ast1 - Math 4124 Monday, February 14 Solutions to Sample...

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

View Full Document Right Arrow Icon
Math 4124 Monday, February 14 Solutions to Sample First Test 1. Since | S 3 × S 5 | = 3!5! = 6! and | S n | = n !, we see that the only possible n for which | S 3 × S 5 | can be isomorphic to S n is 6. Now we look at the orders of the elements. The order of (( 123 ) , ( 12345 )) is the lowest common multiple of 3 and 5, which is 15; thus S 3 × S 5 has an element of order 15. The cycle shapes of elements of S 6 are (1), (12), (123), (1234), (12345), (123456), (12)(34), (12)(34)(56), (12)(345), (12)(3456), (123)(456), and these have orders 1,2,3,4,5,6,2,2,6,4,3 respectively; in particular S 6 has no elements of order 15. Therefore S 3 × S 5 is not isomorphic to S 6 , and the result follows. 2. (i) First G 6 = /0; this is obvious because II t = I which tells us that I G . (ii) Now let A , B G . We next show that AB G . The definition of G tells us that AA t = BB t = I , and we need to establish that ( AB )( AB ) t = I . But ( AB )( AB ) t = ABB t A t = AIA t = AA t = I , as required. (iii) Finally we must show that
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the 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.

Page1 / 2

ast1 - Math 4124 Monday, February 14 Solutions to Sample...

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