Could any one give me a hint how to show that if the kernel of a group homomorphism from $S^1\times S^1$ to itself is finite then it must be cyclic subgroup of $S^1\times S^1$?
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Denote the kernel $K=\ker\varphi$ of a homomorphism of $S^1\times S^1$ to itself. Suppose it is finite.
There exist two projection maps $ S^1\times S^1\to S^1$ of the $1$st and $2$nd coordinates respectively, say $\pi$ and $\rho$. Since $K$ is finite, $\pi(K)$ and $\rho(K)$ are finite, and hence cyclic. Notice that $K$ must be a subgroup of $\pi(K)\times \rho(K)$ (since it is a subset), i.e. a subgroup of a direct product of two cyclic groups. Hence $K$ is either itself cyclic or a direct product of just two cyclic groups.
Both possibilities are realizable; using the circle group $\Bbb T$ ($z\in \Bbb C$ with $|z|=1$ under multiplication) as $S^1$, we have that the map $\varphi:(z,w)\mapsto (z^n,w^m)$ has kernel isomorphic to $C_n\times C_m$ (note $C_k$ denotes the cyclic group of order $k$); choosing one of $n,m$ to be $1$ will result in $K$ isomorphic to a single cyclic group.
(Aside: to see that finite subgroups of $S^1$ are cyclic, say one has order $n$ and, using $\Bbb T$ again, notice it must be a subgroup of the $n$th roots of unity; the subgroups of a cyclic group are cyclic.)