# I'm writing a group theory essay

#### cuBerBruce

##### Member
Hm, so I have a conjecture which I'm not keen to focus on because it's not very interesting, but just out of curiosity is it true that the group of all the states reachable by peeling off the stickers and sticking back on them anywhere is isomorphic to S56/S9? I think the cardinality of that group is right and there's what appears to me to be an intuitive isomorphism but I haven't tried proving it at all.

FWIW by S56 I mean the Symmetric Group on 56 letters.
If you consider the unstickered cubies and their orientations to be distinguishable, I believe there are $$54!/(9!^6)$$ possible stickerings. If you consider the orientation of the cube not to matter, it would be approximately 24 times fewer (although I believe the exact number to be rather tricky to calculate). As with the 4x4x4 cube with its indistinguishable pieces, these arrangements of indistinguishable stickers do not form a group. Also S56/S9 (or S54/S9, if that's what you meant) is not a group, since S9 is not a normal subgroup of S56 (or S54). Even though those coset spaces are not groups, we may still be able to talk about an isomorphism, provided the the number of elements matched (although that is not the case here either).

It might also be a somewhat interesting question to determine the total number of possible stickerings if we allow stickerings to be considered equivalent if one can be converted to the other by some element of <U,x,y>. One might also allow stickerings to be considered equivalent if one can be converted to the other by disassembling and reassembling the cube. (I am assuming the stickers to have square symmetry, are a solid color with no markings, and are divided into 6 sets of 9 indistinguishable stickers.)

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#### Jude

##### Member
If you consider the unstickered cubies and their orientations to be distinguishable, I believe there are $$54!/(9!^6)$$ possible stickerings. If you consider the orientation of the cube not to matter, it would be approximately 24 times fewer (although I believe the exact number to be rather tricky to calculate). As with the 4x4x4 cube with its indistinguishable pieces, these arrangements of indistinguishable stickers do not form a group. Also S56/S9 (or S54/S9, if that's what you meant) is not a group, since S9 is not a normal subgroup of S56 (or S54). Even though those coset spaces are not groups, we may still be able to talk about an isomorphism, provided the the number of elements matched (although that is not the case here either).

It might also be a somewhat interesting question to determine the total number of possible stickerings if we allow stickerings to be considered equivalent if one can be converted to the other by some element of <U,x,y>. One might also allow stickerings to be considered equivalent if one can be converted to the other by disassembling and reassembling the cube. (I am assuming the stickers to have square symmetry, are a solid color with no markings, and are divided into 6 sets of 9 indistinguishable stickers.)
Excellent post, everything there makes perfect sense and sounds spot on to me! Cheers!

#### TMOY

##### Member
If you consider the unstickered cubies and their orientations to be distinguishable, I believe there are $$54!/(9!^6)$$ possible stickerings. If you consider the orientation of the cube not to matter, it would be approximately 24 times fewer (although I believe the exact number to be rather tricky to calculate).
Actually it's not that hard. Since there are 9 undistinguishable sticers of each color and 24 possible orientations, the order of the group of symmetries of any given position must divide gcd(9,24)=3, hence the only possible symmetries are invariances by 120-degree rotation around an axis going through two opposite corners. A position satisfying that condition is entirely determined by the choice of a pair of opposite corners and the way some given pair of opposite faces is stickered; moreover, such a pair of faces must contain exactly 3 stickers of each color, hence $$18!/(3!^2)$$ possiblities. Now if you consider positions which are undistinguishable by rotation, the pair of corners doesn't matter and any position is undistinguishable only from its image by the action consistiong of switching opposite faces, which is always reachable by some rotation (for example x2 y if the pair of corners is UBR-DFL) and always different from the starting position. The number of such positions is then $$18!/(2*3!^2)$$, and we must add twice that number to $$54!/(24*9!^6)$$ to gat the total number of positions undistinguishable by rotation on the cube.

#### bselena

##### Member
I have also thought of writing my own take on how speed cubing could be and rather than talk about technique, they practically base things on carefully plotted patterns that they just have to follow.

And while a lot of people find that interesting and great, there is actually a side to it that considers it as a plain and straight up puzzle without that much of its sleeves.

#### blah

##### brah
protip: use backticks () for open single quote, like
Code:
this'
Edit: Good read. But don't you find it awkward to describe angles of rotation in radians?

Also, protip2: $$n \equiv 0 \pmod 3$$
Code:
n \equiv 0 \pmod 3     % stands for "parentheses mod"

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#### macky

I'm amused that you cited Demazure and Gabriel for the definition of semi-direct product.
For references in math you should use MathSciNet to get BibTeX entries!

Some more tips. If you \usepackage{enumerate} in the preamble, you can use

\begin{enumerate}[(a)]
\item ...
\item ...
\end{enumerate}

to generate (a), (b), etc.
align and align* environments are also useful.
DeclareMathOperator{\sgn}{sgn} and same for \im and \ker to make those straight.
Write out "for all" using mbox instead of using the symbol.

Also, protip2: $$n \equiv 0 \pmod 3$$
Code:
n \equiv 0 \pmod 3     % stands for "parentheses mod"
Lots of real pros seem to like \pod for congruences in Z and \bmod for more general congruences modulo an ideal.

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#### Jude

##### Member
does anyone have a copy of this by any chance? lost my version, would be great if someone could re-up!

oh and thanks for the feedback guys

#### Christopher Mowla

does anyone have a copy of this by any chance? lost my version, would be great if someone could re-up!

oh and thanks for the feedback guys
I'm sorry for not responding sooner. I thought someone would have done it by now. Maybe someone already emailed you a copy, but your post asking for a copy was still here.

Anyway, I didn't think I had a copy, but I just realized that I did. I have uploaded it on the web. You can download it from here. As soon as you confirm that you've downloaded it, I will remove it from my archive (because it's your work).

I renamed it "Jude's essay" when I saved it to my computer (if that wasn't its name already) because I have several group theory documents on twisty puzzles.