I was wondering, how can I prove that all cyclic subgroups generated by a permutation, has the same order as the permutation?
For example, cyclic subgroup $\langle(---)\rangle$ will have order 3. So far, my text book haven't given a proof for this, actually, it haven't even stated it. I just seems like this is the case.
Is my conjecture correct? And how would I go about proving it?
Just to be clear, I understand why the order of a cyclic group is $n$. As per the answer below. What I dont't understand is, why does a cyclic group generated by a permutation of a given order, end up with the same order as the permutation which generated it.
For example, the permutation (--)(--) have order 2, since the lowest common multiple of the two cycle lengths are 2. Now, $\langle(--)(--)\rangle$ will be a cyclic subgroup of order 2. So basically, the cycle subgroup order seems to be equal to the permutation order.