model5 - ca = r a ( g ) = b . Multiplying both sides of the...

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

View Full Document Right Arrow Icon
MODEL ANSWERS TO THE FIFTH HOMEWORK 11. The solutions of the equation x 2 = e are precisely the elements of G which are their own inverses. The function i : G -→ G is a bijection, since i is its own inverse. The elements which are not their own inverses therefore come in pairs, so the number of elements which are their own inverses has the same parity as | G | . If | G | is even, then the number of elements which are their own inverses is therefore even. Since e is its own inverse, there is at least one other element of G which is its own inverse. 13. Let G = { a, e } . DeFne e · e = e · a = e and a · e = a · a = e . Then e is a left identity, and e is a right inverse of both e and a , but G is not a group, as a does not have a left inverse. 14. Given a G , deFne a function l a : G -→ G by the rule l a ( g )= ag . Suppose that l a ( g )= l a ( h ). Then ag = ah , so that g = h . But then l a is an injective function. As G is Fnite, it follows that l a is surjective. In particular we may Fnd e G such that ae = l a ( e )= a . Pick b G . DeFne a function r a : G -→ G by the rule r a ( g )= ga . Arguing as before, r a is a bijection. But then we may Fnd c G such that
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the 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 = aea-1 H . Thus aHa-1 is non-empty. Now suppose that b and c are elements of aHa-1 . Then we may Fnd h and g H such that b = aga-1 and c = aha-1 . In this case bc = ( aha-1 )( aga-1 ) = ah ( a-1 a ) ga-1 = a ( gh ) a-1 aHa-1 . 1 Thus aH a-1 is closed under products. Similarly b-1 = ( aha-1 )-1 = ah-1 a-1 aHa-1 . Thus aH a-1 is closed under inverses. 2...
View Full Document

Page1 / 2

model5 - ca = r a ( g ) = b . Multiplying both sides of the...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online