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

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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 = /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 G is closed under taking inverses, so let A G . We want to prove that A - 1 G , in other words A - 1 ( A - 1 ) t = I . But A G implies AA t = I , implies A t = A - 1 , implies A = ( A -

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