Show that in a finite cyclic group G of order n, writtten multiplicatively, the equation x(adsbygoogle = window.adsbygoogle || []).push({}); ^{m}= e has exactly m solutions x in G for each positive integers m that divides n.

Attempt...

Proof:

Let G be a finite cyclic group of order n, and suppose m is a positive integer that divides n. Let x be an element in G. Then, x generates a cyclic subgroup of G that contains (n/m) elements since the gcd(n,m)=m [we know this by a theorem presented in the section0.

Hence,

(1) (x^{m})^{n/m}= x^{n}= e.

The positive integer m was picked arbitrarily. Thus, equation (1) will hold for each m that divides n. Hence, x^{m}=e has exactly m solutions.

QED??

Does this hold or am I missing something?

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# Homework Help: Another problem involving cyclic groups.

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