10Ma5aHw3Sol

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

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 03/03/2011 for the course MATH 5A taught by Professor Dinakarramakrishnan during the Fall '10 term at Caltech.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online