Let f=(x1,x2,...,xr) be in Sn. Show that o(f) = r Solution consider f^r. That is

Question

Let f=(x1,x2,...,xr) be in Sn. Show that o(f) = r

Explanation / Answer

consider f^r. That is, (x1,x2,...,xr)(x1,x2,...,xr)...(x1,x2,..… r times. This is really a composition of mappings. In other words, f: x1 -> x2. For the first function, we get f: x1 -> x2. For the second, we get f: x2 -> x3. We repeat this process. We see that the kth application of the function maps xk to x(k+1) (if need be, you can prove this by induction) The rth application would then take xr -> x(r+1), but there is no x(r+1), so it loops around from xr to x1. Therefore, an rth application will map x1 -> x1, x2 -> x2, etc. Thus, we know that f^r = e. So, we know that r is a multiple of the order of f. Suppose h is a natural number such that h x(h+1) But since there are r elements in (x1,x2,...,xr) and h

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