomials in two variables of degrees 0, 1, ..., 6, of respective dimensions 1, 2, ..., 7. Here -I acts trivially in dim 1, 3, 5, 7 and gives the desired irreducibles for $PSL_2$. Also, I should add that the group over the prime field is often called $L_2(7)$ but is isomorphic to $L_3(2)$ (the name used by Conway et al. in their book An Atlas of Brauer Characters). –  Jim Humphreys Aug 20 '12 at 0:46
Thank you very much ! –  Alexander Chervov Aug 20 '12 at 17:05

Well, the simple groups ${\rm SL}(2,2^{n})$ ( $n >1$) certainly have $2$-dimensional irreducible representations in characteristic $2$, and in fact $A_{5} \cong {\rm SL}(2,4)$ (to see this, just note that ${\rm SL}(2,4)$ has order $(4+1)(4^{2}-4) = 60 ).$ The standard argument to show that a non-Abelian finite simple group can't have a $2$-dimensional irreducible representation does not work in characteristic $2$, since it relies on the fact that an involution (which must be non-central using simplicity) must have determinant $-1$, since one of its eigenvalues must be $1$ and the other $-1.$ But again since $G$ is non-Abelian simple, every representation has trivial determinant, a contradiction. But this line argument does not work over a field of characteristic $2$.

Note for historical interest, related to comments in Qiaochu's answer: Since a finite group of odd order is solvable, and since the dimension of absolutely irreducible modules in any characteristic divide the group order when the group is solvable, it follows that a finite group of odd order can't have a $2$-dimensional absolutely irreducible module in any characteristic. However, this requires the rather heavyweight Feit-Thompson theorem, so may be regarded as a little unsatisfactory in the context of this discussion. So I outline a different argument. This still requires some relatively difficult group theory, namely the Hall-Higman theorem, or its descendants, a Theorem of E. Shult and the theory of $p$-stability. The fact that we need from this theory is that if $p$ and $q$ are odd primes, and a $p$-group $P$ normalizes a $q$-group $Q$ and the group $PQ$ has a faithful linear representation in characteristic other than $q$, then every element of $p$ which acts with minimum polynomial of degree less than $p$ centralizes $Q.$ Hence if a finite group $G$ of odd order has a faithful $2$-dimensional representation in some finite characteristic $p$, then it follows that for each prime divisor $q$ of $|G|$ other than $p$, we have that $N_{G}(Q)/C_{G}(Q)$ is a $q$-group for each $q$-subgroup $Q$ of $G.$ By a Theorem of Frobenius, $G$ has a normal $q$-complement. It follows that $G/O_{p}(G)$ is nilpotent, where $O_{p}(G)$ is the largest normal $p$-subgroup of $G.$ Hence (without using the classification of finite simple groups, or the odd order theorem), we see that a non-Abelian simple group of odd order can't have a $2$-dimensional irreducible representation in any characteristic.

share|improve this answer

I think it is the characteristic. $A_5$ naturally sits in $\text{SO}(3)$ as the group of rotational symmetries of the icosahedron. Pulling $A_5$ back along the double cover $\text{SU}(2) \to \text{SO}(3)$ gives a double cover of $A_5$, the binary icosahedral group $G$, and this group is therefore equipped with a $2$-dimensional complex representation.

By this MO question, this representation is defined over $\mathcal{O}_K$ for a number field $K$, so we can reduce it $\bmod 2$; the double cover $G \to A_5$ has kernel $\{ \pm I \}$ which is killed when working $\bmod 2$, so we get a $2$-dimensional representation of $A_5$ in characteristic $2$.

I am not sure exactly what proof you're referring to of the fact that a simple group $G$ can't have an irreducible representation of dimension $2$. I hope it is this one:

  • The dimension of an irreducible representation divides the order of $G$.
  • By Cauchy's theorem, $G$ has an element of order $2$.
  • By averaging, $G$ is a subgroup of $\text{U}(2)$. By simplicity, it is a subgroup of $\text{SU}(2)$.
  • But the only element of order $2$ in $\text{SU}(2)$ is $-I$, which is central; contradiction.

I don't know if the first step works in characteristic $2$, and the third step is problematic also (we should work with $\text{SL}_2$), but the last step definitely fails (as seen above).

share|improve this answer
The first step does not usually work in prime characteristic, but does work for solvable groups by the fairly difficult Fong-Swan theorem. –  Geoff Robinson Aug 7 '12 at 0:40
@Geoff: thanks for the comment. The proof I'm familiar with fails but I didn't know if a different technique could work in positive characteristic. –  Qiaochu Yuan Aug 7 '12 at 1:15
I am probably the only person who got confused, but just in case anyone else was also wondering: am I right in saying that the argument avoids Feit-Thompson by saying "assume G has an irrep of degree 2, then G must have even order, then G must have an involution," etc? –  Yemon Choi May 7 '13 at 0:17
Also, Qiaochu, do you know of a reference for the argument you give, or was it just something that emerged when you were learning representation theory? For a short note I'm typing up, I would like to both cite the result and indicate (since the note will probably be read by functional/harmonic analysts) that the proof doesn't require Feit-Thompson. –  Yemon Choi May 7 '13 at 0:19
@Yemon: yes, and I don't remember. I think parts of the argument come up in the McKay correspondence. –  Qiaochu Yuan May 7 '13 at 2:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.