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 ² , we may assume that G is cyclic. In this case the intersection ± i =0 ± a i ² 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, Inverse function, right inverse

Click to edit the document details