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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.

JustDanyul
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  • What is the definition of order of a permutation? What is the definition of order of a group? – Hagen von Eitzen Nov 23 '14 at 18:41
  • The definition of the order of a permutation is its the lowest common multiple of its cycles. And the definition of the order of a cyclic group is that is how many time you need to compose the generator with itself to obtain the identity. – JustDanyul Nov 23 '14 at 18:50
  • Is that really the definition or rather a proven method to quickly calculate it? DonÄt you have a definition for order of $g$ where $g$ is an element of an arbitrary group $G$? – Hagen von Eitzen Nov 23 '14 at 21:17

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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.

Thomas
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