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Describe all group homomorphisms from $\mathbb Z$ to $D_8$.

So $\mathbb Z=\langle1\rangle$, and $f(1)$ determines the homomorphism. So every homomorphism is determined by $f(1)=\sigma$, and for some $m \in Z$, $f(m)=\sigma^m$. How can I find the number of homomorphisms from here though? What should I look at?

Thanks in advance!

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    $\begingroup$ $f(1)$ determines the homomorphism. But conversely, given any $\sigma$, there's a unique $f$ such that $f(1) = \sigma$... $\endgroup$ – Najib Idrissi Jun 25 '14 at 7:55
  • $\begingroup$ Can you please explain why is that correct? $\endgroup$ – Hestag Jun 25 '14 at 8:14
  • $\begingroup$ Let $\sigma \in D_8$, then define $f(m) = \sigma^m$. Then you can check immediately that this is a homomorphism. $\endgroup$ – Najib Idrissi Jun 25 '14 at 8:17
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You know $f$ is determined by the value on $f(1)\in D_8$.

So, are there any restrictions on the value of $f(1)$?

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  • $\begingroup$ Doesn't seem so. Is that a good enough explanation to why every $f(1)=\sigma$ determines a unique homomorphism? If so, there are 8 homomorphisms... $\endgroup$ – Hestag Jun 25 '14 at 8:01
  • $\begingroup$ @Hestag That's correct. Since $f(1)$ determines $f(n)$ for all $n\in\Bbb Z$, the value of $f(1)$ determines a unique homomorphism. In fact, $|\hom(\Bbb Z,G)|=|G|$ for any group $G$ by the same argument. $\endgroup$ – blue Jun 25 '14 at 8:18

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