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Date: Mon, 23 Oct 1995 11:42:32 -0400
From: michael reid
To: boland@sci.kun.nl, cube-lovers@ai.mit.edu
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