mar02 - N G . Prove that | G / HN | and | N / N H | divide...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 4124 Wednesday, March 2 March 2, Ungraded Homework Exercise 4.1.4 on page 116 Let S 3 act on the set Ω of ordered pairs: { ( i , j ) | 1 i , j 3 } by σ (( i , j )) = ( σ ( i ) , σ ( j )) . Find the orbits of S 3 on Ω . For each σ S 3 find the cycle decomposition of σ under this action (i.e., find its cycle decomposition when σ is considered as an element of S 9 – first fix a labelling of these nine ordered pairs). For each orbit O of S 3 acting on these nine points pick some a O and find the stabilizer of a in S 3 . Thus, for example, ( 1 2 )( 3 , 2 ) = ( 3 , 1 ) . There are two orbits under S 3 , namely { (1,1), (2,2), (3,3) } and { (1,2), (2,3), (3,1), (1,3), (2,1), (3,2) } . The stabilizer of (1,1) in S 3 is h ( 2 3 ) i , and the stabilizer of (1,2) in S 3 is 1. The cycle decompositions are given as follows. (1) ( ( 1 , 1 ) ) (12) ( ( 1 , 1 )( 2 , 2 ) )( ( 1 , 2 )( 2 , 1 ) )( ( 1 , 3 )( 2 , 3 ) )( ( 3 , 1 )( 3 , 2 ) ) (23) ( ( 2 , 2 )( 3 , 3 ) )( ( 2 , 3 )( 3 , 2 ) )( ( 2 , 1 )( 3 , 1 ) )( ( 1 , 2 )( 1 , 3 ) ) (13) ( ( 1 , 1 )( 3 , 3 ) )( ( 1 , 3 )( 3 , 1 ) )( ( 1 , 2 )( 3 , 2 ) )( ( 2 , 1 )( 2 , 3 ) ) (123) ( ( 1 , 1 )( 2 , 2 )( 3 , 3 ) )( ( 1 , 2 )( 2 , 3 )( 3 , 1 ) )( ( 1 , 3 )( 2 , 1 )( 3 , 2 ) ) (132) ( ( 1 , 1 )( 3 , 3 )( 2 , 2 ) )( ( 1 , 2 )( 3 , 1 )( 2 , 3 ) )( ( 1 , 3 )( 3 , 2 )( 2 , 1 ) ) Let G be a finite group, let H G and let
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: N G . Prove that | G / HN | and | N / N H | divide | G / H | . Note that HN G , because N G (we need one of the subgroups in HN to be normal for HN to be a subgroup). Then | G / HN | = | G | / | HN | = ( | G | / | H | ) / ( | HN | / | H | ) = | G / H | / | HN / H | . Since | HN / H | is a positive integer, we deduce that | G / HN | divides | G / H | . By one of the isomorphism theorems, we have HN / N = H / H N and hence | HN | / | N | = | H | / | H N | . Therefore | N / N H | = | N | / | N H | = | HN | / | H | = ( | G | / | H | ) / | G | / | HN | = | G / H | / | G / HN | . Since | G / HN | is a positive integer, we conclude that | N / N H | divides | G / H | ....
View Full Document

This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.

Ask a homework question - tutors are online