Michael Artin's Algebra, chapter 10 (both unstarred, and complex representations)
M.8 Prove that a finite simple group that is not of prime order has no nontrivial representation of dimension 2.
M.14 Let $\rho\colon G\to GL(V)$ be a two-dimensional representation of a finite group $G$, and assume that $1$ is an eigenvalue of $\rho_g$ for every $g$ in $G$. Prove that $\rho$ is a sum of two one-dimensional representations.
All these exercises are closely related to $GL_2(\mathbb C)$, and I think it's closely related to the property of $U_2$, the unitary group, therefore they go together.
We can simplify both questions in nearly the same way.
The first one:
It's not hard to show the correctness of abelian case, therefore we discard this case for now. Suppose there's a nontrivial 2D representation $\rho$ of a finite simple group $G$. Since $\rho$ is nontrivial and $G$ is simple, $\ker\rho$ is trivial, and $G$ embeds as a subgroup of $GL_2$. By Maschke's theorem, WLOG, we can suppose that $G\subset U_2$. Moreover, consider the mapping $\det\colon G\to\mathbb C$, we have $\ker\det$ is nontrivial, since $G$ isn't abelian, therefore by the normality of $G$, the image is trivial, and $G\subset SU_2$, the special unitary group.
The second one:
We can only consider the image of $\rho$. It's a finite group whose matrices have eigenvalue $1$. We'd only show that these matrices are simultaneously diagonalizable, therefore $\rho$ is a direct sum of two 1D representations. WLOG, suppose that the image is contained in $U_2$, by Maschke's theorem.
Both problems are simplified as a property of $U_2$ (the first one reduces a bit more). The first one says that there's no simple subgroup of composite number order, the second one says that if they all have eigenvalue $1$, then they're simultaneously diagonalizable.
How can we proceed? I need some insight of $U_2$ or $SU_2$. Thanks!
EDIT: I think my previous question is also related.
Best Answer
Tobias Kildetoft has suggested that I add this as answer, but it still feels too hard:
M.8 Note that since $G$ is simple and non-abelian (else its of prime order) we know that any one-dimensional representation of $G$ is trivial. Now, suppose that $\rho:G\to\text{GL}_2(\mathbb{C})$ is non-trivial. Then, evidently $\rho$ is faithful. But, since it also can't be the sum of one-dimensional reps (since those are all trivial) it must also be irreducible. This implies, in particular, that $2\mid|G|$ so that $G$ has some element of order $2$, say $g$. Note then that $\rho(g)$ must have eigenvalues $\pm 1$ (since it's order $2$). But, as you've already noted, it must also have determinant $1$. This forces the eigenvalues to both be $-1$, and so, in particular, $\rho(g)$ is a scalar matrix, and so in the center of $\text{GL}_2(\mathbb{C})$. By faithfulness, this means that $g$ is a non-trivial element of $Z(G)$, which by simplicity forces $G=Z(G)$. This is a contradiction.