In this answer, it is mentioned that the binary octahedral group can be realized as $\mathrm{GL}_2(\mathbb{F}_3)$, with "certain elements replaced with scalar multiples in $\mathrm{GL}_2(\mathbb{F}_9)$." (Apparently this has been called the "fake $\mathrm{GL}_2(\mathbb{F}_3)$" by Marty Isaacs.) What are the details of that construction?

In particular, I'd like to see an analogy to (or at least a spin-off of) the following construction that we have for the binary tetrahedral group. The isometries of the regular tetrahedron can be given by permutations of its four vertices. If we look at the action by $\mathrm{SL}_2(\mathbb{F}_3)$ on $\mathbb{P}^1(\mathbb{F}_3)$, we can interpret it as giving a map $\mathrm{SL}_2(\mathbb{F}_3)\rightarrow S_4$ by seeing what it does to the points $\{0,1,2,\infty\}\in\mathbb{P}^1(\mathbb{F}_3)$ (represented here by $\begin{pmatrix} 0\\ 1 \end{pmatrix}$, $\begin{pmatrix} 1\\ 1 \end{pmatrix}$, $\begin{pmatrix} 2\\ 1 \end{pmatrix}$, $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ respectively). This map is surjective and $2$-to-$1$ with kernel $\bigg\{\begin{pmatrix} \pm 1 & 0\\ 0 & \pm 1 \end{pmatrix}\bigg\}$, showing that $\mathrm{PSL}_2(\mathbb{F}_3)$ is isomorphic to the tetrahedral group. This gives a cool way of seeing why $\mathrm{SL}_2(\mathbb{F}_3)$ is called the binary tetrahedral group.

We have a similar thing for the binary icosahedral group and $\mathrm{SL}_2(\mathbb{F}_5)$, but how do we see something similar for the binary octahedral group?

  • $\begingroup$ @AlexanderGruber Do you happen to know the answer to this? The answer of yours that I linked is the only place I've seen the idea mentioned. $\endgroup$ – j0equ1nn Feb 23 '17 at 21:28

I think the concept you are looking for is that of a group extension.

In the question, you describe $PSL_{2}(3)$ as the quotient of $SL_{2}(3)$ by scalar matrices. More generally, for a given group $G$ which has a normal subgroup $N$, there is a unique quotient $G/N$. The isomorphism theorems provide various pieces of information about the quotient in terms of $G$.

The converse problem is: given groups $K$ and $N$, find all groups $G$ such that $G$ contains a subgroup $N'$ isomorphic to $N$, and such that the quotient $G/N'$ is isomorphic to $K$. Such a group $G$ is called an extension of $K$ by $N$. With appropriate conditions on $N$ and $N'$, all possible groups $G$ are classified by the second cohomology group $H^{2}(K,N)$.

As an illustration: it's an easy exercise to see that both the quaternion group of order $8$ and the dihedral group of order $8$ are extensions of the Klein $4$-group by a cyclic group of order $2$. (You can take $N'$ to be the centre in each case.)

To address your question: both the binary octahedral group and $GL_{2}(3)$ are extensions of the symmetric group $S_{4}$ by a cyclic group of order $2$. So they look similar, but they are not isomorphic. (For example, the Sylow 2-subgroups have different numbers of involutions.) My guess is that this is why Isaacs called the binary octahedral group a fake $GL_{2}(3)$: it can be described as an extension of $PGL_{2}(3)$ by $Z(GL_{2}(3))$, but it's not the group you expect.

  • $\begingroup$ I'm with you on this, but I still don't see how the desired extension is attained via replacing some elements of $\mathrm{GL}_2(\mathbb{F}_3)$ with "scalar multiples in $\mathrm{GL}_2(\mathbb{F}_9)$," or even what that means exactly. $\endgroup$ – j0equ1nn Mar 3 '17 at 4:32
  • $\begingroup$ This is a only a guess, but you could check with some computer algebra package: the binary octahedral group could have a conjugacy class of faithful representations in $GL_{2}(9)$, none of which are contained in $GL_{2}(3)$. Perhaps Isaacs found a nice representative of this conjugacy class, where the matrices look almost like $GL_{2}(3)$, but with some matrix elements drawn from the larger field? $\endgroup$ – Padraig Ó Catháin Mar 3 '17 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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