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HW8_Solutions

# HW8_Solutions - Math 541 Solutions to HW#8 The following...

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Math 541 Solutions to HW #8 The following are from Gallian, Chapters 7 and 9. The numbering is first given for the 7th edition, and is then followed by the 6th edition’s numbering (if the numbering is different). # 7.15 : Suppose | G | = pq with p and q prime numbers. By Lagrange’s Theorem, the possible subgroups of G have orders 1, p , q , or pq . For a subgroup H of G to be proper, it must have order less than G , so we can immediately eliminate pq . In the case that | H | = 1, H contains only e so it is clearly cyclic. In the case that | H | = p or | H | = q , then H has prime order, so it must be cyclic. This completes the argument. # 7.20 : Since H and K are both subgroups of G , we know that H K must contain e , so it is nonempty. Suppose there is some other element, g , that is in H K . Then g H and g K , so | g | > 1 must divide the order of H and of K , since both H and K are themselves groups. But, since | H | = 12 and | K | = 35 have no common divisors greater than 1, we have derived a contradiction. Thus, the only element in H K is e , so | H K | = 1. # 7.39 (resp. # 7.35 ): Since G has elements of orders 1 through 10, it’s clear that if | G | = k , then 10 | k , 9 | k , ..., 2 | k . Thus, the least possible order of G is lcm (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10) which is 2 3 · 3 2 · 5 · 7 = 2520. # 9.1 : No, H is not normal in S 3 . Consider that (13) H = { (13) , (123) } 6 = H (13) = { (13) , (132) } . # 9.7 : Suppose that H has index 2 in G . Then, for any a not in H , we have that aH H = and aH H = G . Therefore, the complement of H in G is simply aH . The exact same argument where Ha

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