From:

~~~ Subject:

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