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
Unformatted text preview: ca = r a ( g ) = b . Multiplying both sides of the equation ae = a on the left by c , we see that be = b , so that e is a right inverse. inally, as l b is surjective, we may Fnd b such that bb = l b ( b ) = e . But then b is a right inverse of b , and so G is a group. 1. Done in class. 2. Suppose that G contains an element of inFnite order a . Replacing G by a , we may assume that G is cyclic. In this case the intersection i =0 a i is { e } . 4. (a) Note that e = aea1 H . Thus aHa1 is nonempty. Now suppose that b and c are elements of aHa1 . Then we may Fnd h and g H such that b = aga1 and c = aha1 . In this case bc = ( aha1 )( aga1 ) = ah ( a1 a ) ga1 = a ( gh ) a1 aHa1 . 1 Thus aH a1 is closed under products. Similarly b1 = ( aha1 )1 = ah1 a1 aHa1 . Thus aH a1 is closed under inverses. 2...
View Full
Document
 Spring '08
 LONG
 Algebra

Click to edit the document details