Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $S^1\times S^1$?

share|improve this question
    
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 –  El Angel Exterminador Jun 22 '12 at 8:28
4  
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? –  El Angel Exterminador 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

1 Answer 1

up vote 4 down vote accepted

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.)

share|improve this answer
    
Will it be $S^1$ in your 2nd line ? –  El Angel Exterminador Jun 22 '12 at 8:59
    
Whoops, typo! ${}$ –  anon Jun 22 '12 at 9:04
    
Why $K$ must be a subgroup of $\pi(K)\times \rho(K)$? –  El Angel Exterminador 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.