Two dimensional complex group representations 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.
 A: 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.
A: M.14: We may assume that $\rho$ is faithful (just replace $G$ by $G/\ker(\rho)$ and neither the hypothesis nor the conclusion changes). In particular, no non-identity element is in the kernel of $\det$ since $\det(g)$ is precisely the non-one eigenvalue of $g$. Thus $\det:G/\ker(\rho) \to \mathbb{C}^\times$ is an embedding, and $G/\ker(\rho)$ is finite abelian, so every representation is a direct sum of one-dimensional representations.
A: Another solution for M.14 (this is probably easier for you to come up with than Jack's answer):
Suppose in contradiction that $\rho$ is not a sum of two one-dimensional representations, it must be irreducible. Since one of the eigenvalues is 1, the matrix for $\rho(g)$ must look like $$\begin{pmatrix}1 & 0\\\ 0 & \lambda_g\end{pmatrix}$$hence $\chi(g) = 1 + \lambda_g$.
Now, take the inner product of $\chi$ with the unit representation. (As a side note, this is a standard trick in similar exercises where you know something about the sum of the characters.) Because a two dimensional representation can never be isomorphic to a one dimensional representation, this inner product must equal to zero (as they are both irreducible). Thus $$\frac{1}{|G|}\sum_{g \in G} \chi(g)= 0 = 1+\frac{1}{|G|}\sum_{g \in G} \lambda_g$$ By the triangle inequality, $$|\sum_{g \in G} \lambda_g| \le \sum_{g \in G} |\lambda_g| = |G| $$ with equality taken when all $\lambda_g$'s are equal and as the sum is -|G|, they would all have to be -1. But the norm of this representation is 0 $\neq$ 1, so our assumption fails and $\rho$ should not be irreducible, hence a sum of two one-dimensional representations.
A: As Frank notes for M8, $G \leq SU_{2} (\mathbb{C})$. There is a 2-to-1 group homomorphism:
$$SU_{2} (\mathbb{C}) \to SO_{3} (\mathbb{R})$$
This is the universal cover of $SO_{3} (\mathbb{R}) \cong \mathbb{R}P^{3}$ (Artin covers this-I believe- in chapter 9). Since $G$ is simple, this means $G \leq SO_{3} (\mathbb{R})$ also, which means that $G$ is a finite subgroup of what Artin calls the ``rotation group" (at the end of chapter 6) meaning a group acting faithfully on the sphere via rigid motion. Artin classifies all such finite groups. There are only five possibilities for $G $ according to this classification:


*

*$\mathbb{Z}/n$

*The dihedral group $D_{n}$

*The tetrahedral group $A_{4}$

*The octahedral group $S_{4}$ 

*The icosahedral group $A_{5}$
The only group in this list which is not of prime order and is simple is $A_{5}$, which has 5 irreducible complex representations. One of these is the trivial one and the remaining 4 have dimension > 2. So, it has no non-trivial representations of dimension 2. Hence, M8. This may have been what Artin had in mind, although I like Alex's argument better!
