# Find the number of homomorphisms between cyclic groups.

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)?

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

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.

• Is it 10 for (b) and (c) as well? – Vivek Dec 18 '12 at 13:40
• No i think its $10$ for (b) and $5$ for (C). – Kns Jul 2 '13 at 15:33
• @Kns, there are only 2 for (c), i.e. $f(1)=$ and $f(1)=$. 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. – Valentin Jun 8 '16 at 8:12