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In each of the following examples determine the number of homomorphisms between the given groups:

$(a)$ from $\mathbb{Z}$ to $\mathbb{Z}_{10}$;

$(b)$ from $\mathbb{Z}_{10}$ to $\mathbb{Z}_{10}$;

$(c)$ from $\mathbb{Z}_{8}$ to $\mathbb{Z}_{10}$.

Could anyone just give me hints for the problem? Well, let $f:\mathbb{Z}\rightarrow \mathbb{Z}_{10}$ be homo, then $f(1)=[n]$ for any $[n]\in \mathbb{Z}_{10}$ will give a homomorphism hence there are $10$ for (a)?

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    $\begingroup$ Yes, that's exactly right. f(1) is all that matters, because 1 generates the whole group. $\endgroup$ – Billy Dec 17 '12 at 20:49
  • $\begingroup$ This is a related post. $\endgroup$ – Bijesh K.S Jul 10 '17 at 9:44
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Hint:

A homomorphism on a cyclic group is completely determined by its value on a generator of the group.


Edit:

Your thoughts on $(a)$ are indeed correct.

Use similar reasoning, along with the given hint to arrive at answers for $(b)$ and $(c)$. See what you can do, and I'll be happy to follow up in comments.


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  • $\begingroup$ Is it 10 for (b) and (c) as well? $\endgroup$ – Vivek Dec 18 '12 at 13:40
  • $\begingroup$ No i think its $10$ for (b) and $5$ for (C). $\endgroup$ – Kns Jul 2 '13 at 15:33
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    $\begingroup$ @Kns, there are only 2 for (c), i.e. $f(1)=[5]$ and $f(1)=[0]$. The only possibilities to define $f:\mathbb{Z}_n \to \mathbb{Z}_m$ on a generator is to map $1$ to a multiple of $\frac{m}{gcd(m,n)}$. (Try to figure out why). There are thus $gcd(m,n)$ possibilities. $\endgroup$ – Valentin Jun 8 '16 at 8:12

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