# group homomorphism from $S^1\times S^1$ to itself

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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This is false; take the homomorphism with constant value the identity. –  Qiaochu Yuan Jun 22 '12 at 8:26
Thank you I have edited my question –  La Belle Noiseuse Jun 22 '12 at 8:28
I think it may still be false: take the circle group $\Bbb T$ of $\Bbb C$, and consider $(z,w)\mapsto (z^2,w^2)$. The kernel is $\cong C_2\times C_2$, which is not cyclic. –  anon Jun 22 '12 at 8:28
what is $C_2$? could you please write? and how $\mathbb{C}$ comes into the picture? –  La Belle Noiseuse Jun 22 '12 at 8:30
I'm using the set of complex numbers with modulus $1$ under multiplication as the group $S^1$. The kernel of the map I gave is $\{\pm1\}\times\{\pm1\}$. $C_2$ stands for the cyclic group of order $2$. | However, it may be possible to say that the kernel (if finite) is either cyclic or a direct product of just two cyclic groups. –  anon Jun 22 '12 at 8:34

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.)

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Will it be $S^1$ in your 2nd line ? –  La Belle Noiseuse Jun 22 '12 at 8:59
Whoops, typo! ${}$ –  anon Jun 22 '12 at 9:04
Why $K$ must be a subgroup of $\pi(K)\times \rho(K)$? –  La Belle Noiseuse Jun 22 '12 at 9:06
$K$ is a group, $\pi(K)\times\rho(K)$ is a group, and the first is a subset of the latter. For if $(a,b)\in K$, then $a\in\pi(K)$ and $b\in \rho(K)$, and hence $(a,b)\in\pi(K)\times\rho(K)$. –  anon Jun 22 '12 at 9:07