# Sylow Subgroups of the Cube Group?



## Lucas Garron (Mar 23, 2010)

According to the Sylow theorems, there exists subgroups of the cube group of the following orders:

\( 2^{27}, 3^{14}, 5^3, 7^2, 11^1 \).

That's because these are the maximal prime powers in the factorization of |G| = 43252003274489856000.

Subgroups of order 11, 49, and 125 are easy to think of:
11: An 11-cycle, <(UF UL UB UR DF DL DB DR FL BL BR)>
49: Two independent 7-cycles, <(UF UL UB UR DF DL DB), (UFL UBL UBR UFR DFL DBL DBR)>
125: Three independent 5-cycles, <(UF UL UB UR DF) (DL DB DR FL BL), (UFL UBL UBR UFR DFL)>

A "3-subgroup" is also not that hard to construct, although now orientation factors in: 4 3-edge cycles, 2 3-edge cycles, 7 corners of CO freedom, which makes \( 3^{13} \). We can also permute three orbits of three-edge-cycles among each other to get up to \( 3^{14} \).


What I'm really wondering about is the group of order \( 2^{27} \). I think I can get it like this (writing it out by hand to give the idea; forgive small errors):


Spoiler



<
[Flip UF and UL], 
[Flip UF and UB], 
[Flip UF and UR], 
[Flip UF and DF], 
[Flip UF and DL], 
[Flip UF and DB], 
[Flip UF and DR], 
[Flip UF and FL], 
[Flip UF and BL], 
[Flip UF and BR], 
[Flip UF and FR],
(UF DB),
(UL DR),
(UB DF),
(UR DL),
(FL BR),
(BL FR),
(UF UB)(DB DF),
(UL UR)(DR DL),
(FL FR)(BL BR),
(UF UL)(DF DL)(UB UR)(DB DR),
(UFL UBR),
(UBL UFR),
(DBR DFL),
(DFR DBL),
(UFL UBL)(UFR UBR),
(DFL DBL)(DFR DBR),
(UFL DFL)(UBL DBL)(UBR DBR)(UFL DFL),
>


Which has 28 generators and gives us an illegal group whose even permutations form a subgroup of G of order \( 2^{27} \).

Assuming I got that right, every subgroup of order \( 2^{27} \) is conjugate to that, and there are an odd number of such subgroups (Sylow theorems 2 & 3). This doesn't sound as intuitive as I'd like: why should exactly this structure be maximal?
Has anybody else on this forum looked into this before, and has more insight?

(Sylow subgroups were on the syllabus for the 88Q class I took last year, but we didn't cover them, and Joyner doesn't mention them.)


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## Lucas Garron (Mar 23, 2010)

Lucas Garron said:


> Which has 28 generators and gives us an illegal group whose even permutations form a subgroup of G of order \( 2^{27} \).


I should probably mention that I overspecified that group. However, each of its elements is uniquely generated by performing any subset of those generators in order (provided you perform an even number of even permutations).


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