Question

2. In this problem, we will prove the following result: If G is a group of order 35, then G is isomorphic to Zg We will proceed by contradiction, so throughout the following questions, assume that G is a group of order 35 that is not cyclic. Most of these questions can he solved independently I. Show that every element of G except the identity has order 5 or 7. Let Ts be the set of elements of order 5 and let T be the set of elements of order 7. (These are not subgroups!) 2. Show that if H and H2 are two subgroups of order 5, then either Hn H2-e or H H2. Use this to show that the subgroups of order 5 partition Ts into disjoint sets of four elements each. Show similarly that T7 can be parititioned into sets of order 6. Conclude that 4-(# of subgroups of order 5) + 6 . (# of subgroups of order 7)-34. 3. Use the equation above to conclude that there is at least one subgroup of order 5 and at least one subgroup of order 7, 4. Let a be an element of onder 5 and let y be an element of order 7 (ie., acTs.y CT). Show that z and y do not commute. (Proceed by contradiction: assume that ry ya and show that this assumption lcads to G being cyclic.) 5, Fix a y E For each z E T, , let C,-(a, yzy-1, y2 y-2, . , . , y6зу 6} Show that these seven elements are distinct 6. Suppose w z e T Show that Crzy . C, fr any k e Z and that if C nd is not empy then С., C Conclu de that 7. Show that IT| is divisible by 5 and therefore that it is divisible by 30 8. Dcrive a contradiction, showing that G is cyclic , is divisible by 7. in th refore hal il is sible by 28.

Please answer the parts 6 and 7. Thank you.

Answer 1

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