prove that a cycle of length k has order kSolutionIf a1a2ak

prove that a cycle of length k has order k

If =(a1,a2,…,ak), then the order of is k because k=Id.

Denote =(a0,a1,…,ak1)) (to simplify notation later on).

Proposition: for all nN and jZ/kZ it holds that n(aj)=aj+n(modk).

Proof: by induction, since trivally (aj+n(modk))=aj+n+1(modk).

Now note that by our proposition k=Id, and for all j<kj<k we have jId.

If =(a1,a2,…,ak), then the order of is k because k=Id.

Denote =(a0,a1,…,ak1)) (to simplify notation later on).

Proposition: for all nN and jZ/kZ it holds that n(aj)=aj+n(modk).

Proof: by induction, since trivally (aj+n(modk))=aj+n+1(modk).

Now note that by our proposition k=Id, and for all j<kj<k we have jId.