ahw3 - Math 3124 Thursday, September 15 Third Homework...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 3124 Thursday, September 15 Third Homework Solutions 1. First put the position so that there is the space is in the bottom right hand corner; there will be several ways of doing this and it does not matter which you choose. We will choose 12 14 5 7 10 3 9 11 8 2 1 13 6 4 15 . We now obtain the permutation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 12 14 5 7 10 3 9 11 8 2 1 13 6 4 15 We must decide whether this is a product of an even number or an odd number of transpositions (2-cycles). Writing the permutation as a product of disjoint cycles, we obtain ( 1 12 13 6 3 5 10 2 14 4 7 9 8 11 ) This is a 14-cycle, hence a product of 13 transpositions (cf. problem 6.9 on page 40) which is odd. Therefore the position cannot be reached. 2. Problem 6.4 on page 40 Write each of the given permutations as a product of dis- joint cycles. Also determine whether the permutation in S 6 is in A 6 . (a) 1 2 3 4 5 6 1 6 4 5 3 2 This sends 1 1, 2 6 2, 3 4 5 3, so when written as a product of disjoint cycles is (2 6)(3 4 5). A 2-cycle is a product of one transposition and a 3-cycle is a product of two transpositions. Therefore theone transposition and a 3-cycle is a product of two transpositions....
View Full Document

Page1 / 2

ahw3 - Math 3124 Thursday, September 15 Third Homework...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online