model5

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

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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

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

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