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?
update
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.
Best Answer
This is borderline trivial ... if you get your definitions right.
Consider a cyclic group generated by an element $g$. Then the order of $g$ is the smallest natural number $n$ such that $g^n = e$ (where $e$ is the identity element in $G$). Now the group $G$ is exactly all the powers of $G$: $$ G = \langle g \rangle = \{g, g^2, g^3, \dots, g^{n-1}, e = g^n\} $$ This group will have $n$ elements exactly because the order of $g$ is $n$. If it was less than $n$, say $m$, then you would have $g^m =e$. And it certainly isn't greater.