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Let $G$ be a cyclic group of finite order $n$. Then, the order of the cyclic subgroup $\langle g^k\rangle$ is $\frac {n}{\operatorname{gcd}(n,k)}$

Proof:

$G$ cyclic $\Rightarrow \exists g \in G: G = \langle g\rangle$ $\Rightarrow \operatorname{ord}(g) = \lvert G\rvert = n $

We know, because every subgroup of a cyclic group is cyclic, that $\lvert\langle g^k\rangle\rvert = \operatorname{ord}(g^k)$. Therefore, we only need to find the order of $g^k$.

Let $l$ be the order of $g^k$. Then:

$(g^k)^l = e \iff g^{kl} = e \iff n|(kl)$

I'm now trying to show that $l = n/{\operatorname{gcd}(n,k)}$, but this seems difficult for me. I tried to assume that there was an element $l' < n/{\operatorname{gcd}(n,k)}$ that satisfied the conditions, in the hope of reaching a contradiction, but this did not work out. Can anyone help?

tomasz
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  • See also http://math.stackexchange.com/a/2216035/589. – lhf Apr 04 '17 at 10:28
  • Instead of assuming that $G$ is cyclic of order $n$, you should probably assume that $G=\langle g\rangle$ and $g$ has order $n$. Otherwise, the conclusion doesn't make much sense ($g$ is undefined, it could very well be the identity). This makes the first part of your proof moot. – tomasz Apr 04 '17 at 10:37
  • Also in your second paragraph, the fact that order of $\langle g^k\rangle$ is the order of $g^k$ has nothing to do with the fact that a subgroup of a cyclic group is cyclic. $\langle g^k\rangle$ is cyclic by definition, regardless of the group in the background. – tomasz Apr 04 '17 at 10:41

2 Answers2

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From where you are:

$n \vert kl$

Let $m = \gcd(n,k)$. Then (since $\gcd (\frac{n}{m}, \frac{k}{m}) = 1)\mid l \Rightarrow$

$$\frac{n}{m} \leq l $$

But $(g^k)^{\frac{n}{m}} = g^{(n\frac{k}{m})}= e^{\frac{k}{m}}=e \Rightarrow$

$$l \leq \frac{n}{m}$$

We conclude that:

$$l = \frac{n}{m}$$

Bernard
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Pellenthor
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Note that $kl$ must be the least common multiple of $k$ and $n$.

Then make use of: $$nk=\gcd(n,k)\times\text{lcm}(n,k)$$

drhab
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