06.09.2021 Views

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

SHOW MORE
SHOW LESS

Create successful ePaper yourself

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

F.9. PERMUTATIONS 477<br />

smallest k ∈ I n which was not included and begin a cycle starting with this. Continue this<br />

way. Use the convention that (k) is just the identity. This representation is unique up to<br />

order of the cycles which does not matter because they commute. Note that a transposition<br />

canbewrittenas(a, b).<br />

A cycle can be written as a product of non disjoint transpositions.<br />

(i 1 ,i 2 , ··· ,i m )=(i m−1 ,i m ) ···(i 2 ,i m )(i 1 ,i m )<br />

Thus if m is odd, the permutation has sign 1 and if m is even, the permutation has sign<br />

−1. Also, it is clear that the inverse of the above permutation is (i 1 ,i 2 , ··· ,i m ) −1 =<br />

(i m , ··· ,i 2 ,i 1 ) .<br />

Definition F.9.3 A n is the subgroup of S n such that for σ ∈ A n , σ is the product of an<br />

even number of transpositions. It is called the alternating group.<br />

The following important result is useful in describing A n .<br />

Proposition F.9.4 Let n ≥ 3. Then every permutation in A n is the product of 3 cycles<br />

and the identity.<br />

Proof: In case n =3, you can list all of the permutations in A n<br />

( ) ( ) ( )<br />

1 2 3 1 2 3 1 2 3<br />

,<br />

,<br />

1 2 3 2 3 1 3 1 2<br />

In terms of cycles, these are<br />

(1, 2, 3) , (1, 3, 2)<br />

You can easily check that they are inverses of each other. Now suppose n ≥ 4. The<br />

permutations in A n are defined as the product of an even number of transpositions. There<br />

are two cases. The first case is where you have two transpositions which share a number,<br />

(a, c)(c, b) =(a, c, b)<br />

Thus when they share a number, the product is just a 3 cycle. Next suppose you have the<br />

product of two transpositions which are disjoint. This can happen because n ≥ 4. First note<br />

that<br />

(a, b) =(c, b)(b, a, c) =(c, b, a)(c, a)<br />

Therefore,<br />

(a, b)(c, d) = (c, b, a)(c, a)(a, d)(d, c, a)<br />

= (c, b, a)(c, a, d)(d, c, a)<br />

and so every product of disjoint transpositions is the product of 3 cycles. <br />

Lemma F.9.5 If n ≥ 5, then if B is a normal subgroup of A n , and B is not the identity,<br />

then B must contain a 3 cycle.<br />

Proof: Let α be the permutation in B which is “closest” to the identity without being<br />

the identity. That is, out of all permutations which are not the identity, this is one which<br />

has the most fixed points or equivalently moves the fewest numbers. Then α is the product<br />

of disjoint cycles. Suppose that the longest cycle is the first one and it has at least four<br />

numbers. Thus<br />

α =(i 1 ,i 2 ,i 3 ,i 4 , ··· ,m) γ 1 ···γ p

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

Saved successfully!

Ooh no, something went wrong!