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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?

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  • $\begingroup$ @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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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This gives a cool way of seeing why $\mathrm{SL}_2(\mathbb{F}_3)$ is called the binary tetrahedral group.

I'm going to disagree with this. While the double cover $S^3\to \mathrm{SO}(3)$ restricts to a double cover $2T\to T$, and projection gives a double cover $\mathrm{SL}_2(\mathbb{F}_3)\to\mathrm{PSL}_2(\mathbb{F}_3)$, what's missing here is an a priori reason to expect these to be the same double cover. Indeed, if the situation with $O$ is any indication, what's probably happened here is that the law of small numbers was strong enough for $2T,\mathrm{SL}_2(\mathbb{F}_3)$ to be the same double covers of $T\cong\mathrm{PSL}_2(\mathbb{F}_3)$ and $2I,\mathrm{SL}_2(\mathbb{F}_5)$ to be the same double cover of $I\cong\mathrm{PSL}_2(\mathbb{F}_5)$, it was not quite strong enough to guarantee $2O,\mathrm{GL}_2(\mathbb{F}_3)$ to be the same double cover of $O\cong\mathrm{PGL}_2(\mathbb{F}_3)$.

Note $I$ is also $\mathrm{PSL}_2(\mathbb{F}_5)\cong\mathrm{PSL}(\mathbb{F}_4)=\mathrm{PGL}_2(\mathbb{F}_4)$, but $\mathrm{SL}_2(\mathbb{F}_4)$ and $\mathrm{GL}_2(\mathbb{F}_4)$ are not $2I$ (they are $1$-fold and $3$-fold covers respectively). This is because $|\mathbb{F}_4^{\times}|=3$ though, whereas $|\mathbb{F}_3^{\times}|=2$ so the failure of $2O$ to be the same as $\mathrm{GL}_2(\mathbb{F}_3)$ is indeed less obvious.

The action $\mathrm{PSL}_2(\mathbb{F}_4)\curvearrowright\mathbb{P}^1(\mathbb{F}_4)$ matches that of $A_5\curvearrowright\{1,\cdots,5\}$ but the action of $\mathrm{PSL}_2(\mathbb{F}_5)$ on the projective line $\mathbb{P}^1(\mathbb{F}_5)$ gives an exotic, transitive copy of $A_5$ within $S_6$. (Recall the usual copy of $A_5\subset S_6$ has a fixed point so isn't transitive.) It is the image of the usual $A_5$ under the unique nontrivial outer automorphism in $\mathrm{Out}(S_6)\cong\mathbb{Z}_2$.

...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?

The binary covers of symmetry groups are represented by quaternions, which may be represented using $2\times 2$ complex matrices, as $S^3\cong\mathrm{SU}(2)$. Moreover, the components (of the complex matrices) are algebraic numbers for the group elements, which means we ought to be able to mod out by certain primes to get elements of matrix groups over finite fields.

With the quaternion group $Q_8$ if we add in the elements $\frac{1}{2}(\pm1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k})$ we get the binary tetrahedral group $2T$, the vertices of the $24$-cell, and then if we add in the normalized vertices of the dual $24$-cell, permutations of $\frac{1}{\sqrt{2}}(\pm\mathbf{i}\pm\mathbf{j})$, we get the binary octahedral group $2O$ of order $48$. The field $\mathbb{F}_3$ is the smallest field in which $\pm1$ are distinct, so it is a good choice to mod the corresponding $2\times2$ complex matrices by $3$.

Of course, $-1=\frac{1}{2}$ has no square root in $\mathbb{F}_3$, so we must go to $\mathbb{F}_{3^2}=\mathbb{F}_3(i)$ to represent these matrices; checking by hand I get $\overline{2T}=\mathrm{SU}_2(\mathbb{F}_{3^2})$. The space $\mathbb{F}_{3^2}$ admits a sesquilinear form using complex conjugation and there is thus a special unitary group, just like over $\mathbb{C}$. In fact, the containment $\overline{2T}\subseteq\mathrm{SU}_2(\mathbb{F}_{3^2})$ should be clear, because the equations defining special unitary matrices are preserved under modding; checking sizes and the kernel proves it's an equality.

[We can also take the elements of $2I$, turn them into $2\times2$ complex matrices in the standard way (since $S^3\cong\mathrm{SU}(2)$) and then mod by $5$ to get elements of $\mathrm{SL}_2(\mathbb{F}_5)$. Since $-1=2^2$ in $\mathbb{F}_3$, the $i$s in the matrices become $2$, and the determinant is preserved by modding. (Note $5$ is the smallest prime we can mod by in which $0,\pm1,\pm i$ are all distinct.) Also the $\sqrt{5}$s in the golden ratio just drop out. By checking sizes and kernels we see that $\overline{2T}=\mathrm{SL}_2(\mathbb{F}_5)$. Since the kernel of $2T\to T$ (modded out) matches the kernel of $\mathrm{SL}_2(\mathbb{F}_5)\to\mathrm{PSL}_2(\mathbb{F}_5)$, we may also conclude $I\cong\mathrm{PSL}_2(\mathbb{F}_5)$.

We can't do anything similar to this or the below for $I\cong\mathrm{PSL}_2(\mathbb{F}_4)$ since $2$ is not invertible in $\mathbb{F}_4$.]

Notice the $(2,0)$ and $(1,1)$ signature forms over $\mathbb{F}_{3^2}$ are equivalent since $-1=1^2+1^2$, so

$$ \overline{x_1}x_2+\overline{y_1}y_2=\overline{x_1}x_2-\overline{(1+i)y_1}\,(1+i)y_2 $$

Therefore we may conclude

$$ \tau\mathrm{SU}_2(\mathbb{F}_{3^2})\tau^{-1}=\mathrm{SU}_{1,1}(\mathbb{F}_{3^2}), \qquad \tau=\begin{bmatrix} 1 & 0 \\ 0 & 1+i \end{bmatrix} $$

Moreover, due to the so-called Cayley transform $g$ we know

$$ \mathrm{SL}_2(\mathbb{F}_3)=g^{-1}\mathrm{SU}_{1,1}(\mathbb{F}_{3^2})g, \qquad g=\begin{bmatrix} 1 & -i \\ 1 & i \end{bmatrix}$$

Thus, $\overline{2T}=\mathrm{SU}_2(\mathbb{F}_{3^2})$ is conjugate to $\mathrm{SL}_2(\mathbb{F}_3)$ within $\mathrm{GL}_2(\mathbb{F}_{3^2})$ via $g^{-1}\tau$.

The elements of $\overline{2O}$ unaccounted for by $\overline{2T}$ are a single other coset; a representative e.g. $\frac{1}{\sqrt{2}}(a+b\mathbf{i})$ of $2O\setminus 2T$ would correspond (note $\frac{1}{\sqrt{2}}\in\mathbb{F}_{3^2}$) to

$$X=\begin{bmatrix} a+bi & 0 \\ 0 & -a+bi \end{bmatrix}.$$

If we conjugate this (with $a=b=1$) by $g^{-1}\tau$ we get the matrix

$$ (g^{-1}\tau)X(g^{-1}\tau)=i\begin{bmatrix} 1 & -1 \\ 1 & 1\end{bmatrix}=iY $$

Notice $\det Y=-1$ so we may conclude

$$ \begin{array}{ccccr} \mathrm{GL}_2(\mathbb{F}_3) & = & \mathrm{SL}_2(\mathbb{F}_3) & \sqcup & [\mathrm{GL}_2(\mathbb{F}_3)\setminus\mathrm{SL}_2(\mathbb{F}_3)], \\ \overline{2O} & \sim & \mathrm{SL}_2(\mathbb{F}_3) & \sqcup & i[\mathrm{GL}_2(\mathbb{F}_3)\setminus\mathrm{SL}_2(\mathbb{F}_3)]. \end{array} $$

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  • $\begingroup$ As $\overline{2T}=\mathrm{SU}_2(\mathbb{F}_{3^2})$, I wonder if we can describe $\overline{2O}$ in some standard Lie-type way. $\endgroup$ – runway44 May 13 '20 at 16:31

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