$\mathrm{𝑆𝐿}(2,{𝔽}_p)$ has no $1$ dimensional irreducible representations over ${𝔽}_p$

Lemma.

The group $G\mathrm{=}\mathrm{SL}\mathit{}\mathrm{(}\mathrm{2}\mathrm{,}{\mathrm{F}}_p\mathrm{)}$ has no non-trivial $\mathrm{1}$ dimensional irreducible representations over ${\mathrm{F}}_p$.

Proof.

Notice that a $1$ dimensional representations over ${𝔽}_p$ is just a homomorphism:

$$\rho :\mathrm{SL}(2,{𝔽}_p)⟶{𝔽}_p^{\times }.$$

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

$$\overline{\rho }:\mathrm{PSL}(2,{𝔽}_p)⟶{𝔽}_p^{\times }/\{\pm \mathrm{𝐼𝑑}\}$$

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

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

$$\rho :\mathrm{SL}(2,{𝔽}_p)⟶\{\pm \mathrm{𝐼𝑑}\}$$

satisfies $\rho (-\mathrm{𝐼𝑑})=\mathrm{𝐼𝑑}$. Let $S\in \mathrm{SL}(2,{𝔽}_p)$ be the matrix:

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

Title	$\mathrm{𝑆𝐿}(2,{𝔽}_p)$ has no $1$ dimensional irreducible representations over ${𝔽}_p$
Canonical name	mathitSL2mathbbFpHasNo1DimensionalIrreducibleRepresentationsOvermathbbFp
Date of creation	2013-03-22 15:09:56
Last modified on	2013-03-22 15:09:56
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	7
Author	alozano (2414)
Entry type	Theorem
Classification	msc 20G15