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_{n}that satisfy the equation x

^{n}=e.

well, if m|n and x

^{m}=e then x

^{n}=e, hence, if we raise x (which is of order m and m divides n) to the exponent n we'll have x

^{n}=e. so the answer will be this:

S={all cycles of the length m s.t. m|n}

Is that a true conclusion?

As an example, in S4, since divisors of 4 are 1,2 and 4 the solutions will be all cycles with lengths 1,2 and 4. namely, {e,(1 2),(1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3 4), (4 3 2 1), (1 3)(2 4), (1 2)(3 4), (1 4)(2 3)}

Is there anything I'm missing?