HW6notes - HOMEWORK 6:GRADERS NOTES AND SELECTED SOLUTIONS...

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Unformatted text preview: HOMEWORK 6:GRADERS NOTES AND SELECTED SOLUTIONS page 84, no. 55 If a = 24 and b = 10 find the possibilities for a b . Note that every element x of a b is also an element of the cyclic groups a , b , x divides a = 24 and b = 10, since the only common divisors of 10 and 24 are 1 and 2 then it follows x = 1 or 2. Here is a fine point that many students did not really notice/consider in there solutions: There could, concievably be more than one element of a b with order 2, let x,y be distinct elements of order 2 in a b , but by corollary 2 from page 76 then x and y would generate the same cyclic subgroups of a , b , but since both are of order 2, then x = { e,x } = y = { e,y } so x = y , contradicting the assumption that there could exist distinct elements of a b of order 2. So the only possibilities are that a b = 1 or 2. page 114, no. 6 Show that A 18 contains an element of order 15 ....
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This note was uploaded on 10/27/2009 for the course MATH 330 taught by Professor Staff during the Spring '08 term at Ill. Chicago.

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HW6notes - HOMEWORK 6:GRADERS NOTES AND SELECTED SOLUTIONS...

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