$\\mathit{SL}(2,\\mathbb{F}_{p})$ 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.∎

𝑆𝐿⁢(2,𝔽p) has no 1 dimensional irreducible representations over 𝔽p

Lemma.

Proof.

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