Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{Z}_{13}$).

share|cite|improve this question
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
up vote 2 down vote accepted

Yes. Explicitly one has:

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

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


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


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

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 $b=(1,2,3)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.