From mreid@ptc.com Mon Oct 23 11:20:02 1995 Return-Path: Received: from ptc.com (poster.ptc.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08395; Mon, 23 Oct 95 11:20:02 EDT Received: from ducie.ptc.com by ptc.com (5.x/SMI-SVR4-NN) id AA23979; Mon, 23 Oct 1995 11:15:45 -0400 Message-Id: <9510231515.AA23979@ptc.com> Received: by ducie.ptc.com (1.38.193.4/16.2.nn) id AA15699; Mon, 23 Oct 1995 11:42:32 -0400 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