How many elements of order 15 does a cyclic group of order 3

How many elements of order 15 does a cyclic group of order 35 have? Justify your answer. No element has order 15 in a cyclic group of order 35.

Solution

Assume there are no elements of order 3535.

As 35=5735=57, by Sylow\'s theorem there must be 1(mod5)1(mod5) subgroups of order 55. And 1(mod7)1(mod7)of order 77.

Now if there are 88 subgroups of order 77, then each one contains 66 distinct elements ++ id, so there would be 68=4868=48 distinct elements in the group. This is clearly false, so there can only be 11subgroup of order 77. So there are 66 elements of order 77.

Aside from the identity all the other elements must have order 55. So there are 3561=283561=28elements of order 55. We can split these into distinct subgroups each containing 44 distinct elements ++ id. So there are 77 subgroups of order 55.

But 7=2(mod5)7=2(mod5) so this contradicts Sylow\'s theorem. So there must be an element of order 3535and GG is cyclic.