Some time ago I had an idea to write a book about symmetry, it seemed a cool thing to write a book, and symmetry looked like an easier challenge than most things, but also very visual and attractive.The idea was sort of forgotten in the great scheme of things until after a long pause away from the Internet.
Upon my return and thinking about building a web-page found symmetry isn't quite as easy as first appeared, where to start, what knowledge can be assumed known about mathematics for instance.
Let me illustrate the problem with the following: A group G is a non empty set and a binary law of composition satisfying some assumptions called axioms, symmetry of shapes have associated with them a group called the dihedral group. So it could be central to any discussion about symmetry to also talk about their groups. But I mentioned a set earlier, and should that also be explained. You will appreciate some starting point has to be decided upon and if you look up 'Zermelo-Fraenkel' axiomatic set theory in 'Wikipedia' you will see how complicated this subject can very quickly get.
Naive set theory only has to be known to make some sort of progress, but I'm a perfectionist and so will define what I mean by a set in this context.
X is a set, if it is a collection of objects called members, where an object can be anything, (different people have different ideas about objects, for instance programming in C++ where the objects are classes with methods), when we can decide X is not a member of itself (avoiding Russell's paradox) and non-empty X has a member different to itself.
When I was at university and for that matter school, sets were explained in the context of diagrams and Venn drawings and this will suffice, at least for the time being.
A group G is a non-empty set and a binary law of composition called multiplication (juxtaposition or '.' ) or addition '+' with the following properties:
The bulletted points are axioms of a group and nothing else is allowed to be assumed without stating the assumptions. A group is an example of a mathematical structure, all truth about groups is derived from these basic assumptions. Other examples of mathematical structures are monoids, rings, fields, lattices and many more. You may have noticed that left and right identities and inverses are not assumed to be the same. A much more general definition of a group exists where only left identities and left inverses are stated, the proof the definitions are equivalent may be difficult and evades me at the moment, in any case this definition is a lot easier to deal with.You can of course use any letter to represent unity, the list is not exhaustive.
The chief forms of beauty are order and symmetry and definiteness, which the mathematical sciences demonstrate in a special degree.(Aristotle: Philosopher Scientist [384 BC-322 BC])
Anything displaying some sort of pattern has symmetry. Consider a decimal number between (0,1), their exists a number 0.a1a2a3... such that
a(∑n=1n=kn) = a(∑n=1n=k+1n) digit.
Then this number has a symmetry, but for the purpose of this discussion I am going to restrict myself to talking about regular polyhedra, the five so called 'Platonic Solids' and mainly polygons, regular n-sided two dimensional shapes.
Starting with the equilateral triangle; a shape is symmetric if after performing some real or imaginary action to it, its appearance and the space it occupies remains unchanged.
So we have three rotations re r1 r2 and three reflections through lines through each of its vertices bisecting its opposite side, m1 m2 m3 which are permutations of the letters a b c .
An anti-clockwise positive rotation through 120o labelled by r1 can be written as a permutation of three symbols (a b c) which is read as:
When combining elements by juxtaposition called multiplication it's usual to drop elements that remain fixed. So r2r1 working from right to left means first rotate through 120o then 240o which should give re.
Now it turns out that multiplication by this method is a perfectly reasonable binary law of composition and the rotations and reflections are perfectly reasonable members of a set G, and this set combined with the law of composition satisfies the axioms for a group.
Choosing re=(a)(b)(c) r1=(a b c) r2=(a c b) m1=(a b) m2=(b c) m3=(a c) then choosing to work by row then column we get the following multiplication table.
| Symmetry of an equilateral triangle | ||||||
|---|---|---|---|---|---|---|
| re | r1 | r2 | m1 | m2 | m3 | |
| re | re | r1 | r2 | m1 | m2 | m3 |
| r1 | r1 | r2 | re | m3 | m1 | m2 |
| r2 | r2 | re | r1 | m2 | m3 | m1 |
| m1 | m1 | m2 | m3 | re | r1 | r2 |
| m2 | m2 | m3 | m1 | r2 | re | r1 |
| m3 | m3 | m1 | m2 | r1 | r2 | re |
Using the above table you can check that it is a group, the group is called D3 the dihedral group acting on three elements. The above group is not commutative (abelian) meaning not all ab=ba a,b ε G (a,b belongs to G) with m1r1≠r1m1.
A permutation is a re-arrangement of letters or numbers; (1,2,3) and (1,3,2) are permutations of the numbers 1,2,3. For nεN(a whole number) letters their are
n! (n*(n-1)*...*1) permutations; considering an n-tuple (α1,α2,....,αn) we have n ways to choose the first member, for each first member a remaining (n-1) ways to choose the second member and so on. For the three letters 1,2,3 gives (3*2*1)=6 as above.
We can combine permutations with the above binary law of composition to obtain Sn , the group of permutations of n letters with n! elements.
Successfully divorcing S3 from the physical action of rotation or reflection, we can think of D3 as acting on the equilateral triangle, in the sense that if δ represents an equilateral triangle re(δ) does nothing and m2(reδ)=(m2re)δ means reflect δ in an axis.
All reflections are self inverses and combined with the identity (unity) form subgroups; {re,m2} for instance.
The table illustrates another subgroup:
| Symmetry of an equilateral triangle | ||||||
|---|---|---|---|---|---|---|
| re | r1 | r2 | m1 | m2 | m3 | |
| re | re | r1 | r2 | m1 | m2 | m3 |
| r1 | r1 | r2 | re | m3 | m1 | m2 |
| r2 | r2 | re | r1 | m2 | m3 | m1 |
| m1 | m1 | m2 | m3 | re | r1 | r2 |
| m2 | m2 | m3 | m1 | r2 | re | r1 |
| m3 | m3 | m1 | m2 | r1 | r2 | re |
Focusing on the set R of rotations {re,r1,r2} we can see it is a subgroup and divides the order of D3, the group of symmetry of an equilateral triangle.
Postmultiplying every member of R by a reflection gives a reflection and as you can check divides the table of D3 into four blocks. The set mR={mre,mr1,mr2} is called a left coset of R. The subscript of 'm' is dropped because the reflections images yield only a permutation of the set {m1,m2,m3}.
| Symmetry of an equilateral triangle | ||||||
|---|---|---|---|---|---|---|
| re | r1 | r2 | m1 | m2 | m3 | |
| re | re | r1 | r2 | m1 | m2 | m3 |
| r1 | r1 | r2 | re | m3 | m1 | m2 |
| r2 | r2 | re | r1 | m2 | m3 | m1 |
| m1 | m1 | m2 | m3 | re | r1 | r2 |
| m2 | m2 | m3 | m1 | r2 | re | r1 |
| m3 | m3 | m1 | m2 | r1 | r2 | re |
| R | mR |
| mR | R |
Obviously a rotation of a polygon will not leave any point fixed. The permutation (1 2 3 ... n) of an n-gon is a rotation and (1 2 3 ... n) is called a cycle of length n. Not all rotations of n-gons are cycles of length n, lettering the vertices of a square 1,2,3,4, gives the rotation rπ=(1 3)(2 4). A cycle of length 2 is called a transposition and the product of transpositions (1 2)(3 4) is a reflection in D4 the symmetry of the square. Clearly it will be hard to distinguish between rotations or reflections by consideration of permutations alone. Things get harder still in 3-space, for instance you can rotate a pyramid in the plane of its square base leaving fixed the point at the top.
The cycle of length n=(1 2 3 ... n) or any general cycle can be expressed as a product of transpositions, (1 n)(1 (n-1)) ...(1 3)(1 2) and the cycle is either an odd number of transpositions or an even number of transpositions, not both. (1 3 2)=(1 2)(1 3) is an even number of transpositions and any other combination equal to (1 3 2) will also be even. An elegant proof will follow later.
In the table above it didn't matter which reflection we chose to multiply R by to obtain the left coset mR. This in general will not always be the case and if {e}<H<G, H a proper subgroup of G and gH a left coset of H we would like multiplication of cosets, in the sense of multiplying representatives of the cosets, to be well defined. To mean if g' ε gH and k' ε kH then g'k' belongs to the same left coset as gk does. Put another way, it doesn't matter which elements we choose to multiply, their image will always be in the same set. It may be useful to note that g ε gH since e ε H and ge=g.
In order to achieve our goal, to ensure the cosets of a subgroup form a group equivalence relations are useful. They ensure a partition of a set (into disjoint non overlapping subsets) and will help determine if two cosets are equal, apart from shaking a stick at them.
~ is an equivalence relation defined between elements a, b, c ε X ↔ (if and only if)
Let a ε A and b ε B then divide X by an equivalence relation a ~ b → a ε B
(transitivity) if c ε A → c ~ a and a ~ b → c ~ b then c ε B → A ≤ B (A is contained in B)
(symmetry) a ~ b → b ~ a and b ε A and by a symmetric argument B ≤ A → A=B
The converse is easily achieved starting from a partition of X. qed
Since a set cannot be both equal to and not equal to another set an equivalence relation partitions a set into disjoint equivalence classes.
Let H be a non trivial subgroup of G and define the equivalence relation x,y εG x ~ y ↔ x-1y ε H:
Now if x' ε xH and y' ε yH
then x-1x' ε H and y-1y' ε H → x ~ x' and y ~ y'. We want x'y' to belong to the same coset as xy or xyH and therefore xy ~ x'y' and (xy)-1x'y' ε H.We want
y-1x-1x'y' ε H or y-1hy' ε H (since x-1x' ε H) or y-1Hy = H and Hy = yH
If x-1hx ε H for all h ε H and all x ε G then H is called a normal subgroup of G, and we have found exactly what we need for cosets to form a group, that is for them to be commutative or abelian. In the triangle example mR = Rm.
Let the notation |X| denote the number of members of X and the union of X and of Y, X U Y to be the set x ε X or x ε Y or both.
Proof: Partition G by the left cosets gH with equivalence relation g ~ g' ↔ g-1g' ε H.
Then the union of these cosets are G and their are finitely many of them, k say.
If gh = gh' where h,h' ε H then g-1gh = g-1gh' (g has an inverse) → h = h' → |gH| ≥ |H|
but gH = {gh1,gh2,gh3,...,ghn} obviously cannot have more members than H
→ |H| ≥ |gH| → |H| = |gH|.
Then |G| = k|H| and |H| divides |G| qed.
If the cosets (left or right) of H a subgroup of G form a group as in the R,mR example table above; then it is called a factor group with the convenient notation G/H; convenient because if G is finite the order of G/H or |G/H| called the index of G/H is just |G|/|H|. And we know the factor group exists when H is a normal subgroup of G.
D3/R has index 2; is R a normal subgroup of D3? Of course it is, because mR = Rm. What about Dn the symmetry's of an n-gon?
We have n rotations obviously, and these form a subgroup of Dn and so |Dn| = k|Rn|, but we only have rotations and reflections to consider, so the number of reflections must be a multiple of the number of rotations m ≥ 1 say.
If n is odd, a reflection axis passing through 2 vertices would only leave 2 vertices fixed, their must be an odd number of vertices to one side of the axis of reflection and an even number the other, which is impossible # So the axis' of reflection only pass through one vertex and the opposite edge, and their are n of them.
If n is even an axis of reflection can pass through 2 vertices, look at the square, and since their are n vertices their are n/2 axis' of reflection passing through them (we don't count axis' twice). But we know that the number of reflections is a whole number multiple of the number of rotations, so their must be some more and these are the n/2 reflections bisecting opposite sides leaving 0 points fixed.
For a polygon the number of rotations equals the number of reflections. For an n-gon Dn has 2n elements, and 2n = n! only when n=3, therefore Sn ≠ Dn for all n > 3.
Where Dn/Rn exists the index of Dn/Rn = 2. By the statement Dn/Rn exists I mean is a factor group, this can only occure when the product of two reflections is a rotation.
In mathematics different words are used to describe how a proposition fits into its environment. Indeed 'proposition' is sometimes used in place of 'theorem'. As a general guide a 'lemma' is a little theorem, which may or may not be little or easy to prove, but goes a long way to the main result mostly called a 'theorem'; and applications of the 'theorem', which may or may not be 'applications' in the real sense, but immediate consequences, are normally called corollaries, corollary in the singular.
A promised proof of the above theorem due to William I. Miller[ 1971 ] but maybe not displayed this way, is to be split into two parts, the technical and clever part I'm going to call a lemma, and its immediate consequence or corollary, although I'm not going to call it that, will be the theorem above.
Proof Let e be the identity and e=(a1 a2)(a3 a4)...(aj aj+1) say
a general product of transpositions.
Some number m appears in the list above and choose m with the least subscript i in the list
(a1 a2)...(ai ai+1)...(aj aj+1); m cannot be in the final transposition because on multiplying out m will be left fixed, try it and you will see this contradicts our construction of e.
So m must appear somewhere in the middle of the product and you can convince yourself the following are all the possibilities for pairs of transpositions with m in the first element.
Rule
Doing this (inductively) m cannot appear in the last transposition so eventually must arrive at the form of i),repeating for all letters in the product we obtain e=e.e.e.e...e where every e in the product is the result of removing 2 transpositions.So e is an even number of transpositions. qed
Restating the theorem:
Proof Let a be a permutation and ai,bj transpositions.
then a = a1a2a3...ar = b1b2b3...bs (new notation)
and possibly not all ai bj distinct.
A transposition is its own inverse → a-1 = ar-1ar-1-1...a1-1
so e = a-1a = ar-1ar-1-1...a1-1b1b2...bs
which is an even number of transpositions by the lemma. → r+s is even → both r and s even or odd.
Therefore a is either an even permutation or an odd permutation. qed
