10Ma5aHw3Sol

10Ma5aHw3Sol - Ma 5 a HW ASSIGNMENT 3 FALL 2010 SOLUTIONS...

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Unformatted text preview: Ma 5 a HW ASSIGNMENT 3 FALL 2010 SOLUTIONS The exercises are taken from the text, Abstract Algebra (third edi- tion) by Dummitt and Foote. Page 45, 23 . Lets call the set of 3 pairs of opposite faces of the cube: S = {{ a 1 ,a 2 } , { b 1 ,b 2 } , { c 1 ,c 2 }} , and denote by G the group of rigid motions (of the cube). Now the element in G that rotates by about an axis connecting the center of 2 faces fixes S . Therefore G is not faithful. Suppose g G is in the kernel. Then g fixes S . If g fixes 2 pairs of faces, then g fixes the third pair, since it is a rigid motion. Therefore, g can fix exactly 3, when it is the identity, or 1 pair of faces. Thus there are 4 elements in the kernel. Page 48, 6 . Let H = { g G | | g | < } . Then since e has trivial order, H is nonempty. Take g and h in H with orders a , b < . Then since G is abelian, ( gh ) ab = ( g a )( h b ) = 1. Therefore gh also has finite order, and H is closed under products. Suppose g- 1 is the inverse of g , then 1 = ( gg- 1 ) a = g a ( g- 1 ) a = ( g- 1 ) a . Therefore g- 1 also has finite...
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This note was uploaded on 03/03/2011 for the course MATH 5A taught by Professor Dinakarramakrishnan during the Fall '10 term at Caltech.

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10Ma5aHw3Sol - Ma 5 a HW ASSIGNMENT 3 FALL 2010 SOLUTIONS...

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