# character degrees of finite simple groups: an elementary question

I have a simple question on characters of simple groups; stated as exercise in Isaacs' character theory book.

(2.3) Let $\chi$ be a character of $G$. If $\mathfrak{X}$ is a representation corresponding to $\chi$, define $\det\chi(x):=\det\mathfrak{X}(x)$. Show: $\det\chi$ is well-defined linear character of $G$.

This is an easy exercise; consider the next one.

(3.3) Show that no simple group can have irreducible character of degree 2.

Hint: Problem 2.3 is relevant.

Question: My question is not about solution of (3.3) since I have solved it in some different way. I didn't click how (2.3) can be used for this?

Other way for (3.3): $|G|$ is divisible by $2$=degree of irreducible character. If $|G|=2.(odd)$ then it is well-known from basic group theory that $G$ can't be simple (Actually it is solvable). So $4$ should divide $|G|$. Consider subgroup $H$ of order $4$ in $G$. Let $\mathfrak{X}:G\rightarrow {\rm GL}_2(\mathbb{C})$ be irreducible with $G$ simple (non-abelian). We can show that some element $x\in H$, $x\neq 1$ goes to $\pm I$. So $Z(\chi)=\{g\in G: |\chi(g)|=\chi(1)\}$ is non-trivial normal subgroup. q.e.d.

## 1 Answer

To use 2.3, suppose $G$ is simple with irreducible character $\chi$ of degree $2$. If $\rho\colon G\to GL_2(\mathbb{C})$ affords $\chi$, then $\ker\rho\unlhd G$ must be trivial, since $\rho$ is not the trivial rep, so $\rho$ is an injection.

Now $G$ can't be abelian, so $[G,G]\neq 1$, so by simplicity, $G=[G,G]$. The linear characters of $G$ are in bijection with $G/[G,G]$, which has order $1$, so the only linear character is the trivial character. But by 2.3, $\det\chi$ is a linear character, hence it must be trivial. So $\det\rho(g)=1$ for all $g\in G$.

Now the degree of an irreducible character divides $|G|$, so $2$ divides $|G|$. By Cauchy, there exists $g\in G$ of order $2$. Then $\rho(g)$ has order $2$, and is similar to a $2\times 2$ diagonal matrix whose diagonal entries are 2nd roots of unity, i.e., $\pm 1$. So necessarily $\rho(g)$ is similar to $-I_2$, since it must have determinant $1$, hence actually $\rho(g)=-I_2$. Then $\rho(g)\rho(x)=\rho(x)\rho(g)$ for all $x\in G$, so since $\rho$ is injective, $g$ is central in $G$. So finally $\langle g\rangle\unlhd G$, and is proper, since $|G|\neq 2$, else $G$ would be abelian. But this contradicts simplicity.