Question: What is the number of two-dimensional irreducible representations of a finite group ? How it can be expressed in groups-theoretic terms ? (Number of 1-dimensional irreps is |G/[G,G]| ).

The question is somewhat naive and actually I heard it from our teachers when I was an undergrad many years ago, it was always outspoken with some kind of mysterious flavour - "nobody knows, but may be...".

Some analogies:

There is a paper by V. Drinfeld 1981, which title is "Number of two-dimensional irreducible representations of the fundamental group of a curve over a finite field".

The main theorem express the number via the zeta function of the curve over F_q. (Russian pdf is for free - main formula can be seen from there). Of course, it is very specific class of the groups, however may be something can be done ?

Another analogy which comes to mind is related to topological quantum field theories, quantization of Wess-Zumino and Chern-Simons models. One consider the moduli space of d-dimensional representatation of the fundamental group. It is natarally symplectic manifold and its VOLUME is somewhat an anologue of the "number" of irreps for finite group. The volume can be calculated and is related to the famous Verlinde formula.

So, it is of course, both considerations are related to fundamental (=Galois) groups of CURVES.

Question: WHY? What makes fundamental (=Galois) groups of curves so specific ? Can it be somehow generalized to other classes of curves ?

  • 1
    $\begingroup$ Which field are you working over? I.e., you want to count homomorphisms to $GL_2(K)$ for which fields $K$? $\endgroup$
    – Ian Agol
    Sep 21, 2012 at 17:46
  • $\begingroup$ @Agol complex numbers, any other field is also welcome. $\endgroup$ Sep 21, 2012 at 18:38
  • 2
    $\begingroup$ The case of $K=F_2$ is slightly nontrivial. $GL(2,F_2) \cong S_3$, and any nontrivial homomorphism is irreducible, although you want to count homomorphisms up to conjugacy. Transitive homomorphisms correspond to subgroups of $G$ of index $3$ and nontransitive nontrivial homomorphisms correspond to subgroups of index $2$. $\endgroup$ Sep 21, 2012 at 22:36
  • $\begingroup$ I seem to recall Noah Snyder thinking about such a question group-theoretically. Paging @Noah Snyder? (Is that how you do it?) $\endgroup$
    – Marty
    Aug 29, 2017 at 5:10
  • $\begingroup$ Let $V$ be the variety (in the sense of universal algebra) generated by finite subgroups of $\mathrm{GL}_2(\mathbf{C})$ (one should be able to give a small list of generators of this variety — notably the only simple groups therein are abelian and Alt$_5$). It should be small enough to completely classify groups within it, and classify in a sort of explicit way their 2-dim reps. Then for an arbitrary group $G$, it has a largest quotient $G_V$ in $V$, and this would make an answer. But this would require some work anyway. $\endgroup$
    – YCor
    Jul 22, 2021 at 19:48

1 Answer 1


This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4},S_{4}$ or $A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index $2.$ On the other hand, any finite group with an Abelian normal subgroup of index $2$ has all its irreducible representations of degree $1$ or $2,$ so the number of $2$-dimensional irreducible representations is easily calculated.

A more careful analysis of the primitive case shows that if $G$ has a faithful $2$-dimensional primitive complex representations, then (as has been known since the late 19th century), $G = Z(G)E,$ where $ E \cong {\rm SL}(2,3), {\rm GL}(2,3),{\rm SL}(2,5)$ or the binary octahedral group (also, a double cover of order $48$ of $S_{4},$ (as ${\rm GL}(2,3)$ is), but with a generalized quaternion Sylow $2$-subgroup).

Now let $G$ be any finite group, and let $K$ be the intersection of the kernels of the irreducble complex representation of $G$ of degree at most $2$. The above discussion means that the only possible non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated if it appears. The answer to your question only depends on the structure of $G/K,$ so we may reduce to the case that all composition factors of $G$ are cyclic or $A_{5}.$ Also, by Clifford theory, we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$

Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it.

  • $\begingroup$ The result, mentioned in Prof. Robinson's post. is due to F. Klein. $\endgroup$
    – yakov
    Jul 8, 2016 at 13:46

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