# Message #43

From: mahdeltaphi <mark.hennings@ntlworld.com>

Subject: Re: [MC4D] Orientations of the centre cubes …

Date: Wed, 10 Sep 2003 11:36:13 -0000

>Working on a uniformly coloured 3x3x3 cube or tesseract is not

>solving the full symmetry group. The subgroup of the full symmetry

>group which fixes the colours, but ignores orientations, is a normal

>subgroup of the full group, and the quotient group of the full

>symmetry group by this normal subgroup is the group that is being

>studied when working with a uniformly coloured cube/tesseract.

>Although nothing like as big as the full symmetry group, the colour-

>preserving subgroup is nonetheless respectably large, and probably

>deserves some consideration.

David’s comment:

>I think there may be a slight ambiguity in the above

>which I would like to understand properly. There are

>two sorts of relevant orientation issues for a given

>hyper-cubie - that which is forced by the stickers on

>it (the obvious orientation issue) and that for the

>axes for which the hyper-cubie has no stickers (the

>unobvious one applying to 1- and 2-color

>hyper-cubies). So there are two levels at which one

>may "ignore" orientations.

Yes, there are two types of orientation, which could be classed

as "visible" and "invisible" (on a uniformly coloured tesseract). A

cubie sharing 2,3 or 4 faces can be in the "correct position" (its

2,3 or 4 colours match the colours of the centre cubies of the faces

to which it belongs, but in an incorrect orientation in that

position. This sort of orientation problem is, of course, a standard

component of solving the tesseract as it currently is. We could call

these "visible orientation" issues. Like David, when solving the

tesseract I first place the cubies in the correct positions, and

then fix their visible orientations (well, I normally do this first

for the 2-face cubies, then the 3-face cubies, and finally for the 4-

face cubies - it is much easier to solve the 2-face cubies if you do

not have to worry about what you are doing to the 3- and 4-face

cubies, and similarly it is simpler to solve the 3-face cubies if

you only have to worry about not mucking up the 2-face cubies, and

do not yet care about the 4-face ones - solving the whole position

plus visible orientation problem for each level of cubie before

proceeding to the next stage makes it easier to see what you are

doing!).

The second type of orientation, the "invisible" type, relates only

to those cubies which share 1 or 2 faces of the tesseract. These

cubies can be in the right position (the 1-face cubies have to be!),

and (in the case of the 2-face cubies) can be in the right "visible"

orientation with respect to the faces to which they belong. However,

they can still be in a number of different orientations themselves.

It is currently impossible to see whether, for example, a 2-face

cubie has been given a quarter twist along the axis running through

the centre cubies of the faces to which it belongs or not.

Similarly, it is possible for the centre cubie of a face of the

tesseract to be rotated, without affecting any other part of that

face.

When I previously referred to "orientations", I meant "invisible

orientations". Thus a colour-preserving permutation of the tesseract

is which which puts all the cubies in their correct positions and in

their correct visible orientations. If a colour-preserving

orientation of the cube is performed then a uniformly coloured

tesseract, which started by looking solved, will still look solved.

Start with a clean cube, perform a colour-preserving permutation

(such as Q F(11) Q^{-1} F(11)^{-1}), and the programme will give you

a congratulatory beep!

There is, as David suggests, yet another subgroup of the full

permutation group, namely the one that comprises all orientation-

adjusting permutations, both visible and invisible. How you choose

to describe it is up to you. We could let G be the full permutation

group of the tesseract, and then let C be the (normal) colour-

preserving subgroup of "invisible orientation" permutations. The

quotient group G/C would then be the permutation group of the

uniformly coloured tesseract. We could also consider the normal

subgroup P of G consisting of those permutations of the tesseract

which fix the positions of all cubies, but can alter both visible

and invisible orientations. The colour-preserving subgroup C is of

course itself a normal subgroup of P. We could then consider the

group G/P, which is the symmetry group of the positions of the

cubies alone. Alternatively, we could consider the quotient group

P/C, which is the permutation group of the visible orientations of

the uniformly coloured tesseract. By the Third Isomorphism Theorem,

P/C is a normal subgroup of G/C, and the quotient group (G/C)/(P/C)

is isomorphic to the group G/P. We would interpret (G/C)/(P/C) as

the positional symmetry group of the uniformly coloured tesseract.

Reassuringly, the Third Isomorphism Theorem tells us that it does

not matter whether we start with a uniformly coloured tesseract or a

fully marked one - if we are only interested in positional issues,

it does not matter if we choose to disregard orientations in two

stages (first the invisible, then the visible) or all in one go!

Although respectably large, the position-fixing permutation group P

is still much smaller than the full symmetry group G and so, as

experience shows when solving the tesseract, dealing with the

positional permutation group G/P takes up the lion’s share of

solution time!

Mark