# $\mathit{SL}(2,\mathbb{F}_{p})$ has no $1$ dimensional irreducible representations over $\mathbb{F}_{p}$

###### Lemma.

The group $G=\operatorname{SL}(2,\mathbb{F}_{p})$ has no non-trivial $1$ dimensional irreducible representations over $\mathbb{F}_{p}$.

###### Proof.

Notice that a $1$ dimensional representations over $\mathbb{F}_{p}$ is just a homomorphism:

 $\rho\colon\operatorname{SL}(2,\mathbb{F}_{p})\longrightarrow\mathbb{F}_{p}^{% \times}.$

Let $\rho$ be as above. Then there exist an induced homomorphism for the projective special linear group:

 $\bar{\rho}\colon\operatorname{PSL}(2,\mathbb{F}_{p})\longrightarrow\mathbb{F}_% {p}^{\times}/\{\pm\mathit{Id}\}$

defined by $\bar{\rho}(A)=\rho(B)\mod\{\pm\mathit{Id}\}$, where $B$ is any lift of $A$ to $\operatorname{SL}(2,\mathbb{F}_{p})$ (this is well defined because $\rho(-\mathit{Id})=\pm\mathit{Id}$). Since $\operatorname{PSL}(2,\mathbb{F}_{p})$ is simple, the image of $\bar{\rho}$ is trivial, and therefore, the image of $\rho$ is contained in $\{\pm\mathit{Id}\}$.

However, $\operatorname{SL}(2,\mathbb{F}_{p})$ does not have subgroups of index $2$ (a subgroup of index $2$ is normal). For our purposes, it suffices to show that:

 $\rho\colon\operatorname{SL}(2,\mathbb{F}_{p})\longrightarrow\{\pm\mathit{Id}\}$

satisfies $\rho(-\mathit{Id})=\mathit{Id}$. Let $S\in\operatorname{SL}(2,\mathbb{F}_{p})$ be the matrix:

 $S=\left(\begin{array}[]{cc}0&-1\\ 1&0\\ \end{array}\right).$

Notice that $S^{2}=-\mathit{Id}$ and so $\rho(S^{2})=(\rho(S))^{2}=\mathit{Id}=\rho(-\mathit{Id})$, as desired. ∎

Title $\mathit{SL}(2,\mathbb{F}_{p})$ has no $1$ dimensional irreducible representations over $\mathbb{F}_{p}$ mathitSL2mathbbFpHasNo1DimensionalIrreducibleRepresentationsOvermathbbFp 2013-03-22 15:09:56 2013-03-22 15:09:56 alozano (2414) alozano (2414) 7 alozano (2414) Theorem msc 20G15