   Date: Mon, 23 Oct 95 11:42:32 -0400   From: michael reid <mreid@ptc.com >
~~~  Subject: Re: Embedding G in a symmetrical group

michiel boland writes

It is clear that the group G of the cube (the one with
4.3252x10^19 elements) can be embedded in a
symmetrical group, e.g. S_48, since each move of the cube can be
seen as a permutation of 48 objects. Hence, there is a smallest
number n such that G can be embedded in S_n. I'm curious to find
out what this number is.

```48.
```

first note that any homomorphism G --> S_n can be factored as

```G  -->  S_m_1 x S_m_2 x ... x S_m_k  >-->  S_n
```
```where  m_1, m_2, ... , m_k  are the sizes of the orbits of  G  acting
on  {1, 2, ... , n},  and thus  m_1 + m_2 + ... + m_k = n.
furthermore, the action of  G  on each  {1, 2, ... , m_i}  is
transitive.
```

transitive G-sets are easy to understand. for any subgroup H of G,
G acts transitively on the cosets G/H by left multiplication.
also, any transitive G-set is of this form. given a homomorphism
G --> S_m with a transitive action, let H be the subgroup of G
that fixes the element 1. then it's easy to see that the cosets G/H
are in one-to-one correspondence with elements in the orbit of 1
(which by hypothesis are all of 1, 2, ... , m) and the action of
G on G/H is isomorphic to the action of G on {1, 2, ... , m}.

the kernel of the homomorphism G --> sym(G/H) is the largest normal
subgroup of G contained in H , which is just the intersection of
all G-conjugates of H.

of course, in this case we have m = (G : H) (index of H in G).
thus michiel's question can be settled by considering all subgroups of
G with index less than 48.

unless i've overlooked some, there are exactly 8 such, up to G-conjugacy.
they are

G itself
G' = commutator subgroup of G = subgroup of positions an even
number of quarter turns from start
C_0 = subgroup where the corner UFR is in place, but may be twisted
C'_0 = commutator subgroup of C_0 = intersection of C_0 and G'
E_0 = subgroup where the edge UR is in place, but may be flipped
E'_0 = commutator subgroup of E_0 = intersection of E_0 and G'
C_1 = subgroup where the corner UFR is in place and is not twisted
E_1 = subgroup where the edge UR is in place and is not flipped.

```for each of these, except the last two, the kernel of  G --> sym(G/H)
contains all elements that only flip edges in place and twist corners
in place.
number of
subgroup   index       kernel                                   conjugates
```
```G         1           G                                           1
G'        2           G'                                          1
C_0       8      {all corners in place, may be twisted}           8
C'_0     16      {all corners in place, may be twisted}           8
E_0      12      {all edges in place, may be flipped}            12
E'_0     24      {all edges in place, may be flipped}            12
C_1      24      {all corners in place, may not be twisted}       8
E_1      24      {all edges in place, may not be flipped}        12
```

thus the only way to get and embedding (i.e. injective homomorphism)
G --> S_n using the subgroups above is

```G --> sym(G/C_1) x sym(G/E_1) >--> S_48
```

which in fact, is just the action of G on the 48 non-center facelets.

i had previously stumbled across this exact same question, so
now i'm curious: why are you interested in this?

mike     