Cyclic subgroups of Torus Consider a torus $T_2:= S^1 \times S^1$. Fix integers $a,b$ such that gcd(a,b)=1. We define the circle $S^1(a,b) \subset T_2$ via the following embedding
\begin{align} 
S^1 &\hookrightarrow S^1 \times S^1 \\
t &\mapsto (t^a,t^b)
\end{align}
Further we define the finite cyclic group $\mathbb{Z}_n(a,b)$ to be the standard $\mathbb{Z}_n \subset S^1(a,b)$ i.e the cyclic group defined via the embedding
\begin{align} 
\mathbb{Z}_n &\hookrightarrow S^1 \times S^1 \\
\zeta &\mapsto (\zeta^a,\zeta^b)
\end{align}
where $\zeta$ is a primitive $\text{n}^{\text{th}}$ root of unity.

Given $\mathbb{Z}_n \subset S^1(a,b)$, under what conditions on $a^\prime, b^\prime$ is $\mathbb{Z}_n \subset S^1(a^\prime, b^\prime) \cap S^1(a,b)$.

I have tried working out several examples with different values for $n,a,b$, and for those values I do get that if $a^\prime \equiv a ~(mod ~n~)$ and $b^\prime \equiv b ~(mod ~n~)$ then
$\mathbb{Z}_n \subset S^1(a^\prime, b^\prime) \cap S^1(a,b)$.

Is $a^\prime \equiv a ~(mod ~n~)$ and $b^\prime \equiv b ~(mod ~n~)$ a necessary and sufficient condition for $\mathbb{Z}_n \subset S^1(a^\prime, b^\prime) \cap S^1(a,b)$?

If this is indeed true then any help in proving it would be appreciated.
 A: The circle subgroup corresponding to the pair $(a,b)$ is an $(a,b)$-torus knot. Say we want to compute the intersection of the $(a,b)$- and $(c,d)$-torus knots. Write
$$ (\zeta^a,\zeta^b)=(\xi^c,\xi^d). $$
With $\zeta=\exp(2\pi i s)$ and $\xi=\exp(2\pi i t)$ (where $s,t\in\mathbb{R}$) this becomes
$$ \begin{cases} as\equiv ct \\ bs\equiv dt \end{cases} \mod1 ​$$
(that is, as elements of $\mathbb{R}/\mathbb{Z}$). We can do $d$ times the first minus $c$ times the second to obtain $(ad-bc)s\in\mathbb{Z}$, and similarly $a$ times the second minus $b$ times the first for $(ad-bc)t\in\mathbb{Z}$, which implies $\zeta,\xi$ must be $n$st roots of unity, where $n=ad-bc$.
The homomorphisms $(x^a,x^b)$ and $(x^c,x^d)$ have the same image when restricted to the $n$th roots of unity. To see this, it suffices to show we may solve the system
$$ \begin{cases}
am \equiv c \\ bm\equiv d
 \end{cases}
\mod n $$
By Bezout's we may pick $x,y$ so that $ax+by=1$. Then if we add $x$ times the first equation and $y$ times the second we get $m\equiv cx+dy\mod n$. This may be verified a valid solution, e.g.
$$ \begin{array}{ll}
 am & =a(cx+dy) \\
 & = c(ax)+(ad)y \\
 & = c(1-by)+(ad)y \\
 & = c+(ad-bc)y \\
 & \equiv c
\end{array} $$
and similarly $bm\equiv d$ mod $n$. Moreover, if $\zeta$ is a primitive $n$th root then $(\zeta^a,\zeta^b)$ is also a primitive $n$th root, again by Bezout's. Therefore, the intersection is cyclic of order $n=ad-bc$.
This shows $a\equiv c,b\equiv d$ mod $n$ is sufficient for the intersection to contain a cyclic subgroup of order $n$, but it is not necessary, e.g. for $(a,b)=(2,-1)$, $(c,d)=(1,2)$, $n=5$.
