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It is known that

For a cyclic group $C_n = \langle g \rangle$ of order $n$, we have $\langle g^k \rangle = \langle g^{(k, n)} \rangle$, where $k \in \mathbb{Z}$.

I am able to verify this result. Now, I try to derive it by showing something like

If $\langle g^k \rangle = \langle g^l \rangle$, then we have $l = (k, n)$ (in terms of "mod n").

$\langle g^k \rangle = \langle g^l \rangle$ means that $g^k \in \langle g^l \rangle$ and $g^l \in \langle g^k \rangle$.

That is, $\exists t \in \mathbb{Z}: g^k = g^{lt}$ and $\exists s \in \mathbb{Z}: g^l = g^{ks}$.

Therefore, $k \equiv lt \; (\text{mod}\; n)$ and $l \equiv ks \; (\text{mod}\; n)$.

How to proceed with this argument?

hengxin
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  • You can proceed like here. – Dietrich Burde Mar 16 '19 at 09:09
  • @DietrichBurde Thanks. Because $\frac{n}{(k,n)} = \frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $\langle g^k \rangle = \langle g^l \rangle$? Could you please show me more hints? – hengxin Mar 16 '19 at 09:41

1 Answers1

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If $d\mid n$, then the order of $\langle g^d\rangle$ is $n/d$. Thus, the order of $\langle g^k\rangle=\langle g^{(n,k)}\rangle$ is $n/(n,k)$, and similarly, the order of $\langle g^l\rangle=\langle g^{(n,l)}\rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $\langle g^k\rangle=\langle g^l\rangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.

W-t-P
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