Show that a finite simple group cannot have a 2-dimensional irreducible representation over \(\mathbb C\).
Hint: the determinant might prove useful.
Solution
Suppose, for contradiction, that \(G\) is finite simple and that \[ \rho:G\longrightarrow \GL(V),\qquad \dim_{\mathbb C}V=2, \] is irreducible.
First, \(ho\) is faithful. Indeed, \(\ker\rho\trianglelefteq G\). Since \(G\) is simple, \(\ker\rho\) is either \(1\) or \(G\). The latter would make the representation trivial, hence reducible in dimension \(2\). Thus \(\ker\rho=1\).
Consider \[ \det\rho:G\longrightarrow \mathbb C^\times. \] Its kernel is normal. If its kernel were trivial, then \(G\) would embed in the abelian group \(\mathbb C^\times\), so \(G\) would be abelian. A finite simple abelian group is cyclic of prime order, and every irreducible complex representation of an abelian group is \(1\)-dimensional, a contradiction. Hence \[ \det\rho(g)=1\qquad\text{for every }g\in G. \] Thus \(ho(G)\subseteq \SL_2(\mathbb C)\).
We now justify that \(2\mid |G|\). We use the standard character-degree divisibility argument, included here for completeness. Let \(\chi\) be an irreducible complex character of degree \(d=\chi(1)\). For each conjugacy class \(C\) with representative \(c\), the class sum \[ z_C=\sum_{g\in C}g\in Z(\mathbb Z[G]) \] acts on the irreducible representation by the scalar \[ \omega_C=\frac{|C|\chi(c)}{d}. \] Because multiplication by \(z_C\) is an integral endomorphism of the finite free abelian group \(Z(\mathbb Z[G])\), Cayley–Hamilton shows that \(\omega_C\) is an algebraic integer. Character values are algebraic integers as well. Character orthogonality gives \[ |G|=\sum_{g\in G}\chi(g)\chi(g^{-1}) =d^2+\sum_{C\ne\{1\}}|C|\chi(c)\chi(c^{-1}). \] Dividing by \(d\), \[ \frac{|G|}{d} =d+\sum_{C\ne\{1\}}\omega_C\chi(c^{-1}). \] The right side is an algebraic integer, while the left side is rational; hence \(|G|/d\in\mathbb Z\). Therefore \(d\mid |G|\). Applying this with \(d=2\) gives \(2\mid |G|\).
By Cauchy’s theorem, \(G\) has an element \(t\) of order \(2\). Since \[ \rho(t)^2=I, \] the matrix \(ho(t)\) is diagonalizable with eigenvalues in \(\{1,-1\}\). Also \[ \det\rho(t)=1. \] Therefore its two eigenvalues are either both \(1\) or both \(-1\). Faithfulness excludes \(ho(t)=I\), because \(t\ne1\). Hence \[ \rho(t)=-I. \] But \(-I\) is central in \(ho(G)\). Since \(ho\) is faithful, \(t\in Z(G)\). Thus \(Z(G)\ne1\). Simplicity forces \(Z(G)=G\), so \(G\) is abelian, again contradicting the existence of a \(2\)-dimensional irreducible complex representation.
Hence a finite simple group has no \(2\)-dimensional irreducible complex representation.