Let $m,l\in\mathbb{N}$ be such that $\gcd(m,l)=1$. I denote by $S^1$ the circle (i.e. $S^1 = \{z\in\mathbb{C} : |z|=1\}$) and by $C_n$ the cyclic group of order $n$ (considered as a subgroup of roots of unity of order $n$ in $S^1$).
Consider the following subgroup of $S^1\times S^1$:
$$H_{m,l}:=\{(s,t)\in S^1\times S^1 : s^m = t^l\}$$
I can prove using "abstract" arguments that $H=H_{m,l}$ is isomorphic to $S^1\times C_n$ for some $n\in\mathbb{N}$.
Proof: induced by the standard topology, $H$ is a closed subgroup of $S^1\times S^1$ and therefore it is also a Lie group hence takes one of the following forms
$(1) H\cong S^1\times S^1$
$(2) H\cong S^1\times C_n$ for some cyclic group $C_n$
$(3) H$ is finite.
By elimination we can show that only $(2)$ is possible. Indeed, $(3)$ is impossible because $H$ is infinite. $(1)$ is impossible because $H$ has only $m$ elements of order $m$ as if $s^m=1$ we have $t^l=1$ if in addition $t^m=1$ then $t=1$ (because $\gcd(m,l)=1$). In other words $\{(s,1):s\in C_m\}$ are the only roots of order $m$.
Can you construct a concrete isomorphism $\varphi:S^1\times C_n\rightarrow H$? also what is $n$? (I'd guess that $n=1$ but I'm not so sure)
Indeed $n = 1$. Write $H_{m, l}$ for your subgroup. Consider $S^1$ as $[0, 1)$ with addition modulo 1, and consider $$ \begin{align*} \phi: [0, 1) \to H_{m, l}\\ t \mapsto (e^{2\pi i lt}, e^{2 \pi i mt}). \end{align*} $$ It is clear that this is a group homomorphism whose image is contained in $H_{m, l}$. From $\gcd(m, l) = 1$ we obtain that the kernel is trivial, so $\phi$ is injective. It remains to be shown that $\phi$ is surjective.
Consider $(z_0, z_1) \in H_{m, l}$. Note that the $z_0$ appears as a first coordinate of a point in $H_{m, l}$ exactly $l$ times: the full set of points for which this is true $$ \{(z_0, e^{\frac{j}{l} 2 \pi i}z_1) \mid j \in \{0, \ldots, l-1\}\}. $$ In particular, there are precisely $l$ points in $H_{m, l}$ whose first coordinate is $z_0$. But it is clear that $\phi$, which is injective, hits precisely a point whose first coordinate is $z_0$ exactly $l$ times. Thus $\phi$ is surjective.