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

Some thoughts about Symmetry

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.

A little about graphs

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 {Trivial Graph}
but the graph {Trivial Graph  Trivial 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


  1. If the graph is non-trivial and there exists a vertex with only one edge,
    remove it along with its edge.
  2. 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 {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):


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.

Theorem: There are exactly five Platonic Solids.

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

Introducing Functions

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=x2 in terms of graphs, the sort you see at school rather than the graph theory above.

picture of a function, computer
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 domain and codomain and 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.

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-1f = ff-1 = i:X → X.

A counter example and examples of a function

First the counter example; let D3 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
D3 : X → X
  D3(x) = Y≤X where xεX and Y is the set {rex, r1x, r2x, m1x, m2x, m3x}.
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 D3 say d<D3, such that
d : X → X
 d(x) = yεX.
For instance choosing the identity i ε D3, i is a subgroup of D3 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 Sn n≥2 is a group easily checked by considering only the equivalent products of transpositions of permutations mentioned earlier. Permutations in Sn are either even or odd products of transpositions, and the subset of even permutations An of Sn 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 Sn.

In order to show |Sn|/|An| = 2 we can set up a bijection between An and Bn the set of all odd permutations of Sn as follows
f : An → Bn
 f(σ) = σ.(a b) for any σ ε An, fixed (a b) ε Sn

Since σ is an even permutation σ.(a b) is odd and belongs to Bn, the mapping has a unique image and therefore is a well defined function.
f is onto since if β ε Bn then β.(a b) ε An ( 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 An then α.(a b) = τ.(a b), Sn is a group and so α = τ.
Therefore f is bijective and |Sn|/|An| = 2.  An is called the alternating group on n letters, a subgroup of Sn.

In the D3 example, A3 is the subgroup of rotations, but it is not in general true that An will be such.

Structure preserving functions

Two spherical morphisms in a vortex of binary numbers, depicting a homomorphism almost

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'

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(y1) = y', h(y2) = y'-1 with h(y1)h(y2) = 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 y2 = 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 g1,g2 ε G, h(g1g2) = h(g1)h(g2). Note the binary law of composition may not be the same in both domain and codomain.

With the above definition we have:

  1. e,g ε G and h(eg) = h(e)h(g) = h(g) and h(e) is the unique identity of G'.
  2. g,g-1 ε G and h(gg-1) = h(g)h(g-1) = h(e) = e' and h(g-1) = h(g)-1.
  3. 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).

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.

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).

function commutativity

The Fundamental Homomorphism Theorem.

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-1g = x and k'k-1g = 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(g1) = f(g2) then [g1] = [g2] and h(g1) = h(g2) some g1,g2 ε G; h(G) is a group and so h(g1)h(g2)-1 = h(g1g2-1) = [e] and g1g2-1 ε K. This looks familiar, from our work in equivalence relations
g1 ~ g2 ↔ g1g2-1 ε K and Kg1 = Kg2.

f is a homomorphism; Kg1Kg2 = Kg1g2; G/K is well defined
f(Kg1g2) = [g1g2] = [g1][g2] = f(Kg1)f(Kg2) 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 Sn 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 A4 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.

Rotational or proper symmetries of the tetrahedron.
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 Hre is the normal subgroup of all π or 180o rotations. Working by row then column, Hr4Hr5 means do Hr5 then Hr4.

Toggle cells to reveal cosets
 S4/H, H normal in S4
HreHreHr4Hr5Hm1 Hm2Hm3
Hr4Hr4Hr5HreHm2 Hm3Hm1
Hr5Hr5HreHr4Hm3 Hm1Hm2
Hm1Hm1Hm3Hm2Hre Hr5Hr4
Hm2Hm2Hm1Hm3Hr4 HreHr5
Hm3Hm3Hm2Hm1Hr5 Hr4Hre
re = (1)(2)(3)(4)
r1 = (12)(34)
r2 = (13)(24)
r3 = (14)(23)
r4 = (123)
r5 = (132)
r6 = (124)
r7 = (142)
r8 = (134)
r9 = (143)
r10 = (234)
r11 = (243)
m1 = (12)
m2 = (13)
m3 = (23)
m4 = (14)
m5 = (24)
m6 = (34)
t1 = (1234)
t2 = (1432)
t3 = (1243)
t4 = (1342)
t5 = (1324)
t6 = (1423)

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-1Hy=Hy-1y. A4 is special since An n >4 is simple, meaning doesn't have any non trivial normal subgroups.

Every element of A4/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 r4, r5; and doing this it is then easy to see A4/H is isomorphic to A3 the rotations of the equilateral triangle; table shown on page 0.

Generating subgroups from elements of a finite group G.

Let a ε G; if we take all products of a; an for n ε N, we get an abelian (xy=yx) subgroup. Call this subgroup Cn which is equivalent to Zn (see modulo arithmetic), where Cn having n elements is the cycle of length n; {an=e, a1, a2,.............,an-1}.

G is finite so an = am  n > m; their are repeat elements or G is infinite. am has an inverse, b say. anb = amb = e, but b is just a product of m copies of a-1; b = a-1←m→a-1, and so anb = an-m = e. By choosing the minimal n such that an = 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 Cp X Cq to be the set of all ordered pairs (a,c), (b,d) with a,b ε Cp, c,d ε Cq 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 Cp X Cq is pq and Cp X Cq is isomorphic to Cq X Cp.

Returning to A4 we see their isn't a cycle of length 6; all cycles are of length 2 or 3, well C2 X C3 has 6 elements but is abelian, and there doesn't exist an isomorphism between C2 X C3 and a subgroup of A4. If we generate subgroups in A4 by taking all products of aiaj fixed i,j; examples of elements are aniamj, akjalia-nj   ai, aj ε A4; 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 A4 is finitely generated (needs proof); A4 does not have a subgroup of order 6.

Let h: Cp1 X Cp2 X............X Cpn → G
 and if h(g1,g2,.......,gn) → g1g2..........gn is an isomorphism then
G is called an inner direct product of Cpi 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 + Za1 + Za2 +............+ Zan where ai is a power of a prime and not all ai distinct.

Here + is just a change of notation to denote the similarity of the arithmetic of abelian groups and addition; Zai is a cycle of length ai addition modulo ai 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 Z2 + Z2 or C2 X C2 with h(r1,r2) = r1r2. You might like to show that r1,r2 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-1y-1. If x and y commute (xy=yx) then xyx-1y-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-1y-1 non trivial.

If x and y don't commute and hence are non-abelian, then xyx-1y-1 is non-abelian.
To see this let z = xy, if z is abelian then y-1z = 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-1e-1 ε C', and for every xyx-1y-1 ε C', (xyx-1y-1)-1 = yxy-1x-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.

C' is normal in G

Proof: To prove this we merely inject gy-1yg-1 = e for any g ε G and of course x,y ε G into;
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-1y-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, S4/A4 is abelian and so A4 must contain all the commutators, but we can take one of those interesting isometries t1, that are not either rotation or reflection, and m1 say; then t1m1 = r8 and m1t1 = r10, and neither t1 or m1 belong to A4.

This ends the introduction and page 2 will expand on your algebraic experience.

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