01.02.2015 Views

Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

UNIT-1 27<br />

In this case the length of the cycle is 3. We can denote this permutation by (a b c).<br />

Definition. A cyclic permutation of length 2 is called a Transposition.<br />

1<br />

For example,<br />

<br />

1<br />

2<br />

3<br />

3<br />

<br />

is a transposition.<br />

2<br />

Definition. Two cycles are said to be disjoint if they have no object in common.<br />

Definition. Two permutations α, β ∈ S n are called disjoint if<br />

for all x ∈ S.<br />

α(x) = x <br />

β(x) ≠ x<br />

α(x) ≠ x β(x) = x<br />

In other words, α and β are disjoint if every x ∈ S moved by one permutation is fixed by the other.<br />

Further, if α and β are disjoint permutations, then αβ = βα. For example, if we consider<br />

then αβ = βα .<br />

1<br />

α =<br />

<br />

2<br />

2<br />

3<br />

3<br />

1<br />

2 3<br />

, β =<br />

1<br />

<br />

<br />

1<br />

2 3<br />

Definition. A permutation α ∈ S n is said to be regular if either it is the identity permutation or it has<br />

no fixed point and is the product of disjoint cycles of the same length.<br />

For example,<br />

1<br />

<br />

2<br />

2<br />

3<br />

is a regular permutation.<br />

3<br />

1<br />

4<br />

5<br />

5<br />

6<br />

6<br />

= (1 2 3) (4 5 6)<br />

4<br />

Theorem 15. Every permutation can be expressed as a product of pairwise disjoint cycles.<br />

Proof. Let S = {x 1 , x 2 ,…, x n } be a finite set having n elements and f ∈ S n . If f is already a cycle, we are<br />

through. So, let us suppose that f is not a cycle. We shall prove this theorem by induction on n.<br />

If n = 1, the result is obvious. Let the theorem be true for a permutation of a set having less than n<br />

elements. Then there exists a positive integer k < n and distinct elements y 1 ; y 2 ,…, y k in {x 1 , x 2 ,…, x n }<br />

such that<br />

f(y 1 ) = y 2<br />

f(y 2 ) = y 3<br />

………..<br />

………..<br />

f(y k−1 ) = y k<br />

f(y k ) = y 1<br />

Therefore (y 1 y 2 …. y k ) is a cycle of length k. Next, let g be the restriction of f to<br />

T = {x 1 , x 2 ,…, x n } − {y 1 , y 2 ,…, y k }<br />

Then g is a permutation of the set T containing n−k elements. Therefore, by induction hypothesis,<br />

g = α 1 α 2 … α m ,<br />

where α 1 , α 2 ,…, α m are pairwise disjoint cycles. But

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!