This is probably very simple, but I am just not getting how to do this how my pr

Question

This is probably very simple, but I am just not getting how to do this how my professor wants.

Prove this corollary
"In a finite cyclic group, the order of an element divides the order of the group"

Use this Theorem to prove it. "Let a be an element of order n in a group and let k be a positive integer. Then <ak>=<agcd(n,k)> and |a|= n/gcd(n,k)." This should be a fairly direct and simple proof.. Use this Theorem to prove it. "Let a be an element of order n in a group and let k be a positive integer. Then <ak>=<agcd(n,k)> and |a|= n/gcd(n,k)." This should be a fairly direct and simple proof..

Explanation / Answer

To state the theorem again, If G is a group and a is an element of G of order n then the subgroup generated by ak is the same as the subgroup generated by agcd(n,k). Further, |ak| = n/gcd(k,n).
If G is a group and a is an element of G of order n then the subgroup generated by ak is the same as the subgroup generated by agcd(n,k). Further, |ak| = n/gcd(k,n).
Suppose a is an element in a cyclic group G. Let a generator for G be ?. Write a = ?k for some integer k>0. (If k = 0 then a = 1 and clearly its order = 1 which divides n) We have <a> = <?k> = <?gcd(k,n)>. Clearly the order of a is the size of this subgroup. Hence |a| = |?k| = n/(gcd(k,n)). In particular, n/|a| = gcd(k,n) which is an integer, so |a| divides n. Suppose a is an element in a cyclic group G. Let a generator for G be ?. Write a = ?k for some integer k>0. (If k = 0 then a = 1 and clearly its order = 1 which divides n) We have <a> = <?k> = <?gcd(k,n)>. Clearly the order of a is the size of this subgroup. Hence |a| = |?k| = n/(gcd(k,n)). In particular, n/|a| = gcd(k,n) which is an integer, so |a| divides n. Suppose a is an element in a cyclic group G. Let a generator for G be ?. Write a = ?k for some integer k>0. (If k = 0 then a = 1 and clearly its order = 1 which divides n) We have <a> = <?k> = <?gcd(k,n)>. Clearly the order of a is the size of this subgroup. Hence |a| = |?k| = n/(gcd(k,n)). In particular, n/|a| = gcd(k,n) which is an integer, so |a| divides n.

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