## Wednesday, March 27, 2013

### Permutation and Group

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A permutation of a permutation, is the product of the two permutations?
Right. The multiplication is not commutative, i.e in general abba. For example $$\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}$$ but $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}=\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}.$$
What happens if I keep permuting using the same permutation?
If you keep permuting with same permutation, you will eventually return to the initial configuration. For example $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix} =\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix} =\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}.$$ The identity permutation, usually denoted by (1), is the permutation that changes nothing. We know that, say, there are n objects, then applying the same permutation for n! times will always end up to be the identity permuation, i.e. $$\tau^{n!}=(1)$$
Can I say τn!-1=1/τ?
Not quite. But you can write τn!-1-1, meaning that it is the multiplicative inverse of τ in the permutation multiplication.

Muliplication with an identity element and inverses. It is a group!?
Right. It is a group.
For every element α of a group G, we define a function Aα(g)=αg. It is not hard to see that Aα is a bijection and so it is a permutation. Indeed we have AαAβ=Aαβ. Hence if we identify α with Aα, we see that each element of a group is indeed a permutation of the group!

## Friday, March 22, 2013

### Permutations

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Permutation of n objects is just a bijection on the set of these n objects?
Right.  If we don't specify the set, we means a permutation on the set {1,2,3,...,n}.

How to present a permutation?
Of course, we can present it as a bijection, e.g.
$$f(x)=\begin{cases}1 & x=1\\3 & x=2\\2 & x=3\end{cases}$$ However, traditionally we have a simpler presentation
$$\left(\begin{matrix}1&2&3\\1&3&2\end{matrix}\right)$$ which is, I believe, self-explanatory. The lower part is the new arrangement.

Why not just write (1 3 2)?
It is because we already use this symbol for a special type of permutations-the cycles. For instance (1 3 2) refers to the permutation f(1)=3, f(3)=2 and f(2)=1, i.e. $$\left(\begin{matrix}1&2&3\\3&1&2\end{matrix}\right).$$
It is easy to see cycles are interesting, mathematically?
First, we have to understand the multiplication of two permutations. If we consider two permutations $f$ and $g$ as two bijections on {1,2,3,...,n}, then their product is just the permutation $(fg)(x)=f\circ g(x)=f(g(x))$, that means if $g(r)=s$ and $f(s)=t$ then $(fg)(r)=s$. For instance, $$\left(\begin{matrix}1&2&3\\1&3&2\end{matrix}\right)\left(\begin{matrix}1&2&3\\3&1&2\end{matrix}\right)=\left(\begin{matrix}1&2&3\\2&1&3\end{matrix}\right).$$ It is known that every permutation is a product of cycles.

So cycles are the primes for permutations?
Not quite. The transpositions (i,j), i.e. the cycles of two elements, are the primes. Every permutation can be written as a product of transpositions. e.g. $$\left(\begin{matrix}1&2&3&4&5\\3&5&4&1&2\end{matrix}\right)=(1\, 3\, 4)(2\, 5)=(1\, 4)(1\, 3)(2\, 5).$$

## Thursday, March 14, 2013

### 1=0!?

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(Growing lazy, not a good sign....)

I have seen a proof, saying that 1=0!?
Which one do you mean '"1=0!"?' or '"1=0"!?', uh?

You got the trick. Why do we use "!" for factorial?
I am not a historian, I don't know why Christian Kramp choose n! to represent factorial.

Why does 0!=1?
Lets first consider the definition of factorial. The n! is the number of ways to permute n objects. In the set theory language, n! is the size of the set of all bijective maps on a set of n elements.
Let A be a set. A bijective map or a bijection or a permutation is a subset F of A×A such that
1. For any elements a∈A, there exists a unique F(a)∈A such that (a,F(a))∈F.
2. For any elements a∈A, there exists a unique F-1(a)∈A such that (a,F-1(a)∈F.
For example the map on {x,y,z} such that F(x)=y, F(y)=z, F(z)=x is a bijection (and hence a permutation), but G(x)=y, G(y)=x, G(z)=x is not.

On {x,y,z}, we have six permutations, right?
Yes. They are
• (F(x), F(y), F(z))=(x, y, z)
• (F(x), F(y), F(z))=(x, z, y)
• (F(x), F(y), F(z))=(y, x, z)
• (F(x), F(y), F(z))=(y, z, x)
• (F(x), F(y), F(z))=(z, x, y)
• (F(x), F(y), F(z))=(z, y, x)
Hence 3!=6.

On empty set ∅, ...?
Note that ∅ equals to and is the only subset of ∅×∅. There is no way we can violate the two conditions, and so ∅ is the bijection on ∅. Therefore 0!=1.

## Thursday, March 7, 2013

### Cardinal Numbers

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(A blog a week, keep my lazy away.... sort of)

Ordinal numbers are about order. Cardinal numbers are about quantities. In the previous post, we know what is an ordinal number. What is a cardinal number?
A cardinal number is a set of sets such that whenever we pick two sets in it, we can pair up their elements.  If two sets lie inside the same cardinal number, we say that they are of the same cardinality.
For example,  {a,b,c} and {x,y,z} are of the same cardinality, as we can pair up their elements as (a,x), (b,y) and (c,z).

A natural number is a cardinal number. Isn't it?
Well, we use a natural number as the representative of its cardinal number.
For example, {a,b,c} and 3={0,1,2} are of the same cardinality, and so we say that {a,b,c} has 3 elements.

So, if the elements of a set can be labelled as 0,1,2,...,n-1, then the set has n elements. Right?
Exactly.  It is just our usual sense of numbers. However, we are used to label the elements with 1,2,3,...,n instead.

The first infinite cardinal number is still ω={0,1,2,3,...} ?
Yeah.  Unlike ordinal numbers, 1+ω=ω=ω+1.
The sum of cardinal numbers are defined by, #A+#B=#(A×{0} ∪ B×{1}).
For example, 3+4=#{a,b,c}+#{x,y,z,w}=#{(a,0),(b,0),(c,0),(x,1),(y,1),(z,1),(w,1)}=7.

So, m+n=n+m?
Right. It is easy to see that A×{0} ∪ B×{1} and A×{1} ∪ B×{0} should have the same cardinality.

I guess the product of two cardinal numbers is (#A)(#B)=#(A×B)?
Good. For example
3×4
=(#{a,b,c})(#{x,y,z,w})
=#{(a,x),(a,y),(a,z),(a,w),(b,x),(b,y),(b,z),(b,w),(c,x),(c,y),(c,z),(c,w)}
=12.
Again, we have m×n=n×m.