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Q&A Canonical Homomorphism $G_{2\times 2} \to \mathrm{PGL}(3,2)$?

posted 5mo ago by clemens‭ · edited 5mo ago by clemens‭

Answer

#4: Post edited by $user avatar$ clemens‭ · 2026-04-18T22:57:45Z (5 months ago)

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At the other website, Jim Belk [writes](https://math.stackexchange.com/a/563194/) that the symmetry group of the $2×2×2$ cube can be defined thus (in GAP):
```
cube := Group( (2,14,23,9)(3,6,22,17)(7,15,16,8), (3,11,24,16)(4,19,23,8)(9,10,18,17), (20,14,16,18)(19,13,15,17)(21,22,23,24), (1,2,3,4)(5,7,9,11)(6,8,10,12), (1,7,22,20)(2,15,21,12)(5,6,14,13), (1,10,24,13)(4,18,21,5)(11,19,20,12) );
```
The command
```
StructureDescription(cube)
```
gives
```
(C3 x C3 x C3 x C3 x C3 x C3 x C3) : S8
```
in other words the group is a semidirect product $C_3^7 \rtimes S_8$.
Revised ending
---
Jim Belk found that
```
Stabilizer(cube,[1,2,3,4,21,22,23,24],OnSets))
```
is isomorphic to $S_8$.
Hence the Rubik's-cube group contains a large simple subgroup isomorphic to $A_8$, which we can find by squaring the elements of the group we found above:
```
a8_instance := Group(List(Elements(Stabilizer(cube,[1,2,3,4,21,22,23,24],OnSets)), x -> x^2));
```
(Again, use `StructureDescription` to verify this.)
Clearly this subgroup has only the trivial homomorphism to $PGL(3,2)$ (as the domain and codomain are simple). So all the elements of `a8_instance` must be mapped to $e$. But this means that all the elements of `NormalClosure(cube,a8_instance)` must be mapped to $e$. But since the normal closure of `a8_instance` is just the index-2 subgroup of the Rubik's-cube group (i.e. $C_3^7 \rtimes A_8$) we see that $C_3^7 \rtimes A_8$ is in the kernel of any homomorphism $C_3^7 \rtimes A_8 \rightarrow PGL(3,2).$
One can also see this easily without GAP from a statement that Jim Belk made: namely that the normal subgroup $C_3^7$ of the Rubik's-cube group is the unique $7$-dimensional subgroup of $C_3^8$ that is stabilized by $S_8$ under the obvious action. After thinking a little it becomes apparent that there is no way to map any elements of this $C_3^7$ to non-identity elements of $PGL(3,2)$.
Hence (as we saw) the index-2 subgroup of the Rubik's-cube group must be mapped to the identity; its coset can be mapped to any non-identity involution or of course to the identity itself. And this suffices to classify the homomorphisms from the Rubik's-cube group to $PGL(3,2)$.
StructureDescription(NormalClosure(cube,a8_instance)));
Original ending (excised)
---
(My answer originally ended thus. Unfortunately I found it answers a different question, about finding a homomorphism from the simple group of order 168 to the Rubik's-cube group, rather than the other way around. I'm still keeping this part of the answer because the latter problem is somewhat interesting and is more likely for people to want to know.)
Since $PGL(3,2)$ is simple a homomorphism to $C_3^7 \wr S_8$ therefrom must factorize into $C_3^7 \wr S_8 \xleftarrow{f} S_8 \xleftarrow{g} PGL(3,2)$. But I believe there is no "canonical" way to choose either $f$ or $g$.

At the other website, Jim Belk [writes](https://math.stackexchange.com/a/563194/) that the symmetry group of the $2×2×2$ cube can be defined thus (in GAP):
```
cube := Group( (2,14,23,9)(3,6,22,17)(7,15,16,8), (3,11,24,16)(4,19,23,8)(9,10,18,17), (20,14,16,18)(19,13,15,17)(21,22,23,24), (1,2,3,4)(5,7,9,11)(6,8,10,12), (1,7,22,20)(2,15,21,12)(5,6,14,13), (1,10,24,13)(4,18,21,5)(11,19,20,12) );
```
The command
```
StructureDescription(cube)
```
gives
```
(C3 x C3 x C3 x C3 x C3 x C3 x C3) : S8
```
in other words the group is a semidirect product $C_3^7 \rtimes S_8$.
Revised ending
---
Jim Belk found that
```
Stabilizer(cube,[1,2,3,4,21,22,23,24],OnSets))
```
is isomorphic to $S_8$.
Hence the Rubik's-cube group contains a large simple subgroup isomorphic to $A_8$, which we can find by squaring the elements of the group we found above:
```
a8_instance := Group(List(Elements(Stabilizer(cube,[1,2,3,4,21,22,23,24],OnSets)), x -> x^2));
```
(Again, use `StructureDescription` to verify this.)
Clearly this subgroup has only the trivial homomorphism to $PGL(3,2)$ (as the domain and codomain are simple). So all the elements of `a8_instance` must be mapped to $e$. But this means that all the elements of `NormalClosure(cube,a8_instance)` must be mapped to $e$. But since the normal closure of `a8_instance` is just the index-2 subgroup of the Rubik's-cube group (i.e. $C_3^7 \rtimes A_8$) we see that $C_3^7 \rtimes A_8$ is in the kernel of any homomorphism $C_3^7 \rtimes A_8 \rightarrow PGL(3,2).$
One can also see this easily without GAP from a statement that Jim Belk made: namely that the normal subgroup $C_3^7$ of the Rubik's-cube group is the unique $7$-dimensional subgroup of $C_3^8$ that is stabilized by $S_8$ under the obvious action. After thinking a little it becomes apparent that there is no way to map any elements of this $C_3^7$ to non-identity elements of $PGL(3,2)$.
Hence (as we saw) the index-2 subgroup of the Rubik's-cube group must be mapped to the identity; its coset can be mapped to any non-identity involution or of course to the identity itself. And this suffices to classify the homomorphisms from the Rubik's-cube group to $PGL(3,2)$.
StructureDescription(NormalClosure(cube,a8_instance)));
Original ending (excised)
---
(My answer originally ended thus. Unfortunately I found it answers a different question, about finding a homomorphism from the simple group of order 168 to the Rubik's-cube group, rather than the other way around. I'm still keeping this part of the answer because the latter problem is somewhat interesting and is more likely for people to want to know.)
Since $PGL(3,2)$ is simple a homomorphism to $C_3^7 \rtimes S_8$ therefrom must factorize into $C_3^7 \rtimes S_8 \xleftarrow{f} S_8 \xleftarrow{g} PGL(3,2)$. But I believe there is no "canonical" way to choose either $f$ or $g$.

#3: Post edited by $user avatar$ clemens‭ · 2026-04-18T19:45:34Z (5 months ago)
Corrected to classify all homomorphisms from the Rubik's-cube group to PGL(3,2).

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(Edit: sorry, this answers a different question, about finding a homomorphism from the simple group of order 168 to the Rubik's-cube group, rather than the other way around. I'm still keeping this answer because the latter problem is somewhat interesting and is more likely for people to want to know.)
At the other website, Jim Belk [writes](https://math.stackexchange.com/a/563194/) that the symmetry group of the $2×2×2$ cube can be defined thus (in GAP):
```
cube := Group( (2,14,23,9)(3,6,22,17)(7,15,16,8), (3,11,24,16)(4,19,23,8)(9,10,18,17), (20,14,16,18)(19,13,15,17)(21,22,23,24), (1,2,3,4)(5,7,9,11)(6,8,10,12), (1,7,22,20)(2,15,21,12)(5,6,14,13), (1,10,24,13)(4,18,21,5)(11,19,20,12) );
```
The command
```
StructureDescription(cube)
```
gives
```
(C3 x C3 x C3 x C3 x C3 x C3 x C3) : S8
```
in other words the group is a semidirect product $C_3^7 \wr S_8$.
Since $GF(3,2)$ is simple a homomorphism to $C_3^7 \wr S_8$ therefrom must factorize into $C_3^7 \wr S_8 \xleftarrow{f} S_8 \xleftarrow{g} GF(3,2)$. But I believe there is no "canonical" way to choose either $f$ or $g$.

At the other website, Jim Belk [writes](https://math.stackexchange.com/a/563194/) that the symmetry group of the $2×2×2$ cube can be defined thus (in GAP):
```
cube := Group( (2,14,23,9)(3,6,22,17)(7,15,16,8), (3,11,24,16)(4,19,23,8)(9,10,18,17), (20,14,16,18)(19,13,15,17)(21,22,23,24), (1,2,3,4)(5,7,9,11)(6,8,10,12), (1,7,22,20)(2,15,21,12)(5,6,14,13), (1,10,24,13)(4,18,21,5)(11,19,20,12) );
```
The command
```
StructureDescription(cube)
```
gives
```
(C3 x C3 x C3 x C3 x C3 x C3 x C3) : S8
```
in other words the group is a semidirect product $C_3^7 \rtimes S_8$.
Revised ending
---
Jim Belk found that
```
Stabilizer(cube,[1,2,3,4,21,22,23,24],OnSets))
```
is isomorphic to $S_8$.
Hence the Rubik's-cube group contains a large simple subgroup isomorphic to $A_8$, which we can find by squaring the elements of the group we found above:
```
a8_instance := Group(List(Elements(Stabilizer(cube,[1,2,3,4,21,22,23,24],OnSets)), x -> x^2));
```
(Again, use `StructureDescription` to verify this.)
Clearly this subgroup has only the trivial homomorphism to $PGL(3,2)$ (as the domain and codomain are simple). So all the elements of `a8_instance` must be mapped to $e$. But this means that all the elements of `NormalClosure(cube,a8_instance)` must be mapped to $e$. But since the normal closure of `a8_instance` is just the index-2 subgroup of the Rubik's-cube group (i.e. $C_3^7 \rtimes A_8$) we see that $C_3^7 \rtimes A_8$ is in the kernel of any homomorphism $C_3^7 \rtimes A_8 \rightarrow PGL(3,2).$
One can also see this easily without GAP from a statement that Jim Belk made: namely that the normal subgroup $C_3^7$ of the Rubik's-cube group is the unique $7$-dimensional subgroup of $C_3^8$ that is stabilized by $S_8$ under the obvious action. After thinking a little it becomes apparent that there is no way to map any elements of this $C_3^7$ to non-identity elements of $PGL(3,2)$.
Hence (as we saw) the index-2 subgroup of the Rubik's-cube group must be mapped to the identity; its coset can be mapped to any non-identity involution or of course to the identity itself. And this suffices to classify the homomorphisms from the Rubik's-cube group to $PGL(3,2)$.
StructureDescription(NormalClosure(cube,a8_instance)));
Original ending (excised)
---
(My answer originally ended thus. Unfortunately I found it answers a different question, about finding a homomorphism from the simple group of order 168 to the Rubik's-cube group, rather than the other way around. I'm still keeping this part of the answer because the latter problem is somewhat interesting and is more likely for people to want to know.)
Since $PGL(3,2)$ is simple a homomorphism to $C_3^7 \wr S_8$ therefrom must factorize into $C_3^7 \wr S_8 \xleftarrow{f} S_8 \xleftarrow{g} PGL(3,2)$. But I believe there is no "canonical" way to choose either $f$ or $g$.

#2: Post edited by $user avatar$ clemens‭ · 2026-04-18T18:48:08Z (5 months ago)
Corrected directions of morphisms

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At the other website, Jim Belk [writes](https://math.stackexchange.com/a/563194/) that the symmetry group of the $2×2×2$ cube can be defined thus (in GAP):
```
cube := Group( (2,14,23,9)(3,6,22,17)(7,15,16,8), (3,11,24,16)(4,19,23,8)(9,10,18,17), (20,14,16,18)(19,13,15,17)(21,22,23,24), (1,2,3,4)(5,7,9,11)(6,8,10,12), (1,7,22,20)(2,15,21,12)(5,6,14,13), (1,10,24,13)(4,18,21,5)(11,19,20,12) );
```
The command
```
StructureDescription(cube)
```
gives
```
(C3 x C3 x C3 x C3 x C3 x C3 x C3) : S8
```
in other words the group is a semidirect product $C_3^7 \wr S_8$.
Since $GF(3,2)$ is simple a homomorphism from $C_3^7 \wr S_8$ thereto must factorize into $C_3^7 \wr S_8 \xrightarrow{f} S_8 \xrightarrow{g} GF(3,2)$. But I believe there is no "canonical" way to choose either $f$ or $g$.

(Edit: sorry, this answers a different question, about finding a homomorphism from the simple group of order 168 to the Rubik's-cube group, rather than the other way around. I'm still keeping this answer because the latter problem is somewhat interesting and is more likely for people to want to know.)
At the other website, Jim Belk [writes](https://math.stackexchange.com/a/563194/) that the symmetry group of the $2×2×2$ cube can be defined thus (in GAP):
```
cube := Group( (2,14,23,9)(3,6,22,17)(7,15,16,8), (3,11,24,16)(4,19,23,8)(9,10,18,17), (20,14,16,18)(19,13,15,17)(21,22,23,24), (1,2,3,4)(5,7,9,11)(6,8,10,12), (1,7,22,20)(2,15,21,12)(5,6,14,13), (1,10,24,13)(4,18,21,5)(11,19,20,12) );
```
The command
```
StructureDescription(cube)
```
gives
```
(C3 x C3 x C3 x C3 x C3 x C3 x C3) : S8
```
in other words the group is a semidirect product $C_3^7 \wr S_8$.
Since $GF(3,2)$ is simple a homomorphism to $C_3^7 \wr S_8$ therefrom must factorize into $C_3^7 \wr S_8 \xleftarrow{f} S_8 \xleftarrow{g} GF(3,2)$. But I believe there is no "canonical" way to choose either $f$ or $g$.

#1: Initial revision by $user avatar$ clemens‭ · 2026-04-11T22:43:54Z (5 months ago)

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At the other website, Jim Belk [writes](https://math.stackexchange.com/a/563194/) that the symmetry group of the $2×2×2$ cube can be defined thus (in GAP):

```
cube := Group( (2,14,23,9)(3,6,22,17)(7,15,16,8), (3,11,24,16)(4,19,23,8)(9,10,18,17), (20,14,16,18)(19,13,15,17)(21,22,23,24), (1,2,3,4)(5,7,9,11)(6,8,10,12), (1,7,22,20)(2,15,21,12)(5,6,14,13), (1,10,24,13)(4,18,21,5)(11,19,20,12) );
```

The command

```
StructureDescription(cube)
```

gives

```
(C3 x C3 x C3 x C3 x C3 x C3 x C3) : S8
```

in other words the group is a semidirect product $C_3^7 \wr S_8$.

Since $GF(3,2)$ is simple a homomorphism from $C_3^7 \wr S_8$ thereto must factorize into $C_3^7 \wr S_8 \xrightarrow{f} S_8 \xrightarrow{g} GF(3,2)$. But I believe there is no "canonical" way to choose either $f$ or $g$.

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