5 clarification

Well, the simple groups ${\rm SL}(2,2^{n})$ ( $n >1$) certainly have $2$-dimensional irreducible representations in characteristic $2$, and in fact $A_{5} \cong {\rm SL}(2,4)$ (to see this, just note that ${\rm SL}(2,4)$ has order $(4+1)(4^{2}-4) = 60 ).$ The standard argument to show that a non-Abelian finite simple group can't have a $2$-dimensional irreducible representation does not work in characteristic $2$, since it relies on the fact that an involution (which must be non-central using simplicity) must have determinant $-1$, since one of its eigenvalues must be $1$ and the other $-1.$ But again since $G$ is non-Abelian simple, every representation has trivial determinant, a contradiction. But this line argument does not work over a field of characteristic $2$.

Note for historical interest, related to comments in Qiaochu's answer: Since a finite group of odd order is solvable, and since the dimension of absolutely irreducible modules in any characteristic divide the group order when the group is solvable, it follows that a finite group of odd order can't have a $2$-dimensional absolutely irreducible module in any characteristic. However, this requires the rather heavyweight Feit-Thompson theorem, so may be regarded as a little unsatisfactory in the context of this discussion. So I outline a different argument. This still requires some relatively difficult group theory, namely the Hall-Higman theorem, or its descendants, a Theorem of E. Shult and the theory of $p$-stability. The fact that we need from this theory is that if $p$ and $q$ are odd primes, and a $p$-group $P$ normalizes a $q$-group $Q$ and the group $PQ$ has a faithful linear representation in characteristic other than $q$, then every element of $p$ which acts with minimum polynomial of degree less than $p$ centralizes $Q.$ Hence if a finite group $G$ of odd order has a faithful $2$-dimensional representation in some finite characteristic $p$, then it follows that for each prime divisor $q$ of $|G|$ other than $p$, we have that $N_{G}(Q)/C_{G}(Q)$ is a $q$-group for each $q$-subgroup $Q$ of $G.$ By a Theorem of Frobenius, $G$ has a normal $q$-complement. It follows that $G/O_{p}(G)$ is nilpotent, where $O_{p}(G)$ is the largest normal $p$-subgroup of $G.$ Hence (without using the classification of finite simple groups, or the odd order theorem), we see that a non-Abelian simple group of odd order can't have a $2$-dimensional irreducible representation in any characteristic.

4 typo

Well, the simple groups ${\rm SL}(2,2^{n})$ ( $n >1$) certainly have $2$-dimensional irreducible representations in characteristic $2$, and in fact $A_{5} \cong {\rm SL}(2,4)$ (to see this, just note that ${\rm SL}(2,4)$ has order $(4+1)(4^{2}-4) = 60 ).$ The standard argument to show that a non-Abelian finite simple group can't have a $2$-dimensional irreducible representation does not work in characteristic $2$, since it relies on the fact that an involution (which must be non-central using simplicity) must have determinant $-1$, since one of its eigenvalues must be $1$ and the other $-1.$ But again since $G$ is non-Abelian simple, every representation has trivial determinant, a contradiction. But this line argument does not work over a field of characteristic $2$.

Note for historical interest, related to comments in Qiaochu's answer: Since a finite group of odd order is solvable, and since the dimension of absolutely irreducible modules in any characteristic divide the group order, it follows that a finite group of odd order can't have a $2$-dimensional absolutely irreducible module in any characteristic. However, this requires the rather heavyweight Feit-Thompson theorem, so may be regarded as a little unsatisfactory in the context of this discussion. So I outline a different argument. This still requires some relatively difficult group theory, namely the Hall-Higman theorem, or its descendants, a Theorem of E. Shult and the theory of $p$-stability. The fact that we need from this theory is that if $p$ and $q$ are odd primes, and a $p$-group $P$ normalizes a $q$-group $Q$ and the group $PQ$ has a faithful linear representation in characteristic other than $q$, then every element of $p$ which acts with minimum polynomial of degree less than $p$ centralizes $Q.$ Hence if a finite group $G$ of odd order has a faithful $2$-dimensional representation in some finite characteristic $p$, then it follows that for each prime divisor $q$ of $|G|$ other than $p$, we have that $N_{G}(Q)/C_{G}(Q)$ is a $q$-group for each $q$-subgroup $Q$ of $G.$ By a Theorem of Frobenius, $G$ has a normal $q$-complement. It follows that $G/O_{p}(G)$ is nilpotent, where $O_{p}(G)$ is the largest normal $p$-subgroup of $G.$ Hence (without using the classification of finite simple groups, or the odd order theoremtheorem), we see that a non-Abelian simple group of odd order can't have a $2$-dimensional irreducible representation in any characteristic.

3 added direct elimimination of odd order $G$

Note for historical interest, related to comments in Qiaochu's answer: Since a finite group of odd order is solvable, and since the dimension of absolutely irreducible modules in any characteristic divide the group order, it follows that a finite group of odd order can't have a $2$-dimensional absolutely irreducible module in any characteristic. However, this requires the rather heavyweight Feit-Thompson theorem, so may be regarded as a little unsatisfactory in the context of this discussion. So I outline a different argument. This still requires some relatively difficult group theory, namely the Hall-Higman theorem, or its descendants, a Theorem of E. Shult and the theory of $p$-stability. The fact that we need from this theory is that if $p$ and $q$ are odd primes, and a $p$-group $P$ normalizes a $q$-group $Q$ and the group $PQ$ has a faithful linear representation in characteristic other than $q$, then every element of $p$ which acts with minimum polynomial of degree less than $p$ centralizes $Q.$ Hence if a finite group $G$ of odd order has a faithful $2$-dimensional representation in some finite characteristic $p$, then it follows that for each prime divisor $q$ of $|G|$ other than $p$, we have that $N_{G}(Q)/C_{G}(Q)$ is a $q$-group for each $q$-subgroup $Q$ of $G.$ By a Theorem of Frobenius, $G$ has a normal $q$-complement. It follows that $G/O_{p}(G)$ is nilpotent, where $O_{p}(G)$ is the largest normal $p$-subgroup of $G.$Hence (without using the classification of finite simple groups, or the odd order theorem,we see that a non-Abelian simple group of odd order can't have a $2$-dimensional irreducible representation in any characteristic.

2 added 80 characters in body
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