Let $G$ be a finite, non-abelian simple group. How would I go about proving that any $2$-dimensional representation of $G$ is trivial? If it helps, I know how to do it when we're considering $1$-dimensional representations.


3 Answers 3


Okay, we're going to have to use some heavy artillery to start off, but I can't think of another way to begin.

Suppose $\rho: G\to \text{GL}_{2} (\mathbb{C})$ is nontrivial. Observe that since $G$ is simple and the representation is nontrivial, we must have $\text{ker} \, \rho =\text{ker}\, \chi = (e)$ (where $\chi$ is the character of this representation). The Feit-Thompson Theorem (!!!) tells us $|G|$ is even. By Cauchy's Theorem, $G$ must have an element $x$ of order $2$.

Now, define $$\hat{\rho}: G \to \text{GL}_{1} (\mathbb{C}) \cong \mathbb{C}^{\times}$$ by $\hat{\rho}(g) = \text{det} (\rho(g))$. Evidently, $\hat{\rho}$ is a homomorphism, hence it gives a degree 1 representation of $G$. We know this representation must be trivial. In other words, $\text{det} (\rho(g)) = 1$ for all $g\in G$. That said, we also know that $\rho(x)^2 = \text{Id}$. The set of eigenvalues of $\rho(x)$ is either $\{1, 1\}$, $\{1,-1\}$, or $\{-1,-1\}$. The first possibility is out of the question, since $\text{ker} \chi = (e)$. The second possibility cannot occur, since then $\text{det} (\rho(x)) = -1$. Thus, the eigenvalues of $\rho(x)$ are $\{-1, -1\}$. The characteristic polynomial of $\rho(x)$ is $(X+1)^2$, and $\rho(x)$ also satisfies $X^2 - 1$. Since the minimal polynomial of $\rho(x)$ must divide both of these, it follows $\rho(x)$ satisfies $X+1$, i.e. $\rho(x) = -\text{Id}$.

Lastly, since $\rho(x)$ is a scalar multiple of the identity, it commutes with any matrix. In particular, for any $g\in G$, we have $$\rho(g) \rho(x) = \rho(x) \rho(g) \implies \rho(gxg^{-1} x^{-1}) = \text{Id}$$ Triviality of $\text{ker} \, \rho$ implies $gxg^{-1} x^{-1} = e$ for all $g\in G$, hence $x\in Z(G)$. Accordingly, $Z(G)$ is a nontrivial normal subgroup of $G$, so it must equal $G$. But $G$ is non-abelian by assumption.

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    $\begingroup$ I think you can avoid Feit-Thompson to show the order is even. The representation is faithful, and there are no non-trivial $1$-dimensional representations, so the representation must be irreducible, so the degree divides the order of the group. $\endgroup$ Dec 20, 2014 at 5:17
  • $\begingroup$ Oh hey, oops. That's much better. $\endgroup$ Dec 20, 2014 at 5:19
  • $\begingroup$ Alternatively, you can "avoid" talking about minimal polynomials by saying that for every $g \in G$, there is a basis $B$ of $\Bbb C^2$ such that the matrix of $\rho(g)$ w.r.t. $B$ is diagonal. Having $\det = 1$, your analysis about eigenvalues shows that $\rho(x) = ±I_2$, which is a scalar matrix. $\endgroup$
    – Watson
    Aug 23, 2016 at 19:04
  • $\begingroup$ Nitpicking: $\operatorname{GL}_1(\Bbb C)\cong \Bbb C^{\color{red}\times}$ $\endgroup$ Jun 15, 2018 at 15:45
  • $\begingroup$ Fixed nit, thanks! $\endgroup$ Jun 15, 2018 at 17:38

Using a lot less machinery:

Such a representation must land in $SL_2(\mathbb C)$, and finite subgroups of $SL_2(\mathbb C)$ are completely classified (they are either cyclic, dihedral, tetrahedral, octahedral or isocahedral). None of these groups are simple, unless cyclic of prime order.

One can prove this classification by letting a finite subgroup $G$ of $SL_2(\mathbb C)$ act on the Riemann sphere $\mathbb P^1$. One can compare the Riemann-Hurwitz formula for the morphism $\mathbb P^1 \to \mathbb P^1/G$ to the class equation of $G$ to deduce bounds on the individual terms appearing in the class equation, thus narrowing down the possible groups.

  • $\begingroup$ The classification is done nicely in Dolgashev's notes on the Mackay correspondence, which Google finds in a sec, among many other places. $\endgroup$ Dec 20, 2014 at 5:13

This is exercise 3.3 in Character Theory of Finite Groups by Isaacs. It is easily deduced from the following facts:

  • Because $G$ is perfect, $\chi$ has image in $SL(2,\mathbb{C})$;
  • Because $\chi$ is irreducible, $\chi(1)$ divides $|G|$;
  • The only element of order $2$ in $SL(2,\mathbb{C})$ is $-I$ (which is central).

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