Algebra and Group Theory  Application to puzzles 

1. Sets and functions 
A set is a collection of things
called elements, and each of these elements occurs at most once in a
set. A (finite) set can be given explicitly as a list inside a pair of curly
brackets, for example {2,4,6,8} is the set of even positive
integers below 10. It has four elements. There are infinite sets  sets that
have an infinite number of elements  such as the set of integers, but
I will generally discuss only finite sets (and groups) here.

On puzzles, we will often consider the set
of puzzle positions.

A function f:AB from
set A to set B is a pairing that assigns to each element in A some element of
B. Set A is called the domain and set B the range. If the
function f assigns element b to a then we say that f maps a to b, and call b
the image of a. We could write this as f(a)=b, but I prefer to
write it as af=b. Writing f on the right like this will also be
more natural in the cube context later, but be aware that in many books it
is written on the left instead.

A move on a puzzle is usually a function on
the set of positions. A move applied to one position results in another
position. This assumes however that a move can always be applied to any
position  in some puzzles such as the Square1 this is not the
case and makes such a puzzle much harder to analyse.

A function is injective or onetoone
if the function maps different elements in A to different elements in B.
A function is surjective or onto if every element in B is
mapped to by the function.
A function is called a bijection if it is both onetoone and onto.
Lemma 1.1: For a bijection f:AB we can uniquely
define a function g:BA, such that if a is any element of A and
b is its image in B under f (i.e. af=b) then bg=a.
Proof: Let b be an element of B. Since f is onto, there must be some a
in A such that af=b. Since f is onetoone, this element a is
uniquely determined. Therefore it is actually possible to simply define g as
that pairing that maps b to a whenever f maps a to b.
The function above is called the inverse
function of f, and is usually denoted by f ^{1}.
Thus every bijection f:AB has an inverse f ^{1}:BA
such that whenever af=b for some a in A and b in B, then bf ^{1}=a.
Lemma 1.2: The inverse of a bijection is also a bijection.
Proof: Let f:AB be a bijection, and f ^{1}:BA
its inverse. Let a be any element in A, and let b=af. Then a=bf ^{1}.
Therefore f ^{1} is surjective. Again let a be any
element of A. There is a unique b in B such that af=b. In other
words there is a unique b in B such that bf ^{1}=a. Thus
f ^{1} is onetoone.
Note that if f:AB is a
bijection, then f uniquely pairs each element of A with an element of B and
vice versa. If A and B and finite, they will therefore have the same number
of elements.

On nearly all puzzles, moves have inverses.
A move can usually be undone, taking you back to the position you had before.
For example on the cube, the move R, a clockwise turn of the right face, can
be undone by an anticlockwise turn of the right face. This can be denoted by
R^{1}, the inverse of R, though R' is more common.

The composition of two functions f:AB
and g:BC, is the function that is the result of applying first
f and then g. Thus to an element a in A we map the element c in C given by c=(af)g.
This function is denoted fg.
Lemma 1.3: If f:AB and g:BC are
onetoone, then fg is also onetoone.
Proof: Suppose a_{1} and a_{2} are
elements of A, such that a_{1}(fg)=a_{2}(fg).
Then (a_{1}f)g=(a_{2}f)g and since g is
onetoone we must have a_{1}f=a_{2}f. Since f
is onetoone, we get a_{1}=a_{2}. Therefore fg
never maps different elements to the same image element, in other words it is
also onetoone.
Lemma 1.4: If f:AB and g:BC are
onto, then fg is also onto.
Proof: Let c be an arbitrary element of C. There must be some element
b in B such that bg=c since g is onto. There must be an a in A such that af=b
since f is onto. Therefore c=bg=(af)g=a(fg) which means that fg
is onto.
Lemma 1.5: If f:AB and g:BC are
bijections, then fg is also a bijection.
Proof: Follows directly from previous two lemmas.

Moves can be combined into move sequences.
Instead of having to keep track of what each move in a move sequence does to
a position, you instead think only of what the sequence as a whole does to
it. A move sequence has become a single entity in itself, which is applied to
a position.
Any move sequence is a composition of
functions. For example, if we apply the move sequence FRB to a position p, we
can do the moves one by one, which gives position ((pF)R)B, or
we can consider it as a single function FRB, which gives
position p(FRB).

The identity function on A is the function
i_{A}:AA
defined by ai=a for every a in A. It is easily seen to be a
bijection.
Lemma 1.6: If f:AB is a function, then fi_{B}=i_{A}f=f.
Proof: For any a in A, we have a(fi_{B})=(af)i_{B}=af=(ai_{A})f=a(i_{A}f).
The three functions f, fi_{B}, and i_{A}f
all have the same effect on any element, they must therefore be the same
function.
Lemma 1.7: If f:AB is a bijection, then ff ^{1}=i_{A}
and f ^{1}f=i_{B}.
Proof: For any a in A, we have an element b in B such that b=af.
Then bf ^{1}=a which gives a(ff ^{1})=(af)f ^{1}=bf ^{1}=a=ai_{A}.
Therefore ff ^{1}=i_{A}. Conversely for any b
in B we can find a in A such that bf ^{1}=a and then b(f ^{1}f)=(bf ^{1})f=af=b.
Therefore f ^{1}f=i_{B}.

Note that if two different move sequences
always have the same effect on the cube, then they are simply different ways
of writing down the same function. Most of the theory developed here is
concerned only with the effect, not with how the function is represented by a
move sequence.
The identity on the cube is the move
sequence that contains no moves at all, or any other move sequence that does
nothing such as F2B2L2R2F2B2L2R2.
The move sequences U2D2 R'L FB'
and RL'FB'RL'F'B are inverses. You can check this by doing the
first and then the second sequence on a solved cube to again find the cube
solved. This automatically means that the second sequence followed by the
first will also do nothing to the cube.

Lemma 1.8: Composition is associative, i.e. for
functions f, g, and h we have f(gh)=(fg)h.
Proof: Let a be any element of the set on which f is defined, then a(f(gh))=(af)(gh))=((af)g)h=(a(fg))h=a((fg)h).
Thus f(gh) and (fg)h have the same effect on any
element, and are therefore the same function.
Lemma 1.9: (fg)^{1}=g^{1}f ^{1}.
Proof: (fg)(g^{1}f ^{1})=((fg)g^{1})f ^{1}=(f(gg^{1}))f ^{1}=(fi)f ^{1}=ff ^{1}=i.

To undo or invert the move sequence FRB, we
must undo the moves starting with the last. The inverse of the function FRB
is B^{1}R^{1}F ^{1}, the inverse of
move sequence FRB is B'R'F'.


2. Groups and homomorphisms 
A group is a set together with an
operation usually called multiplication, which satisfies the following
conditions:
 The group is closed under
multiplication, i.e. if you multiply together any two elements of the
group, this results in another element of the group. The result from
multiplication of element g by element h is denoted gh.
 There is an identity element, i.e. an
element e in the group such that for any element g in the group we have ge=eg=g.
 Every element has an inverse, i.e. if g
is an element of the group then there is an element h in the group such
that gh=hg=e. The inverse of element g is denoted by g^{1}.
 The multiplication is associative, i.e.
if a, b and c are elements of the group, then (ab)c=a(bc).
Lemma 2.1: The identity is unique.
Proof: If e_{1} and e_{2} are both identity elements,
then e_{1}=e_{1}e_{2}=e_{2}.
Lemma 2.2: Inverses are unique.
Proof: If h_{1} and h_{2} are both inverses of element
g, then h_{2}=h_{2}e=h_{2}(gh_{1})=(h_{2}g)h_{1}=eh_{1}=h_{1}.

Consider the set of all possible move
sequences, which are essentially strings of the letters F, L, U, B, R, D.
You can combine any two sequences to give a third, simply by concatenating
the two strings of letters, and possibly doing some trivial cancellations.
Such sequences form a group: There are inverses (simply undo moves, i.e. invert
each move and reverse the order), and there is an identity (simply do nothing,
i.e. an empty string). Associativity is obvious. Unfortunately there are
infinitely many move sequences, so this is an infinite group. Many move
sequences have the same effect on the cube. It is therefore much better to
consider only the effects of a move sequence. The effects of move sequences on
the cube (i.e. the functions used in the first section) also form a group, and
this is called the cube group. Thus any two movesequences that have the
same effect on a cube represent the same element of the cube group.

A group is finite if it has a finite
number of elements. The number of elements in a group G is called the order
of the group, and is denoted by G.

There are only finitely many positions of
the cube, and therefore only finitely many functions on that set of
positions. The cube group is therefore also finite.
There is another group that is interesting,
namely the group of rotations of the cube as a whole. We can hold the cube
with any one of the six faces at the top, and then have four choices as to
which of the adjacent faces should go at the front. There are therefore 24
possible orientations of the cube. A rotation of the whole cube is simply a
change between two of such orientations. It is easily seen that these form a
group, and there are 24 such rotations. If we also allow reflections, then we
get a group of 48 elements, the group of symmetries of the cube.

The order of an element g in a group
is the smallest positive integer n for which g^{n}=e (if
such an n exists), where the exponent notation g^{n} has
the natural meaning of repeated multiplication. If no such number exists,
then g is said to have infinite order. The order of g is usually denoted o(g).

In the group of rotations of the cube,
there are the following elements:
 6 rotations of order 4 around axes through
opposite centres.
 3 rotations of order 2 around axes through
opposite centres.
 8 rotations of order 3 around axes through
opposite corners.
 6 rotations of order 2 around axes through
opposite edges.
 1 identity, order 1.

Lemma 2.3: In a finite group G, all the elements have finite
order.
Proof: Let g be an element of G. Then g, gg=g^{2},
ggg=g^{3}, ... must all be elements of G. As G is
finite, we must have eventually have a repetition, i.e. we must have g^{k}=g^{m}
for some positive integers k>m. Then multiplying by g^{1}
we get g^{k1}=g^{k}·g^{1}=g^{m}·g^{1}=g^{m1}.
By repeating this trick m times, this reduces eventually to g^{km}=g^{mm}=g^{0}=e,
which shows that the order of g is (at most) km, a finite
number.

The cube group is finite, so if you do any
move sequence, and continuously repeat it, you will eventually get back the
position from which you started.

A subgroup H of a group G is a
subset of G that is a group in its own right using the multiplication
operation of G.
Lemma 2.4: A nonempty finite subset H of group G is a
subgroup if and only if it is closed under multiplication.
Proof: Clearly closed multiplication is a necessary condition to be a group. To show
it is sufficient, we must show that the other three group conditions also
hold. Let h be an element of H, and n the order of h (which is finite because the
proof of lemma 2.3 applies to H as well). Then
h^{n} = e must also be an element of H, so H contains the
identity element. Also h^{n1}·h = h·h^{n1} = h^{n} = e,
so the inverse of h is h^{n1} which is also in subgroup
H. The multiplication operation is obviously still associative on H since it
is associative on the larger set G.
Lemma 2.5: If g is an element of finite order in G, then the
set H = {e, g, g^{2}, g^{3}, ...} is a subgroup
of G, and H = o(g).
Proof: The element g has finite order, so clearly H has at most o(g)
elements. Let g^{n} and g^{m} be
any two elements of H. Multiplying them together gives g^{n+m}
which is also in H. H is finite and closed under multiplication, so by the
previous lemma, it is a subgroup. Suppose H has fewer than o(g) elements.
Then g^{n}=g^{m} for some m<n<o(g),
and therefore g^{nm}=e with 0<nm<o(g).
This contradicts the definition of o(g), so H must have exactly o(g)
elements.
If g is an element of finite order, then the set
H={e, g, g^{2}, g^{3}, ...} is actually
a group, and is called a cyclic group. The element g is its
generator and this group is usually denoted by <g>.
Note that if g has infinite order, then we explicitly require
g^{1}, g^{2}, ... to be
in the set also, so then <g>={..., g^{2},
g^{1}, e, g, g^{2}, g^{3}, ...}.
It is also possible to construct groups
from more than one generator. For example if g and h are elements in G, then <g,h>
is the smallest subgroup of G that contains g and h. All its elements can be
written as a product of g and h (and their inverses), for example gghg^{1}hh.

The cube group G_{c} is generated
by the moves, so G_{c}=<F,L,U,B,R,D>. There are
many subgroups of the cube group G_{c}. Most useful subgroups are
formed by elements with some common feature that is not shared by the rest of
G_{c}. Here are some examples:
 The elements that do not move any corners.
 The elements that do not move any edges.
 The elements that only twist corners.
 The elements that only flip edges.
 The elements that do not move anything in the
top two layers.
 The elements generated by F and R. Also called
the twogenerator group.
 The elements generated by F2 and R.
 The elements generated by F, R and U. Also
called the threegenerator group.
 The elements generated by F2, B2, R2, L2, U2,
D2. Also called the square group.
 The elements generated by FB', RL', UD'. Also
called the slice group.
 The elements generated by FB, RL, UD. Also
called the antislice group.

Given a set A, the Symmetric Group S_{A}
is the set of all bijections from A to A, which forms a group under composition
(by lemma 2.4). If A is the finite set {1,2,3,...,n}, then its symmetric
group is denoted by S_{n} and the elements of this group are called permutations.
This will be dealt with in much more depth later.

If you number the 48 moveable facelets of
the cube, the effect of any move sequence is uniquely specified by how it
permutes those facelets. This shows that the cube group G_{c} occurs
as a subgroup of S_{48}. This is why the effect of a move sequence is
usually also called a permutation.

A homomorphism is a function f:GH
between group G and H which respects the group multiplication, so (g_{1}·g_{2})f=(g_{1})f·(g_{2})f.
Note that the first multiplication is in the group G, while the second is in
group H.
Lemma 2.6: If f:GH is a homomorphism then e_{G}f=e_{H}
and (g^{1})f=(gf)^{1}.
Proof: For any g in G, gf=(e_{G}g)f=(e_{G}f)(gf)
so by multiplying both sides by (gf)^{1} on the right gives e_{H}=e_{G}f.
For the second part, e_{H}=e_{G}f=(gg^{1})f=(gf)(g^{1}f).
The image of a homomorphism is a group, as
it inherits all the necessary properties from the group structure of the
domain of the homomorphism. If a homomorphism from group G to group H is not
surjective, then its image will therefore be a subgroup of H. If a homomorphism
is bijective, then it is called an isomorphism and the G and H are
called isomorphic groups. Isomorphic groups are in a sense identical 
they have the same order and same structure, even though the two groups might
occur in a different context.

There is a homomorphism from the normal
Rubik's cube to the 2×2×2 cube. Any move sequence on the normal cube can also
be performed on the smaller cube. Thus any permutation on the large cube maps
to some permutation on the small cube. This mapping is a homomorphism but is
not onetoone, so it shows that the small cube has a group, which is a sense
a simplified version of the group of the larger cube. .

The Symmetric Group S_{n} for some
positive integer n is usually considered to be the group of bijections on the
set {1,2,3,...,n}. The group of permutations of any finite set
of elements A is then isomorphic to the group S_{n} where n=A. This
is a really obvious isomorphism  first we label the elements of A as 1 to n
in any way by some bijection f:{1,2,..,n}A, and then
bijectively map S_{n} to S_{A} using that labelling.

Actually, earlier I really should have said
that G_{c} is isomorphic to a subgroup of S_{48}.

Lets extend the multiplication of elements
of G to multiplication of sets of elements of G: If K and H are
subsets of G, then let KH be the set of all elements kh with k in K and h in
H. Note that K and H need not be subgroups, they merely have to lie in some
larger group G so that there is a way of multiplying their elements. I
introduce this set multiplication because it will simplify proofs later on.
Lemma 2.7: Set multiplication is associative, i.e. A(BC)=(AB)C
for subsets A,B,C of a group G.
Proof: Let x be an element of A(BC). Then x=ad for some a in A and d
in BC. Also, for d we must have d=bc for some b in B, c in C. Therefore
x=a(bc)=(ab)c by associativity of the elementwise multiplication in G. But ab
is in AB, and so (ab)c is in (AB)C. Thus A(BC) is a subset of (AB)C. The
converse, that (AB)C is a subset of A(BC), is proved in a similar way and the
result follows.
Note that we can also multiply a single
element x to a set A to get xA, and the result is simply the set of all
elements xa with a in A. This is the same as multiplying {x} by A, so
associativity still holds.
Lemma 2.8: If g is an element of group G, then gG=G.
Proof: Let h be any element of G. Then h=eh=(gg^{1})h=g(g^{1}h)
lies in gG. Conversely, any element in gG must be in G since G is closed
under multiplication. Therefore G=gG.



3. Actions and orbits 
An action of a group G on a set A is
a function f:A×GA which satisfies (a,e)f=a for all
a in A and also ((a,g)f,h)f=(a,gh)f. What this means is that any
element g of the group G will correspond to a function f_{g}:AA
in such a way that f_{e} is the identity and f_{g}f_{h}=f_{gh}.
Whenever it is clear what group action that
we are dealing with, we might as well write ag instead of af_{g}, as
this can make things much more readable. We can do this because the function
that assigns f_{g} to g is a homomorphism, so the group structure is
carried through. In particular, g and g^{1} are inverses in the
group, and their associated functions on A are also inverse functions. If we
have an action of group G on set A we say that G acts on A. Also for
elements g in G and a in A we say g acts on a, and gives ag (an
element of A).
The orbit of an element a in A is
the set containing ag for all g in G. In other words it is the image of a
when it is acted on by all the group elements. This set will be denoted aG.
Note that the orbit of a will contain a itself since ae=a.
An action is transitive if for any
two elements a and b in the set A, there is some g in G such that ag=b.
This means that the orbit of any element is the whole set, so aG=A for any a
in A.

The cube group G_{c} obviously acts
transitively on the set of positions of the cube, because by definition the
positions are those that can be reached by using cube permutations.
Instead lets consider the positions of the
cube that can be reached by taking the cube apart and reassembling it. The
cube group again acts on this larger set, but this time it is not transitive
as there turn out to be 12 different orbits.

Lemma 3.1: Distinct orbits are disjoint.
Proof: Suppose element z lies in xG, the orbit of x. Then z=xg for
some g in G. The orbit of z is then zG=(xg)G=x(gG)=xG. Therefore if z lies in
two orbits xG and yG, then xG=zG=yG and the orbits are equal. Distinct orbits
then cannot have any elements in common.

Let H be the subgroup of G_{c} with
elements that only move the edges but nothing else. Then H acts on the set of
all positions. Each orbit is a set containing positions that differ only in
their edge arrangement, and therefore all positions in an orbit will have
their corners in the same arrangement. Every possible arrangement of corners
gives rise to such an orbit containing all the positions that share that
corner arrangement. The number of orbits is therefore the number of possible
corner arrangements on the cube.

Theorem 3.2 (Cayley): Every group is isomorphic to a subgroup of some
symmetric group, in other words every group can be considered a group of
permutations on some set.
Proof: Let G act on the set G by right multiplication. By definition
of action, each element g in G gives a function f_{g}:GG
(which in this case if defined by the right multiplication xf_{g}=xg). This
function f_{g} is a bijection on G since it has an inverse, and it therefore
lies in S_{G}, the group of permutations of the set G. The mapping from G
to S_{G}, namely the one that maps g to f_{g}, is easily seen to be a
homomorphism. This mapping is onetoone because for any g and h in G, if we
have f_{g}=f_{h} then
g=eg=ef_{g}=ef_{h}=eh=h. Thus the result
follows.
G acts freely on the set A if any
element g of G leaving one element a of A fixed must leave the whole of A
fixed, i.e. ag=a implies g=e.
Lemma 3.3: If G acts transitively and freely on A, then the set
A is in onetoone correspondence with G.
Proof: Choose any element a in A. Consider the function f:GA
which maps g to ag. The image of this function is obviously aG, but as G acts
transitively this is the whole of A. Thus f is surjective. Also, if gf=hf
then ag=ah, agh^{1}=a and since the action is free we must have gh^{1}=e
and so g=h. This shows that f is onetoone. Therefore f sets up the
correspondence between A and G. Note that the special element a that was
chosen will correspond to the identity in G.

It is a neat fact that there is a
correspondence between cube positions and cube permutations. Every cube
position is reached by some permutation from the solved position, and every
(legal) permutation when applied to the solved position gives some cube
position. Since a nontrivial permutation will always have a visible effect
on any position (i.e. ag is never equal to a), the lemma shows that the
positions and permutations are in an exact onetoone correspondence. This is
not necessarily so on other puzzles.
Any puzzle where all the pieces are
distinct has a onetoone correspondence between its positions and its
permutations. If some pieces are identical this need not be the case, as it
may be so that there is some permutation that when applied to the solved
position swaps identical pieces about, which can't be seen. Nevertheless such
a permutation might have some effect on other positions, and not be the
identity.
We already know how many possible positions
the cube has (approx. 43·10^{18}), and therefore this is the actual
size of the cube group as well.


4. Cosets and coset spaces 
If H is a subgroup of G (though H can be
the whole of G), then H can act on the set G by right multiplication. Any h
in H acts on g in G to give gh in G. The orbits of this action are sets of
the form gH, are called right cosets. We could instead have used
multiplication from the left, to give the orbits Hg which are called left
cosets. (Note: some books call gH a left coset and Hg a right coset,
especially if defined as a multiplication of H by g rather than as an orbit
of a group action.)
Lemma 4.1: Cosets of a finite subgroup H all have the same
number of elements.
Proof: H is itself a coset, because H=eH=He. Let gH be some coset.
Clearly H has as least as many elements as gH by construction, since any h in
H will give an element gh in gH. Conversely, H=eH=(g^{1}g)H=g^{1}(gH)
means that gH has at least as many elements as H by the same argument.
Therefore H=gH. The same arguments show that H=Hg.
Now we come to an extremely important
result.
Theorem 4.2 (Lagrange): If H is a subgroup of finite group G, then G is
divisible by H.
Proof: The (left) cosets of H all have the same size, are disjoint
(because they are orbits and distinct orbits are disjoint), and any g in G
lies in some coset (namely gH). Therefore G is partitioned into sets of size
H and so H divides G.
The number of cosets of H in G is called
the index of H in G and is denoted by (G:H). For finite G, Lagrange's
theorem says that (G:H)=G/H.
Lemma 4.3: If g is an element of finite group G, then the o(g)
divides G.
Proof: The cyclic group <g> is a subgroup of G. Its order is
equal to the order of the generator g, so o(g) divides G by the theorem.

The size of the cube group is roughly 43·10^{18}.
The exact number has no factor of 13 or 17 for example, so for any move
sequence at all, if you repeat it 13 or 17 times on a solved cube it cannot
possibly return to the solved position (unless the move sequence itself did not
have any effect in the first place).

Corollary 4.4: If g is an element of finite group G, then g^{G}=e.
Proof: g^{G}=(g^{o(g)})^{G/o(g)}=e^{G/o(g)}=e.

Any move sequence will have no effect if it
were repeated 43252003274489856000 times (the order of the cube group),
though I would not want to try it.

Corollary 4.5: If G is a prime number, then G is cyclic.
Proof: Let g be an element of G. Then o(g) divides G so o(g)=1 or
o(g)=G. If we choose g to be any element that is not the identity then
o(g)=G. Now the cyclic subgroup <g> has order G so it must be the
whole group, which means that G is cyclic and generated by g.
If group G acts on A, then the stabilizer
of an element a in A is the set of g in G for which ag=a. In other words it
is those group elements that keep a stable, not changing it. It is denoted
Stab_{a}.
Lemma 4.6: The stabilizer of a in A is a subgroup of G.
Proof: If g,h are in the stabilizer, then a(gh)=(ag)h=ah=a, so gh also
lies in the stabilizer. The stabilizer is thus closed under multiplication. Also
ae=a so the stabilizer has the identity. If g is in the stabilizer then
ag^{1}=(ag)g^{1}=a(gg^{1})=ae=a
and so g^{1} is also in the stabilizer. Associativity is inherited
as usual.

Suppose we have a cube on which all the
edges have had their stickers removed so that they are indistinguishable.
This puzzle has fewer positions than a normal cube, but has the same group. The
cube group acts on this smaller set. The stabilizer of the solved position for example
contains all those elements of the group, which move edges only, and
these form a group, too.

If H is a subgroup of G, then the left
coset space, denoted by H\G, is the set of all left cosets, i.e. its
elements are Hx for each x in G. There is a natural group action of G on this
set, by right multiplication. In other words, g in G acts on Hx to give the
left coset (Hx)g=H(xg). It is easily seen to be a transitive action.
Similarly the right coset space is the set of all right cosets and is
denoted by G/H.
Lemma 4.7: The stabilizer of Hx in the left coset space is x^{1}Hx.
Proof: Clearly (Hx)(x^{1}Hx)=H(xx^{1})Hx=HeHx=Hx, so
the set x^{1}Hx is certainly a subset of the stabilizer. Suppose y
lies in the stabilizer, i.e. Hxy=Hx. Then xy lies in Hxy and hence in Hx.
Multiplying on the left by x^{1} shows that y is in x^{1}Hx.
Corollary: The stabilizer of H in the left coset space H\G is H.
Corollary: For any x in G, x^{1}Hx is a subgroup of G

Most solving methods for the cube
implicitly work in coset spaces. Let H be the group of permutations of the
bottom layer. Suppose you were to remove all the stickers of the bottom layer
pieces of a solved cube. Every element in H then looks the same, and so every
element in a coset Hx will also look the same. A coset therefore represents
all those positions of the normal cube that are treated the same way when the
pieces belonging in the bottom layer are ignored. Solving the first two
layers is therefore equivalent to working in H\G. You are applying the moves
(letting G act on H\G) until all only the last layer remains to be done (i.e.
until you reach the coset H). The last layer then solved by working within
the group H itself.

Theorem 4.8: Any transitive action of G on a set A, is
isomorphic to the action of G on the left coset space H\G where H is the
stabilizer of an element a in A.
Proof: Choose any a in A, and let H be its stabilizer. Consider the
function f:H\GA defined by (Hx)f=ax for all x in G. This
definition is welldefined (not dependent on which x is used to represent the
coset) because if Hx=Hy then xy^{1} is in H, axy^{1}=a
and hence ax=ay. The same argument in reverse shows that if ax=ay, then
Hx=Hy, so the function is onetoone. The action is transitive, so for any b
in A we have b=ax for some x in G, in other words the function is onto. This
function shows the correspondence between the sets on which the group action
occurs, and this respects the structure of the action since H(xy)=(Hx)y
and a(xy)=(ax)y.
Corollary 4.9: If group G acts transitively on finite set A, then
A=(G:Stab_{a}) where a is any element in A. In particular, if G is
finite, then Stab_{a}=G/A
Proof: The size of a coset space H\G is the number of cosets, which is
(G:H) by definition. The mapping in the theorem above shows that any
transitive group action is isomorphic to the action of G on the set Stab_{a}\G
for any a in A, and result follows.

Let's consider the group of rotations of
the cube, and its action on the 8 corners. There are 3 rotations that keep
one particular corner in place (120 degree turn, 240 degree turn, and
identity). As there are 8 corners, this immediately shows that the group has
size 3·8=24. There are 6 faces that have order 4 rotations, again showing the
group has size 24. Similarly the 12 edges with their order 2 rotations.

The Orbit Theorem 4.10: Suppose G is a finite group acting on finite set A.
Let c_{g} be the number of fixed points of g's action, i.e. the
number of elements in A such that ag=a. Then the number of orbits of the
group action is equal to the sum of the c_{g} divided by G. In
other words the number of orbits is equal to the average number of fixed
points of the elements g in G.
Proof: Let's count the number of pairs (a,g) with a in A, g in G, for
which ag=a. This total is simply the sum of the c_{g}, but we can
count these pairs in a different way. If S is any orbit, then G acts
transitively on S. From the previous theorem we know that Stab_{a}=G/S,
so every element a in orbit S contributes G/S to the sum of fixed points.
As there are S elements in the orbit S, the total contribution from this
and every other orbit is G. Hence there are G·t pairs, where t is the
number of orbits. Therefore t is the sum of the c_{g} divided by G,
as required.

This is often known as Burnside's lemma,
but has various other names as well. Let's use it to count how many ways the
6 sides of a cube can be coloured if we have n colours to choose from. In our
case G is the group of 24 cube rotations. The number of fixed points of each
type of rotation (colourings that remain the same under that rotation) is:
 6 face axis rotations of order 4: n^{3}.
 3 face axis rotations of order 2: n^{4}.
 8 corner axis rotations of order 3: n^{2}.
 6 edge axis rotations of order 2: n^{3}.
 1 identity, order 1: n^{6}.
This sums to (n^{6}+3n^{4}+12n^{3}+8n^{2})/24
= n^{2}(n+1)(n^{3}n^{2}+4n+8)/24.
Substituting n=1,2,3,... gives the answers 1, 10, 57, 240, 800, 2226, etc. If
you insist on using 6 different colours, then none of these are fixed by any
rotation except for the identity, so there are 6!/24=30 ways to do that
(which can also be thought of as 15=5·3·1 ways of choosing opposite colours,
2 mirror image ways of then colouring the cube).


5. Permutations and parity 
Every finite group is (isomorphic to) a
subset of S_{n} for some integer n by Cayley's theorem. Now we will
now look in more detail at what a permutation does when it acts on the
elements of that set A={1,2,3,...,n}. Even though infinite groups are also
permutations (on an infinite set), we will assume G is finite from now on.
Suppose g in G acts on some distinct
elements a_{1}, a_{2}, ..., a_{n} of A as a permutation
in the following way:
a_{i}g=a_{i+1} for all i<n,
a_{n}g=a_{1}
ag=a for all other elements a in A
then we call g an ncycle, and denote it by (a_{1} a_{2}
... a_{n}). A cycle rotates a number of elements around in a loop,
and does nothing else. In particular, any permutation that just swaps two
elements a and b is a 2cycle denoted (a b), and any permutation that moves
exactly three elements will be a 3cycle. Note that a permutation that moves
exactly 4 elements need not be a 4cycle (a b c d) as it might instead be a
product of two 2cycles such as (a b)(c d) which swaps elements a with b, and
c with d.

The pieces of the cube can also move around
in cycles, for example the move sequence F2RL'U2LR' is a 3cycle
of edge cubies, namely (FD BU FU).

Lemma 5.1: Every permutation is a product of disjoint cycles.
Proof: Let g be any permutation of set A. Take any a in A that is
permuted by g. The elements a, ag, ag^{2} etc. form a cycle because
if ag^{n}=ag^{m} then ag^{nm}=a.
If g does not permute any other elements in A, then g is simply this one
cycle. Otherwise we can take an element b of A that does not lie in this
cycle and form a new cycle (b bg bg^{2} ...). These
cycles are disjoint, because if bg^{n}=ag^{m}
then b=ag^{mn} contrary to our choice of b. This
process can be repeated to enumerate all the disjoint cycles, and it is
obvious that g acts the same on A as the product of these cycles. Therefore g
is the product of these disjoint cycles.

For example, the cycle notation for the
move F is (FU FR FD FL)(FUR FRD FDL FLU), and similarly any move
or move sequence can be written as a product of disjoint cycles to show where
each piece moves to. The orientation of the pieces is not quite captured with
this notation however. Suppose for example that one piece moves from UFL to
URF but that its U facelet ends up in the right face. As we can choose which
facelet to mention first (e.g. FUR or RFU or URF all mean the same piece
position) we say that UFL moves to RFU, thus specifying its twist at the same
time. In the cycle notation for F above it is clear that all the F facelets
stay in the F face. It gets a bit trickier if a cycle itself is twisted, for
example (UFL RFU FLU FUR LUF URF). Here two corners are swapped,
but it is not the same as (UFL RFU) because that would imply that RFU moved
to UFL whereas it should move to FLU. We will denote this by (UFL RFU)^{+},
so that it is clear that it is a 2cycle and also that RFU gets an extra
clockwise twist when it is moved. Pieces that don't move but become flipped
or twisted will be shown as a 1cycle, e.g. (BRU)^{}.

The representation of a permutation as a
product of disjoint cycles is essentially unique, as it can only differ in
the order of the product, and cyclically permuting each cycle.
Lemma 5.2: Every permutation is a product of 2cycles, which
can all be of the form (1 a). In other words, S_{n} is
generated by the permutations (1 a).
Proof: A cycle (a_{1} a_{2} ... a_{k})
can be split as a product (1 a_{1}) (1 a_{2}) (1 a_{3})
... (1 a_{k}) (1 a_{1}). A cycle involving the element 1 can
be split into 2cycles in a similar way. Therefore any product of
cycles is also a product of such 2cycles, and hence any permutation.

If on a puzzle you could swap any two
pieces at will, then it is easy to solve any position  with each swap you
can put (at least) one piece in its correct position. If each swap you make
has to involve the piece at one particular place, then it is still easy.

Lemma 5.3: S_{n} = n! = n·(n1)·...·2·1.
Proof: An informal proof runs as follows:
Each permutation is a bijection on A. Suppose we were to specify were each
element of A is mapped to by a permutation. The element 1 can be mapped to
any of the n elements of A. Element 2 cannot be mapped to the same element as
1, so can be mapped to any of the remaining n1 elements. Similarly 3 is
mapped to any one of the remaining n2 elements, and so on. Clearly there are
n·(n1)·...·2·1. possibilities, each of which specify a unique
permutation. Therefore S_{n} = n!.
Here is a more formal proof:
By definition, S_{n} acts transitively on the set A={1,2,...,n}.
From corollary 4.9 we have Stab_{n} = S_{n}/A.
The group Stab_{n} consists of all those permutations which keep n
fixed, and so Stab_{n} can be considered as the set of all
permutations of {1,2,...,n1}, and is isomorphic to S_{n1}.
Therefore S_{n1} = Stab_{n} = S_{n}/A = S_{n}/n.
From this, and the fact that S_{1}=1, the result follows from
induction.

Any puzzle which has n distinct pieces, and
on which any two pieces can be swapped (though this may not be easy!) will
have n! permutations and hence n! positions.

The parity of a permutation (odd or
even) is the parity of the number of cycles of even length in the disjoint
cycle representation, i.e. if the total number of 2cycles, 4cycles,
6cycles etc. is odd, then the permutation is odd, otherwise the permutation
is even.

A quarter turn of a face of the cube is a
4cycle of corners and also a 4cycle of edges. It is therefore an even
permutation.

Lemma 5.4: Multiplying a permutation by a 2cycle changes its
parity.
Proof: If the 2cycle is disjoint from the other permutation, then it
is obvious. Otherwise there are three cases to consider, depending on where
the two elements in the 2cycle lie in the disjoint cycles of the
permutation. We can relabel the elements to get one of these cases:
(2 3 4 ... n)(1 2) = (1 2 3 ... n)
(1 2 3 ... a ... n)(1 a) = (1 2 3 ... a1)(a a+1 ... n)
(1 2 3 ... a)(a+1 ... n)(a a+1) = (1 2 3 ... a1 a+1 a+2 ... n a )
In all three cases the parity changes (regardless of the parity of n and a)
and hence the parity of the whole permutation changes too.
Corollary 5.5: The parity of a permutation is the parity of the
number of 2cycles when written as a product of them.
Proof: Follows from the previous lemma and the fact that the identity
is even. Note that a permutation can be written as such a product in many
ways of different lengths, but this shows that the parity is fixed.
Parity theorem 5.6: When multiplying permutations, the parities are
added. In other words multiplying even by even gives even, odd by odd gives
even, and multiplying odd by even gives odd, and even by odd also gives odd.
Proof: Follows immediately from corollary above.
The parity theorem is also sometimes given
differently as follows: There exists a homomorphism sgn:S_{n}{1,1}
that is onto.

Every single move on the cube is an even
permutation. Therefore every move sequence is also an even permutation. Thus
a single swap of two pieces is an odd permutation and is impossible to
achieve using moves alone  you would have to take the cube apart.

The Alternating Group A_{n}
is a subgroup of S_{n} containing only the even permutations. The
parity theorem above proves that it is a group.
Lemma 5.7: Every even permutation is a product of 3cycles,
which can all be of the form (1 2 a).
Proof: Write the permutation as a product of 2cycles of the form (1
a), which is possible by lemma 5.2. Each pair of 2cycles is a product
of three cycles of the required form because (1 a)(1 b)=(1 a b)=(1 2
a)(1 2 b)^{2}(1 2 a)^{2}.
Lemma 5.8: A_{n}=n!/2
Proof: Lets consider the left cosets of A_{n} in S_{n}.
For any odd permutation p we have p=p(1 2)(1 2) so it lies in
the coset A_{n}(1 2). The even permutations lie in A_{n}
itself, and so there are only two cosets. By Lagrange's theorem we then have
A_{n}=S_{n}/2.

Many puzzles have a parity restriction, so
that only the even permutations of pieces can be achieved and its group is A_{n}.
This means that such a puzzle will have only half as many positions as it
would otherwise have, namely n!/2 instead of n! (assuming its pieces are
distinct so that there is a onetoone correspondence between permutations
and positions).


6. Symmetry and Platonic solids 
If a threedimensional geometric shape
stays unchanged by a rotation around an axis by an angle less than 360° then
that is said to be a rotational symmetry of that shape. In this section
we will enumerate all the different types of finite rotational symmetry there
can be in a threedimensional object.

Although this section is not directly
related to permutation puzzles, many puzzles are based on threedimensional
symmetry. Also, if you wish to design new threedimensional permutation
puzzles then a familiarity with the possible symmetry types is necessary.

The set of rotational symmetries of a shape
clearly form a group: Combining any two rotational symmetries will still
leave the geometrical shape unchanged and so is also rotational symmetry.
Throughout this section G will denote any finite such group of rotational
symmetries. The order of a rotational symmetry is simply its order
when considered as an element of this finite group.
Lemma 6.1: For any bounded geometrical shape the axes of the
rotational symmetries of the shape (if any) intersect in a single point.
Proof: Take any point of the shape and consider its orbit. This orbit
is a set of points that by definition remains unchanged (or rather gets
mapped to itself) by any of the rotational symmetries of the shape. Let O be
the gravitational centre of these points. This single point will therefore
also remain unchanged by any rotational symmetry, and hence must lie on the
axis of every rotational symmetry. The result then follows.
Consider a unit sphere centred on the point O from the lemma
above. Any axis of a rotational symmetry intersects this sphere at a pair of
opposite points, the poles of this symmetry. The order of a pole
is the highest number that occurs as the order of one of its symmetries.

The rotational symmetries of the cube have
been listed already. These correspond to:
 6 poles of order 4 (faces)
 8 poles of order 3 (corners)
 12 poles of order 2 (edges)
Note that although there
are rotational symmetries of order two around the face centres, its poles
have order 4.

Lemma 6.2: There is a group action of the group of rotational symmetries
G on the set of poles A.
Proof: Let p be any pole in A, and let r in G be any one of its
corresponding rotational symmetries. By definition, pr=p. For any rotation s,
ps is a pole of the rotational symmetry s^{1}rs because
(ps)(s^{1}rs)=p(ss^{1}rs)=prs=ps. Thus ps is a
pole for any pole p and any rotational symmetry s, and so this is a group
action.
Corollary 6.3: Any orbit of this group action has poles of the
same order.
Lemma 6.4: If an orbit S of this group action has poles of
order k, then kS=G.
Proof: G acts transitively on S (by the definition of an orbit), so by
Corollary 4.9 we have Stab_{p}=G/S. The stabilizer
of a pole p is the group of the rotational symmetries around this pole. Let r
be the rotational symmetry with pole p with the smallest (nonzero) angle, α.
If s is any other rotational symmetry around pole p of angle β, then a rotation of
βcα is also a rotational symmetry for any integer c. If we choose c to be
the integer nearest to β/α, then α<βcα<α.
As α is the smallest possible nonzero symmetry angle, we get βcα=0,
and so β must be divisible by α, and hence s is multiple of r. The rotations
therefore form a cyclic group generated by r. The order of this group is the
order of r, which is also the order of the pole. The result follows.

In the example of the cube there are 24
rotational symmetries, and we do indeed find that multiplying the number of
poles of each order by their order we indeed get 24 again: 24=6·4=8·3=12·2.

Lemma 6.5: Let n=G and let k_{i} be the order of the
poles in the ith orbit of the Gaction on the set of poles. Then 2(11/n)=Sum(11/k_{i}).
Proof: Let t be the number of orbits. From the orbit theorem 4.10 we
have t=Sum(c_{g})/n. The number of fixed points of any
nontrivial rotation g in G is 2. The identity rotation however fixes all
poles. From lemma 6.4 we know that there are n/k_{i} poles in the
ith orbit. Putting this all together we find that t=( 2(n1) + Sum(n/k_{i}))/n,
where the sum ranges over the orbits from i=1 to t. After some rearranging
the result follows. Note that in this rearranging, the t term is split up
into t ones and brought into the sum.
Theorem 6.6: The only rotational symmetry groups are:
 The cyclic groups, the symmetry of an ngonal
pyramid.
 The dihedral groups, the symmetry of an ngonal
cylinder.
 The tetrahedral group, the symmetry of a
tetrahedron.
 The octahedral group, the symmetry of a cube or
octahedron.
 The icosahedral group, the symmetry of a
dodecahedron or icosahedron.
Proof: We enumerate all possible solutions of the equation from lemma
6.5.
Suppose there are only two orbits. Then 2(11/n)=21/k_{1}1/k_{2}
or 2=n/k_{1}+n/k_{2}. Both terms on the right
must be strictly positive integers (n/k_{i} is the number of poles in
the ith orbit), so k_{1}=k_{2}=n is the only
solution. There are only two poles (both of order n), so this means that
there is only one axis of rotation, and we get the cyclic group of order n.
Suppose there are three orbits. Then 2(11/n)=21/k_{1}1/k_{2}1/k_{3},
where we can assume that the orbits are arranged so that k_{1} is
smallest and k_{3} largest. Since the left hand side is greater than
1, not all k_{i} can be 3 or greater. So we can assume that k_{1}=2.
Suppose that also k_{2}=2. Then k_{3}=n/2. In this case we
have a rotational symmetry of order k around one axis, and k axes of symmetry
of order 2. This gives the dihedral group or order 2k.
Suppose now that k_{2}=3. Then 1/k_{3}1/6=2/n, and we find
that the only further solutions that occur are (k_{3},n)=(3,12),
(4,24), or (5,60). These solutions give the symmetries of the Platonic
solids. The first solution gives tetrahedral symmetry, and has 8 poles of
order 3 (in two orbits of four poles), and 6 poles of order 2. The second
solution has octahedral symmetry, with 8 poles of order 3, 6 of order 4, and
12 of order 2. The last is icosahedral symmetry and has 20 poles of order 3,
12 poles of order 5, and 30 of order 2.
No solutions are possible with k_{2}>3, so there are no more
solutions.
The rotational symmetry groups of the
Platonic solids  tetrahedron, cube/octahedron, and dodecahedron/icosahedron
 are in fact isomorphic to the permutation groups A_{4}, S_{4},
and A_{5}.
Consider the four vertices of a
tetrahedron, which are permuted by any rotational symmetry. It is easily seen
that the two types of rotation are even permutations of the vertices, and
since there are 12 rotations in the group it must in fact be the whole of A_{4}.
In a similar way every (nontrivial)
rotational symmetry of the cube is a (nontrivial) permutation of the four
main diagonals of that cube. As there are 24=S_{4} rotations, the
group must be the whole of S_{4}. It is interesting to note that a
tetrahedron can be embedded inside a cube so that they share vertices. This
also establishes the tetrahedral symmetry group A_{4} as a subgroup
of the cube's symmetry group S_{4}. There are actually two ways of embedding
a tetrahedron in a cube, and this corresponds to the two cosets of A_{4}
in S_{4}.
It is possible to embed five tetrahedra
inside a dodecahedron so that all 20 of its vertices are shared. Every
(nontrivial) rotational symmetry is a (nontrivial) even permutation of the
five tetrahedra, and this shows that the 60 rotations form a group isomorphic
to A_{5}.

The pyraminx is a tetrahedron, and
obviously has tetrahedral symmetry. The pieces are moved by rotations of
order 3 around the corners, or equivalently rotations of order 3 of the
faces. The pyramorphix has the same symmetry, but its moves are around the
poles of order 2, i.e. around the edges of the tetrahedron. If moves are
restricted to half turns, the tetrahedral shape is maintained.
The Rubik's cube has the octahedral
symmetry group, its moves are around the 6 poles of order 4. The Dino cube
also has the same symmetry but its moves are around the 8 poles of order 3,
and similarly the Helicopter Cube uses moves around the 12 poles of order 2,
The octahedral symmetry group (S_{4}) has the tetrahedral symmetry
group (A_{4}) as a subgroup. This is seen in the pyramorphix  its
moves are like the 2x2x2 cube and so in essence have octahedral symmetry, but
its shape does not have that symmetry which is why its shape changes when
mixed. By restricting to half turns the tetrahedral symmetry is maintained,
and so does its shape.
The Megaminx, Impossiball, Alexander's star and the Dogic all have icosahedral
symmetry. In fact they all use moves around the 12 poles of order 5. There are some
less common puzzles that use the other poles, such as the Starminx I, Bauhinia,
Eitan Star (20 poles of order 3) and the Helicopter Dodecahedron (30 poles of order 2).
The Skewb is actually a slightly odd case
because the symmetry of its moves is actually tetrahedral as can be seen when
taking the puzzle apart  it is impossible to exchange the two sets of
corners. The original Skewb (and the Skewb Diamond) has more symmetry than
that in its outward shape. A rather pleasing consequence of this is that when
the Skewb is embedded into a dodecahedron instead (i.e. the Skewb Ultimate)
then the shape does not change because the tetrahedral symmetry group (A_{4})
is established as a subgroup of the dodecahedral symmetry group (A_{5}).
Note that it is possible to make puzzles
that use moves which are based on several types of poles. For example, it is
possible to build a cross between the Dino cube and the 2x2x2 cube, or more
easily a Rhombic Dodecahedron (which has octahedral symmetry) where each
corner can turn, i.e. both those of order 3 and those of order 4.


7. Conjugation and commutation 
If g is an element of the group G, then a conjugate
of g is an element h^{1}gh for some h in G. Note that if x is a
conjugate of y, then y is also a conjugate of x.
Lemma 7.1: Conjugate permutations have the same cycle
structure.
Proof: Suppose x and y are conjugates, i.e. there is some g in the
group such that x=g^{1}yg. Suppose part of the cycle structure of y
is (... a b ...), i.e. ay=b. It easily seen that (ag)x=(ag)g^{1}yg=ayg=bg.
Therefore part of the cycle structure of x is (... ag bg ...).
This extends to all elements on which y acts. If y acts on the elements a_{i},
then x acts on the elements a_{i}g in the exact same way.

Conjugates have the same cycle structure,
so they have the same effect except that it is applied to different pieces.
In a puzzle, this means that if you have a move sequence that
swaps/cycles/flips/twists some particular pieces, then using conjugates of
that move sequence you can swap/cycle/flip/twist any other pieces you want.

Lemma 7.2: If g is any element in G and H is a subgroup, then
the g^{1}Hg is also a group, and is isomorphic to H.
Proof: The map f:Hg^{1}Hg given by hf=g^{1}hg
is an homomorphism because (h_{1}h_{2})f=g^{1}h_{1}h_{2}g=g^{1}h_{1}gg^{1}h_{2}g=h_{1}fh_{2}f.
It is obviously onto. If g^{1}h_{1}g=g^{1}h_{2}g
then h_{1}=h_{2} so f is onetoone. Thus g^{1}Hg
and H are isomorphic groups.

Consider the group of permutations on the Cube,
which do not move the UFL corner. By conjugating this group with the move U,
we get permutations that do not move URF. This is clearly isomorphic to the
first group, which is obvious if you simply rotate the cube as a whole.

The centre C_{G} of a group
G is the set of elements in G that commute with every element in G, i.e. for any
h in C_{G} and g in G we have gh=hg.
Lemma 7.3: The centre is a subgroup.
Proof: Suppose x and y are in the centre. Then for any g in G,
gxy=xgy=xyg so then xy is in the centre. Also ge=g=eg and gx^{1}=x^{1}xgx^{1}=x^{1}gxx^{1}=x^{1}g.
Lemma 7.4: The centre of S_{n} is trivial (contains
only the identity) when n>2.
Proof: Let g be a permutation that is not the identity. Then there
must be some distinct elements a and b of {1,2,3,...,n} such that ag=b. Let c
be another element that is different to a and b. The permutation (a c)g will
map c to b whereas g(a c) maps a to b, so g does not commute with (a c).
Therefore g is not in the centre, and hence no nonidentity element is in the
centre.
Lemma 7.5: The centre of A_{n} is trivial (contains
only the identity) when n>3.
Proof: As above except that the even permutation (a c d) is used
instead of (a c).

Many puzzles have a trivial centre group,
because usually any move sequence that commutes with everything cannot move
any pieces from their position and twists/flips every similar piece the same
amount. On the cube it is impossible to twist all the corners in the same
direction, which leaves only the superflip, the manoeuvre that flips all 12
edges.

Lemma 7.6: S_{n} is generated by the two elements
p=(234...n) and q=(12).
Proof: p^{1}(1 a)p = (1 a+1), so any twocycle (1 a) can be
achieved by conjugation of q by p. By lemma 5.2, these generate all of S_{n}.

The cube group can obviously be generated
by six elements, the six moves that you can do. It can in fact be generated
by two elements. Number the corners 18 and the edges AL
and let p be the following permutation:
p = (2345678)(BCDEFGHIJKL)
Taking powers of p that are multiples of 7 and 11 we find:
p^{56} = (BCDEFGHIJKL)
p^{22} = (2345678)
It is fairly easy to show that if q=(12)(AB) then p and q can
generate all possible positions of the corners and edges of the cube. To get
all orientations as well, we can use q=(12)(3)+(AB)+(C)+
instead. We find that the positions can still be generated in the same way,
but we also have:
q^{4} = (1)+ (2)+ (3)+
q^{6} = (A)+ (B)+
By applying conjugates of these we can generate all orientations of the
pieces. Short examples of such generators are:
p = R2FLD'R' = (URF,BDR,LFD,LDB,LBU,LUF,FRD) (FU,DR,FR,UR,DF,BD,RB,DL,BL,UL,FL)
q = UBLUL'U'B' = (UBR,LUF) (ULB)+ (UR,BU)+
(UL)+
It is even possible generate the cube group using a very different pair p and
q, one of order 2, the other of order 4.

If g and h are elements of the group G,
then the commutator of g and h is the element g^{1}h^{1}gh.
The commutator is a measure of how much the elements g and h commute  if
they do commute then the commutator is the identity. The commutator of g and
h is sometimes denoted by [x,y].

Commutators are a very fruitful source of
useful move sequences. Since a commutator is a kind of measure of how near
two elements commute, taking a commutator of two elements that nearly commute
will give something that moves very little and hence is likely to have a
useful effect.

Lemma 7.7: Every homomorphism f:GH between
groups respects commutators, i.e. [g,h]f=[gf,hf].
Proof: [g,h]f = (g^{1}h^{1}gh)f = (g^{1}f)(h^{1}f)(gf)(hf)
= (gf)^{1}(hf)^{1}(gf)(hf) = [gf,hf].
Corollary 7.8: g^{1}[x,y]g=[g^{1}xg,g^{1}yg].
Proof: Taking conjugates is a homomorphism (see lemma 7.2) so the
result follows, though it can just as easily be proved directly.
The commutator subgroup of G is the
subgroup generated by all commutators in G. Note that the commutator group is
not simply the set of commutators itself but instead is generated by that
set, as this set is not necessarily closed under multiplication and therefore
is not necessarily a group in its own right.

Lets see how far the Cube could be solved
by using only commutators. It is fairly straightforward to find commutators
that flip two edges, or twist two corners. As conjugates of commutators are
still commutators, all the orientations can be solved using only commutators.
A threecycle of edges or corners can also be done by commutation, and so
conjugates give every possible threecycle. By 5.7 every even permutation of
edges and every even permutation of corners is possible. Commutators cannot
generate odd permutations as it is easily seen that every commutator is even.
Therefore the Rubik's Cube can be solved completely using commutators only,
except possibly for a single quarter turn.
On the other hand, the Megaminx only has even
permutations of the pieces. An argument similar to the above shows that the
Megaminx can be solved using only commutators. The same holds for the Dino
cube.


8. Normal groups and quotient groups 
A subgroup H is a normal subgroup of
G if gH=Hg for every element g in G. For normal subgroups, the left and right
cosets are exactly the same for each element.
Lemma 8.1: If H is a normal subgroup of group G, then the
(right) cosets form a group under the multiplication xH·yH=(xy)H.
Proof: First it must be shown that this definition of coset
multiplication is welldefined, in other words that the result (xy)H is
independent of which particular elements x and y that are used to represent
the cosets. Using the set multiplication we have (xH)(yH)=x(Hy)H=x(yH)H=(xy)(HH)=(xy)H.
This is the unambiguous result we want, so the group multiplication of cosets
is in fact the set multiplication we were using all along, and it is indeed
welldefined. The other requirements of a group are inherited from G as it
easily shown that eH is the identity, x^{1}H is the inverse of xH,
and that multiplication is associative.
If H is a normal subgroup of G then the
cosets of H form a group called the quotient group, which is denoted
by G/H. The order of the group equals the number of cosets so G/H=G:H,
the index of H in G. If G is finite, this equals G/H.

The group of permutations that move only
corners, but leaves the edge pieces alone, is an example of a normal subgroup
on the cube. After solving the edges of the cube first, this group is used
when solving the corners.

Lemma 8.2: If H is a normal subgroup of group G, then ghg^{1}
is in H for every element h in H and g in G.
Proof: H is normal, so gH=Hg for any g in G. Therefore (gH)g^{1}=(Hg)g^{1}=H(gg^{1})=H.

Suppose we solve edges first, and then the
corners. Suppose also that we have a move sequence that has some useful
effect on the corners and which does not move the edges. If we want to use a
conjugate of the sequence, in order to apply it to a different set of
corners, then we need not worry about the edges at all, because any conjugate
will not move the edges.

Lemma 8.3: If H is a subgroup of index 2 of group G, then H is
a normal subgroup.
Proof: H has index 2 so there are only two left cosets, one of which
is H itself. The other coset therefore must consist of all elements of G that
are not in H. The same argument works for right cosets. Let g be any element
in G. If g is in H, then gH=H=Hg. If g is not in H, then gH and Hg are both
cosets not identical to H. There is only one such coset, so then also gH=Hg,
and H is therefore normal.
Corollary 8.4: A_{n} is a normal subgroup of S_{n}.
Proof: From lemma 5.8 we know that A_{n} has index 2 in S_{n}.



9. Bibliography and Further Reading 
There are very many books on Algebra. The two algebra books I mainly used while
compiling this page are:
 P.M. Cohn, Algebra (Vol. 1), second edition, John Wiley and Sons, 1982.
 I.N. Herstein, Topics in Algebra, second edition, John Wiley and Sons, 1975.
For section 6 I had some help from the following book, which also has a very
short section on the Rubik's cube:
 P. Neumann, G. Stoy, E. Thompson, Groups and Geometry, Oxford University Press, 1994.

There are few books that delve into the mathematics of the Rubik's cube.
The following are the groundbreaking classics, and any serious cubist should have at least
one of these, if not all:
 Christoph Bandelow, Inside Rubik's Cube and Beyond, BirkHäuser, 1982.
 Alexander Frey, David Singmaster, Handbook of Cubik Math, Enslow Publishers, 1982.
 David Singmaster, Notes on Rubik's 'Magic Cube', Fifth edition, Enslow Publishers, 1980.
This recent book comes highly recommended, and not just because it favourably
mentions this site below a quote by Winston Churchill:
 David Joyner, Adventures in Group Theory, Johns Hopkins University Press, 2002.
