*This preview shows
pages
1–3. Sign up to
view the full content.*

This ** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **150A Algebra Monica Vazirani November 3, 2010 Midterm Name: ID: Section: 1. (10 points) In parts (a)-(b), give a careful definition of the terms in boldface. It is set up in several cases that you can just complete the sentence. a. Let : G G be a homomorphism. Then its kernel , ker = ... Solution: { g G | ( g ) = 1 } b. The order of an element a G is ... Solution: The smallest positive integer m Z > such that a m = 1 . If no such finite m exists, we say the order is infinite. 2. (15 points) Let G be a group. Suppose x G has (finite) order s . Prove that x m = 1 if and only if s divides m . Solution: = If s | m then m = sk for some k Z . Then x m = x sk = ( x s ) k = 1 k = 1 . = By the Division Algorithm, we can write m = ks + r with r < s (and k,r Z ). Because 1 = x m = x sk + r = ( x s ) k x r = 1 x r = x r , by the minimality of s , it must be that r = 0 . But then m = ks showing s | m . Name: ID: Section: 2 3. (20 points) In parts (a) - (d), determine if the given statements are...

View Full
Document