A → B

It is not the case A is true and B is false

(In context '→' is mapped to)

A ← B

It is not the case B is true and A is false

A ↔ B iff or (if and only if)

Combines the above

a ε X

a is a member of X

∑ is a sum

A < B

Is less than or properly contained in

A ≤ B

Is less than equal to or contained in

It is not the case A is true and B is false

(In context '→' is mapped to)

A ← B

It is not the case B is true and A is false

A ↔ B iff or (if and only if)

Combines the above

a ε X

a is a member of X

∑ is a sum

A < B

Is less than or properly contained in

A ≤ B

Is less than equal to or contained in

Earlier I mentioned there are only five 'platonic solid's, regular convex objects with faces of a polygon. This was known to Euclid_{[circa 300 BC]} and a proof appears in the 'Elements', where convex means any two interior points can be joined by an imaginary straight line not intercepting a face, and regular, has congruent or equal faces where every vertex or point of interception between edges, has equal number of faces attached.

It is convenient to consider a face to be a region bounded by edges, so that there *does not exist* points x y interior to the face, such that a line of points **belonging to** the face connecting x y intercepts an edge.

Some group theory was introduced on the previous page and I want to expand on that and other mathematical ideas, but first a proof of the above claim, and to achieve this by what has become known as the 'topological method', I will introduce the Euler_{[ 1707-1783 ]} 'characteristic' χ and 'Schlafli' number.

Graphs are collections of vertices and edges, a vertex is an endpoint of an edge and an edge has two vertices. In my special case of a graph a vertex has at least one edge attached and edges only intercept in vertices. This is nearly the general form, and a triangle with vertices {a,b,c} has graph

{a,b,c,{a b},{b c},{a c}} the set of edges and vertices of the triangle.

A graph is said to be connected if any vertex is linked to any vertex by a path of edges belonging to the graph, and a loop is a path with a common start and end vertex.

A polygon is an example of a connected graph, as is the trivial graph {}

but the graph { } is not connected.

{a,b,c,{a b},{b c},{a c}} the set of edges and vertices of the triangle.

A graph is said to be connected if any vertex is linked to any vertex by a path of edges belonging to the graph, and a loop is a path with a common start and end vertex.

A polygon is an example of a connected graph, as is the trivial graph {}

but the graph { } is not connected.

Thinking of a solid three dimensional object as elastic, stretchable leaving edges and vertices intact, if we poke a hole in one of its faces by removing a point and stretch it out over the plane, two dimensional surface, we get a *planar connected graph.* The faces of the polyhedron are not only those bounded by edges, but also their is a face exterior to the graph corresponding to the face with a point removed. I am going to adopt a convention that any planar graph has at least one face.

This certainly makes things easier from my point of view, I don't want to go into graph theory proper until page three, but I think the following proof is easier to visualise from a graph theory standpoint.

Euler's characteristic 'χ' is a relationship between Faces Edges and Vertices:

F + V - E = χ

Lemma: The characteristic F + V - E of a finite connected planar graph is '2'.

Proof: Starting with any non-trivial connected graph; F + V - E = χ

Rule

- If the graph is non-trivial and there exists a vertex with only one edge,
*remove it along with its edge.* - Otherwise, if the graph is non-trivial,
*remove any edge.*

Following the rule in order, choosing either step i) or ii) not both, at each step,

if we do step i) we get F + (V-1) - (E-1) = χ

and in step ii) we get (F-1) + V - (E-1) = χ

applying the above rule (inductively) the graph is finite so we must end up with the trivial graph {}. Since I assumed every planar graph has at least one face F + V - E = 2. qed

It is now fairly straightforward to prove there exists only five convex regular polyhedra, making use of the Schlafli_{[1814-1895]} symbol (p,q):

(p,q)

- p = number of edges of a face
- q = number of edges or faces meeting at a vertex

A tetrahedron has three triangular faces meeting at each vertex and (p,q) = (3,3), a cube has three square faces meeting at each vertex and (p,q) = (4,3). The q in (p,q) is dropped when considering polygons and a pentagon has Schlafli symbol (5). Higher dimensional polyhedra also have different forms of Schlafli symbol.

From the lemma we know the planar graph of a convex regular polyhedron has

Euler characteristic χ = 2 therefore χ = 2 for a polyhedron.

Proof; Making use of the Schlafli symbol and substitution we have

Fp = 2E = (number of faces)(number of edges of a face)...because two faces meet in a common edge.

Vq = 2E = (number of vertices)(number of edges meeting at each vertex)...because two vertices share a common edge.

Hence F = 2E/p and V = 2E/q , then F + V - E = 2 gives

2E/p + 2E/q - E = 2 and with simple manipulation remembering E>0 gives

1/p + 1/q = 1/2 + 1/E → 1/p + 1/q > 1/2

The minimum number of edges a face can have is three giving Schlafli symbols

(3,3) (3,4) (3,5) as all possibilities with triangular faces since 1/3 + 1/q ≤ 1/2 for q > 5

with four edges to a face we get only (4,3) the cube

with five edges to a face we get only (5,3) and this exhausts all possibilities. qed

Up to now functions have been implied by the context, but they become very important and it would be against the spirit of mathematics to not make them precise.The concept familiar to many is the black box or factory processing inputted raw objects, doing something to them and outputting possibly different objects, and many have seen pictorial representations of functions like y=x^{2} in terms of graphs, the sort you see at school rather than the graph theory above.

To make this idea precise we need two sets X,Y and a rule f, if X is the set of elements we want to apply the rule to it is called the domain, and if Y is the set of elements X is mapped (changed) into, it is called the codomain.

A function f:X → Y is two sets called

domainandcodomainand a rule which maps every x ε X to a unique y = f(x) ε Y. Where → means maps to, in this context.

The set f(X) is called the range or image of f:X → Y, and f(X)≤Y.

The set of all x ε X such that f(x) = yεY (fixed y) is called the *inverse image* of y.

- If f(X) = Y then f is said to be onto or surjective.
- If for all x
_{1},x_{2}εX; f(x_{1}) = f(x_{2}) implies x_{1}= x_{2}then f is said to be 1 - 1 or injective.

Which means for every yεY there exists at most one xεX such that f(x)=y. - If f is both 1 - 1 and onto then it is called a bijection or bijective.

Let f:X → Y, g:Y → Z, we can compose f and g giving gf:X → Z where gf means apply f first then g from right to left and z=g(f(x)). If f:X → Y is bijective, then an inverse exists f^{-1} and f^{-1}f = ff^{-1} = i:X → X.

First the counter example; let D_{3} be the dihedral group of symmetries of an equilateral triangle, X be the graph of the triangle,

X = {a,b,c,{a b},{a c},{b c}} and define the mapping

D_{3} : X → X

D_{3}(x) = Y≤X where xεX and Y is the set {r_{e}x, r_{1}x, r_{2}x, m_{1}x, m_{2}x, m_{3}x}.

**Then this is not a function** since the image of x is **not** *unique*, Y has more than one member. However we can make a function by looking at subsets of D_{3} say d<D_{3}, such that

d : X → X

d(x) = yεX.

For instance choosing the identity i ε D_{3}, i is a subgroup of D_{3} and

i : X → X

i(x) = x

is called the identity function.

Remember, a function has to have a unique image. Digressing from the main theme a little, if we focus on a specific x ε X and G a group with G : X → X, G(x) = Y then with a couple of extra conditions Y is called the *orbit of x under G*.

Another example, consider the mapping

| | : Z → N ( integers map into naturals )

| |(z) = {z iff z ≥ 0 ; -z if z < 0}

this is a function and it is onto but not 1 - 1 since | |(z) = | |(-z). This function has its' own name and notation, it is the modulus function and | |(z) is written |z|.

Bijective or 1 - 1 functions are useful to determine if sets have an equal number of elements. Cantor used an assumed bijection between natural numbers and real numbers in his famous 'diagonal' proof of the uncountability of the reals.

The set of all permutations of n letters S_{n} n≥2 is a group easily checked by considering only the equivalent products of transpositions of permutations mentioned earlier. Permutations in S_{n} are either even or odd products of transpositions, and the subset of even permutations A_{n} of S_{n} is a group: The identity is an even product of transpositions, a product of even permutations is an even product of transpositions, and inverses are just order reversal of the equivalent product of transpositions and are therefore also even. Associativity is inherited from S_{n}.

In order to show |S_{n}|/|A_{n}| = 2 we can set up a bijection between A_{n} and B_{n} the set of all odd permutations of S_{n} as follows

f : A_{n} → B_{n}

f(σ) = σ.(a b) for any σ ε A_{n}, fixed (a b) ε S_{n}

Since σ is an even permutation σ.(a b) is odd and belongs to B_{n}, the mapping has a unique image and therefore is a well defined function.

f is onto since if β ε B_{n} then β.(a b) ε A_{n} ( it is a product of odd permutations ).

f(β.(a b)) = β.(a b).(a b) = β.

f is 1 - 1 since if f(α) = f(τ) for any α,τ ε A_{n} then α.(a b) = τ.(a b), S_{n} is a group and so α = τ.

Therefore f is bijective and |S_{n}|/|A_{n}| = 2. A_{n} is called the *alternating group* on n letters, a subgroup of S_{n}.

In the D_{3} example, A_{3} is the subgroup of rotations, but it is not in general true that A_{n} will be such.

The domain and codomain of a function can have some sort of mathematical structure such as group, ring or topology, in these circumstances functions can be used to show equivalence between structures.

Since groups are our main interest as far as symmetry is concerned, we will restrict our attention to the main structure preserving function with domain a group whose image is also a group.

h:G → G'

h(g)=g'

For h(G) to be a group we require an identity, inverses and associativity:

we want e' ε h(G) such that h(x)=e', x ε G and h(x)h(g) = e'h(g) = h(g),
and for every y' ε h(G),y'^{-1} ε h(G) such that

h(y_{1}) = y', h(y_{2}) = y'^{-1} with h(y_{1})h(y_{2}) = y'y'^{-1} = e'.

With all this and inherited associativity we have a group, but we also want this group to have the same structure as the domain. That is if H ≤ G then h(H) ≤ h(G), h(H) is a subgroup of the image h(G). Well {e} ≤ G therefore we require h(e) = e' and e'g' = h(e)h(g) = h(eg). Keeping the multiplication of functions in mind, some groups have self inverses y = y^{-1} or y^{2} = e. Then H = {e,y} ≤ G and for h(H) to be a subgroup of h(G) we need h(y)h(y^{-1}) = e' = h(y)h(y)^{-1} = h(yy^{-1}). The multiplication h(x)h(y) = h(xy) looks important.

A homomorphism mapping a group into a group is a function defined as follows:

h:G → G'

and for all g_{1},g_{2}ε G, h(g_{1}g_{2}) = h(g_{1})h(g_{2}). Note the binary law of composition may not be the same in both domain and codomain.

With the above definition we have:

- e,g ε G and h(eg) = h(e)h(g) = h(g) and h(e) is the unique identity of G'.
- g,g
^{-1}ε G and h(gg^{-1}) = h(g)h(g^{-1}) = h(e) = e' and h(g^{-1}) = h(g)^{-1}. - G' is a group and is therefore associative, the image h(G) inherits associativity

and is therefore a subgroup of G'.

Earlier I defined the inverse image of a function, not necessarily an inverse function because a single element could be mapped to many elements. The following result I've called a proposition because it's straightforward to prove but normally would be called a theorem by convention.

As a reminder the factor G/H is a group of cosets under coset multiplication, if and only if H is a normal subgroup of H, although this wasn't proved. I stated the cosets of G/H formed a group when Hg = gH for all g ε G. A more useful and equivalent condition is gHg^{-1} = H, meaning ghg^{-1} ε H for all g ε G and all h ε H, H is then called a normal subgroup of G. It turns out that with our definition of a homomorphism this structure is preserved.

Proposition: Let θ be a homomorphism; θ:G → θ(G)≤G'

then H is a normal subgroup of G ↔ θ(H) is a normal subgroup of θ(G).

then H is a normal subgroup of G ↔ θ(H) is a normal subgroup of θ(G).

Proof: Suppose H ≤ G, is normal in G

θ(H) = H' ≤ θ(G) and for every g' ε θ(G) there exists g ε G where θ(g) = g' and θ(g^{-1}) = g'^{-1}

Let h' ε H', θ(h) = h' then g'h'g'^{-1} = θ(g)θ(h)θ(g^{-1}) = θ(ghg^{-1}) = θ(k) some k ε H

implies that θ(k) ε θ(H) and g'h'g'^{-1} ε H', and since this is true for all h' ε H' we have θ(H) is a normal subgroup of θ(G).

Conversely:
Suppose H' is a normal subgroup of θ(G)

Let H be the inverse image of H' and e' ε H' so e ε H since θ(e) = e',

if h ε H then θ(h) ε H' but θ(h)^{-1} ε H' and θ(h)^{-1} = θ(h^{-1}) ε H' so h^{-1} ε H and H is a subgroup.

Choosing any g ε G and for all h ε H

θ(ghg^{-1}) = θ(g)θ(h)θ(g)^{-1} = k' ε H' since H' is normal. The inverse image of k' is a subset of H

ghg^{-1} is in the inverse image of k' and so ghg^{-1} ε H. qed

Corollary: {e'} is a normal subgroup of θ(G)

and its inverse image called the kernel of θ is a normal subgroup of G.

and its inverse image called the kernel of θ is a normal subgroup of G.

Proof follows immediately from the above.

If a homomorphism is also a bijection it is called an isomorphism.

From the above you may have guessed where all this is leading. If h is a homomorphism the inverse image K of unity is a normal subgroup of the domain G, and as stated above if K ≤ G is normal in G then the factor group G/K (G mod K) is well defined and exists. It seems to me intuitively obvious that G/K is bijective with the image of h, h(G).

If h: G → G' is a homomorphism,then G/K is isomorphic to h(G).

Proof: Let f be the canonical mapping f: G/K → h(G), f(Kg) = [g].

f is well defined and onto, choosing [g] ε h(G); [g] = h(g) some g ε G so if y,g ε Kx some x ε G

and g = kx, y = k'x some k,k' ε K; k^{-1}g = x and k'k^{-1}g = y, k'k^{-1} ε K so y ε Kg and our element y ε Kx is mapped to [g]; f(Kg) = [g].

f is 1 - 1 ; suppose that f(g_{1}) = f(g_{2}) then [g_{1}] = [g_{2}] and h(g_{1}) = h(g_{2}) some g_{1},g_{2} ε G; h(G) is a group and so h(g_{1})h(g_{2})^{-1} = h(g_{1}g_{2}^{-1}) = [e] and g_{1}g_{2}^{-1} ε K. This looks familiar, from our work in equivalence relations

g_{1} ~ g_{2} ↔ g_{1}g_{2}^{-1} ε K and Kg_{1} = Kg_{2}.

f is a homomorphism; Kg_{1}Kg_{2} = Kg_{1}g_{2}; G/K is well defined

f(Kg_{1}g_{2}) = [g_{1}g_{2}] = [g_{1}][g_{2}] = f(Kg_{1})f(Kg_{2}) qed.

That last line was a little terse, care has to be taken not to fall into a trap of just writing things down without much care or thought. For instance many will know the chain rule for differentiation with the Leibniz notation dx/dy = (dx/dt)(dt/dy). This is easy to remember because it looks like multiplying fractions, but is actually completely different. Mathematics is not about remembering equations so that with a little arithmetic you can get through examinations, it is much harder than that. For those who are interested, a mathematicians proof of the chain rule can be found in 'Mathematical Analysis II; Burkhill & Burkhill,' but be aware this is set at second year undergraduate level. The last line of the above proof may need a little care, maybe it's just me but I think there is a question of existence despite f being onto, I'm being vague because I'm leaving it up to you to fill the gaps in.

It is indeed a fact accredited to Cayley that every finite group is isomorphic to a subgroup of a permutation group S_{n} some n ε N.

So I want to turn our attention back to symmetry, but this time in three dimensions, with the symmetries of the tetrahedron, the most basic of the 'Platonic Solids.' You will see A_{4} the alternating group on four letters is precisely the group of rotations of a tetrahedron, and does it have a subgroup of order six, by Lagrange *only* if a subgroup exists does it then divide the order of the group.

The table toggle function works in all main browsers but expect a slight delay of several seconds after a cell is clicked. The table for the cell is being downloaded and rendered, hopefully.

I've used right cosets and you can check multiplication is well defined. H whose coset table is under Hr_{e} is the normal subgroup of all π or 180^{o} rotations. Working by row then column, Hr_{4}Hr_{5} means do Hr_{5} then Hr_{4}.

S_{4}/H, H normal in S_{4} | ||||||
---|---|---|---|---|---|---|

Hr_{e} | Hr_{4} | Hr_{5} | Hm_{1} | Hm_{2} | Hm_{3} | |

Hr_{e} | Hr_{e} | Hr_{4} | Hr_{5} | Hm_{1} |
Hm_{2} | Hm_{3} |

Hr_{4} | Hr_{4} | Hr_{5} | Hr_{e} | Hm_{2} |
Hm_{3} | Hm_{1} |

Hr_{5} | Hr_{5} | Hr_{e} | Hr_{4} | Hm_{3} |
Hm_{1} | Hm_{2} |

Hm_{1} | Hm_{1} | Hm_{3} | Hm_{2} | Hr_{e} |
Hr_{5} | Hr_{4} |

Hm_{2} | Hm_{2} | Hm_{1} | Hm_{3} | Hr_{4} |
Hr_{e} | Hr_{5} |

Hm_{3} | Hm_{3} | Hm_{2} | Hm_{1} | Hr_{5} |
Hr_{4} | Hr_{e} |

r_{e} = (1)(2)(3)(4)

r_{1} = (12)(34)

r_{2} = (13)(24)

r_{3} = (14)(23)

r_{4} = (123)

r_{5} = (132)

r_{6} = (124)

r_{7} = (142)

r_{8} = (134)

r_{9} = (143)

r_{10} = (234)

r_{11} = (243)

r

r

r

r

r

r

r

r

r

r

r

m_{1} = (12)

m_{2} = (13)

m_{3} = (23)

m_{4} = (14)

m_{5} = (24)

m_{6} = (34)

t_{1} = (1234)

t_{2} = (1432)

t_{3} = (1243)

t_{4} = (1342)

t_{5} = (1324)

t_{6} = (1423)

m

m

m

m

m

t

t

t

t

t

t

From the table you can see that the cosets of H are well defined under multiplication. You can prove H and more general subgroups are normal if and only if coset multiplication is well defined. Merely by observing what has been said before and Hy^{-1}Hy=Hy^{-1}y. A_{4} is special since A_{n} n >4 is *simple*, meaning doesn't have any non trivial normal subgroups.

Every element of A_{4}/H involves a rotation of x degrees about a face then a half twist resulting in a rotation of x degrees about a different face. Since multiplication is well defined we can choose representatives for cosets from the same face, eg r_{4}, r_{5}; and doing this it is then easy to see A_{4}/H is isomorphic to A_{3} the rotations of the equilateral triangle; table shown on page 0.

Let a ε G; if we take all products of a; a^{n} for n ε N, we get an abelian (**xy=yx**) subgroup. Call this subgroup C_{n} which is equivalent to Z_{n} (see modulo arithmetic), where C_{n} having n elements is the *cycle of length n*; {a^{n}=e, a^{1}, a^{2},.............,a^{n-1}}.

G is finite so a^{n} = a^{m} n > m; their are repeat elements or G is infinite. a^{m} has an inverse, b say. a^{n}b = a^{m}b = e, but b is just a product of m copies of a^{-1}; b = a^{-1}_{←m→}a^{-1}, and so a^{n}b = a^{n-m} = e. By choosing the minimal n such that a^{n} = e we get our identity, cycle of length n, and inverses are obvious.

Consider ordered pairs of cycles, an ordered pair is just an element like (a,b) commonly used to represent a point on a 2-D graph in school mathematics.

We can define the **external direct product** C_{p} X C_{q} to be the set of all ordered pairs (a,c), (b,d) with a,b ε C_{p}, c,d ε C_{q} and multiplication (a,c)(b,d) = (ab,cd), it is obvious and trivial to prove that this is an abelian group. It is easy to see that the order of C_{p} X C_{q} is pq and C_{p} X C_{q} is isomorphic to C_{q} X C_{p}.

Returning to A_{4} we see their isn't a cycle of length 6; all cycles are of length 2 or 3, well C_{2} X C_{3} has 6 elements but is abelian, and there doesn't exist an isomorphism between C_{2} X C_{3} and a subgroup of A_{4}. If we generate subgroups in A_{4} by taking all products of a_{i}a_{j} fixed i,j; examples of elements are a^{n}_{i}a^{m}_{j}, a^{k}_{j}a^{l}_{i}a^{-n}_{j} a_{i}, a_{j} ε A_{4}; the only chance of generating a subgroup of order 6 is to take products of an element of cycle length ≤ 3 and element of cycle length 2. If we take two elements of cycle length 2 we get H which is of order 4, and if one element is of cycle length 3 we get more than 6 elements generated and so by Lagrange must be of order 12. Since every subgroup of A_{4} is finitely generated (needs proof); A_{4} does not have a subgroup of order 6.

Let h: C_{p1} X C_{p2} X............X C_{pn} → G

and if h(g_{1},g_{2},.......,g_{n}) → g_{1}g_{2}..........g_{n} is an isomorphism then

G is called an **inner direct product** of C_{pi} and if we allow infinite groups then

The Fundamental Theorem of Finitely Generated Abelian Groups

Every finitely generated abelian group is isomorphic to:

Z +Z +...+Z + Z_{a1}+ Z_{a2}+............+ Z_{an}where a_{i}is a power of a prime and not all a_{i}distinct.

Here + is just a change of notation to denote the similarity of the arithmetic of abelian groups and addition; Z_{ai} is a cycle of length a_{i} addition modulo a_{i} and Z denotes integers. Proof of this theorem is harder and will be left until later.

H is a subgroup of half twists, is abelian and is isomorphic to Z_{2} + Z_{2} or C_{2} X C_{2} with h(r_{1},r_{2}) = r_{1}r_{2}. You might like to show that r_{1},r_{2} generate H and h is an isomorphism. H has got a special name, it is called the Viergruppe or Klein 4-group.

For this part of the discussion, I want to relax a little, and where I talk about x,y you can assume what kind of group elements I'm talking about from the context.
Consider the product xyx^{-1}y^{-1}. If x and y commute (xy=yx) then xyx^{-1}y^{-1} = e. Only when x and y are non-abelian; to mean they do not commute, although they may commute with other elements; is the **commutator** xyx^{-1}y^{-1} non trivial.

If x and y don't commute and hence are non-abelian, then xyx^{-1}y^{-1} is non-abelian.

To see this let z = xy, if z is abelian then y^{-1}z = zy^{-1} = xyy^{-1} = x → z =yx = xy. Then if x,y do not commute z is non-abelian or contradict the above#. If for these purposes we consider only non-trivial *commutators*, then inductively a product of commutators is non-abelian or the identity.

However, it is not clear that a product of commutators is a commutator, but it is clear that not all non-abelian elements are a product of commutators, and I will demonstrate this shortly.

Letting x,y ε G generate a set C' of all possible products of commutators we get a subgroup;

eee^{-1}e^{-1} ε C', and for every xyx^{-1}y^{-1} ε C', (xyx^{-1}y^{-1})^{-1} = yxy^{-1}x^{-1} ε C'; and since we are taking all possible products the set is closed under multiplication and is therefore a subgroup of G say. It is the smallest subgroup containing all the commutators in G.

The following very important result is easy to verify but much harder to discover the method of proof.

Proof: To prove this we merely inject gy^{-1}yg^{-1} = e for any g ε G and of course x,y ε G into;

g^{-1}xyx^{-1}y^{-1}g

=g^{-1}xyx^{-1}(gy^{-1}yg^{-1})y^{-1}g

=[(g^{-1}x)y(g^{-1}x)^{-1}y^{-1}][yg^{-1}y^{-1}g]

and since this is a product of commutators it is in C'. qed

g

=g

=[(g

and since this is a product of commutators it is in C'. qed

The factor G/C' is therefore a group, and if C'xC'y = C'yC'x then choosing representatives;

xy ε C'yx ↔ xyx^{-1}y^{-1} ε C'. C' is the smallest subgroup containing all the commutators, implies G/C' is the largest abelian factor group.

But unlike the example in the table, where all the half twists were neatly contained in H, in this case not every non-abelian element is a product of commutators: Looking at the table, S_{4}/A_{4} is abelian and so A_{4} must contain all the commutators, but we can take one of those interesting isometries t_{1}, that are not either rotation or reflection, and m_{1} say; then t_{1}m_{1} = r_{8} and m_{1}t_{1} = r_{10}, and neither t_{1} or m_{1} belong to A_{4}.