Multiplicative orders Vs order of a multiplicative group [closed]

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How to demonstrate that all multiplicative orders divide the order (size) of the multiplicative group F of F13. .

Answer 1

You show that the cyclic group <x> generated by an element x is a subgroup of IF* and that "u~v iff u^(-1)*v in <x>" is an equivalence relation that divides the multiplicative group into equivalence classes of equal size.

So that you get

[size of IF*] 
     = [size of <x>] * [number of equivalence classes]

which means that the order of x = [size of <x>] is a divisor of the number of invertible elements, i.e., the size of the multiplicative group of IF

See also the little theorem of Fermat.

Answer 2

Since the group is abelian, the simplest thing is to use that multiplication by any element is a bijection. Let F = {g1, g2, g3, ..., gn} and let h be an arbitrary element. Then also F = {h*g1, h*g2, ..., h*gn}. Hence multiplying all elements together we get g1 * g2 * g3 * ... * gn = h*g1 * h*g2 * ... * h*gn. But the latter equals h^n * g1 * g2 * ... * gn. Now use the cancellation law to conclude that h^n = 1 from which the result follows.