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20. suppose that g is an abelian group of order 35 and every element of g satisfies the equation x = e. prove that g is cyclic. does your argument work if 35 is
20. Suppose that G is an Abelian group of order 35 and every element of G satisfies the equation x = e. Prove that G is cyclic. Does your argument work if 35 is replaced with 33? ont ofC
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Deepak singh 1 Best Answer
If (∀g∈G): x^35=e , then the order of every element of G is 1, 5, 7 or 35. The goal is to prove that some element of G has order 35. Suppose that every element other than e has order 5. Then every element of G∖{e} belongs to some subgroup of order 5. Each such subgroup will consist of 4 elements of order 5 plus e . But there's a problem here: 35 is not of the form 4k+1. So, some element from G∖{e} must have order 7 or 35. And, by the same argument, not all elements from G∖{e} have order 7. Therefore, some a∈G has order 5 or 35 and some b∈G∖{e} has order 7 or 35. If one of them has order 35, we're done: G is cyclic. Otherwise, ab has order 35 and, again, G is cyclic. And, clearly, this argument does not apply to 33 because 33 does not divide 35 .