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HW5notes - c 1 HOMEWORK 5:GRADERS NOTES AND SELECTED...

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c 1 HOMEWORK 5:GRADER’S NOTES AND SELECTED SOLUTIONS Chapter 4, Page 83, No. 21 Let G be a group and let a be an element of G . (1) a 12 = e what can we say about the order of a ? (2) a m = e what can we say about the order of a ? (3) that divides.alt0 G divides.alt0 = 24 and G is cyclic, if a 8 e and a 12 e show that a ⟩ = G . Proof. (1) If a 12 = e then we know that the order of a , divides.alt0 a divides.alt0 divides 12. so divides.alt0 a divides.alt0 = 1,2,3,4,6, or 12, since these are the non-negative divisors of 12. (2) If a m = e then all we can say is that the order of a is a divisor of 12. (3) Suppose that divides.alt0 G divides.alt0 = 24 and G is cyclic, by Corollary 1 on page 2 we know that the order of an element a , divides G , so divides.alt0 a divides.alt0 = 1,2,3,4,6,12, or 24 since these are the divisors of 24. Now since a 8 e a 12 we know, immediatly that divides.alt0 a divides.alt0 ≠ 8 , 12 it follows that divides.alt0 a divides.alt0 cannot be a divisor of 8 or 12 either, suppose that d divides n where n is either 12 or 8. Then we can write
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