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HW7_Solutions

HW7_Solutions - Math 541 Solutions to HW#7 1 Let G be a...

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Math 541 Solutions to HW #7 1. Let G be a group and let a be an element of G with order n . (a) Prove ( bab - 1 ) n = e . ( bab - 1 ) n = bab - 1 · bab - 1 · bab - 1 · ... · bab - 1 | {z } n times = ba · ( b - 1 b ) · a · ( b - 1 b ) · a · ( b - 1 b ) · ... · ( b - 1 b ) · ab - 1 = ba · e · a · e · a · e · ... · e · ab - 1 = ba n b - 1 = beb - 1 = bb - 1 = e (b) Prove that the order of bab - 1 is exactly n . By (a), we know that the order of bab - 1 is no greater than n . So, suppose that the order of bab - 1 is m with m < n . Therefore, it must be true that e = ( bab - 1 ) m = ba m b - 1 . Consequently, b - 1 eb = b - 1 ba m b - 1 b . Reducing both sides, we get that e = a m . But this implies that the order of a is less than n , in violation of our hypothesis. We conclude that the order of bab - 1 must be n . 2. Let G be a group such that a 2 = e for all a G . Prove that G is abelian. [Hint: To show ab = ba use the fact that a 2 = b 2 = ( ab ) 2 = e .] Since ( ab ) 2 = e , we have that abab = e . Multiplying on the left by ba , we get ba · abab = bae ba 2 bab = ba bebab = ba b 2 ab = ba eab = ba ab = ba 3. Let G be a group of size 6 with 5 elements of order 2 and 0 elements of order 3 - i.e. case 2a. (a) Prove that G is abelian. Since G has 5 elements of order 2, the 6th element must be the identity. Thus, a 2 = e for all a G . Applying the result from problem 2, we see that G is abelian. (b) Let a and b be two distinct elements in G with order 2. Prove that ab does not equal e , a or b . We break it into three cases: Suppose ab = e . Since aa = e , we get ab = aa . By canceling (i.e. multiplying by a - 1 on the left), we get b = a which is a contradiction. Suppose ab = a . Cancellation gives b = e , a contradiction. Suppose ab = b . Cancellation gives a = e , a contradiction. 1
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(c) Prove that the 4 element collection { e, a, b, ab } is a subgroup of G .
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