# Projective Tetrahedral Representation

I can embed $A_4$ as a subgroup into $PSL_2(\mathbb{F}_{13})$ (in two different ways in fact). I also have a reduction mod 13 map $$PGL_2(\mathbb{Z}_{13}) \to PGL_2(\mathbb{F}_{13}).$$ My question is:

Is there a subgroup of $PGL_2(\mathbb{Z}_{13})$ which maps to my copy of $A_4$ under the above reduction map?

(I know that one may embed $A_4$ into $PGL_2(\mathbb{C})$, but I don't know about replacing $\mathbb{C}$ with $\mathbb{Z}_{13}$).

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Yes, I believe this follows from "Fong-Swan" in some more generality, since A4 is 13-solvable. There are three problems you run into: does the rep lift to char 0 (yes, by Maschke for instance in this case), does the rep lift to that particular field (Schur indices and trace fields; I think everything is fine here), is there an integral version of the rep (yes, because Z13 is a local PID). Fong-Swan will still work when the characteristic of the field divides the order of the group. – Jack Schmidt Sep 21 '11 at 15:06

Yes. Explicitly one has:

$\newcommand{\ze}{\zeta_3} \newcommand{\zi}{\ze^{-1}} \newcommand{\vp}{\vphantom{\zi}} \newcommand{\SL}{\operatorname{SL}} \newcommand{\GL}{\operatorname{GL}}$

$$\SL(2,3) \cong G_1 = \left\langle \begin{bmatrix} 0 & 1 \vp \\ -1 & 0 \vp \end{bmatrix}, \begin{bmatrix} \ze & 0 \\ -1 & \zi \end{bmatrix} \right\rangle \cong G_2 = \left\langle \begin{bmatrix} 0 & 1 \vp \\ -1 & 0 \vp \end{bmatrix}, \begin{bmatrix} 0 & -\zi \\ 1 & -\ze \end{bmatrix} \right\rangle$$

and

$$G_1 \cap Z(\GL(2,R)) = G_2 \cap Z(\GL(2,R)) = Z = \left\langle\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\right\rangle \cong C_2$$

and

$$G_1/Z \cong G_2/Z \cong A_4$$

This holds over any ring R which contains a primitive 3rd root of unity, in particular, in the 13-adics, $\mathbb{Z}_{13}$. The first representation has rational (Brauer) character and Schur index 2 over $\mathbb{Q}$ (but Schur index 1 over the 13-adics $\mathbb{Q}_{13}$), and the second representation is the unique (up to automorphism of $A_4$) 2-dimensional projective representation of $A_4$ with irrational (Brauer) character.

You can verify that if $G_i = \langle a,b\rangle$, then $a^2 = [a,a^b] = -1$, $a^{(b^2)} = aa^b$, and $b^3 = 1$. Modulo $-1$, one gets the defining relations for $A_4$ on $a=(1,2)(3,4)$ and $b=(1,2,3)$.

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