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I'd like to find all of the homomorphisms $ \varphi: \mathbb{Z}_{15} \to \mathbb{Z}_{6} $.

What I've tried so far:

I know that $ |\text{Im} (\varphi)| $ divides $ \text{gcd}(|\mathbb{Z}_{15}|,|\mathbb{Z}_{6}|) = 3 $. Then, $ |\text{Im} (\varphi)| = 1 $ or $ |\text{Im} (\varphi)| = 3 $.

If $ |\text{Im} (\varphi)| = 1 $ then $ |\text{Ker} (\varphi)| = 15 $, because $ | \mathbb{Z}_{15} | = |\text{Im} (\varphi)|\cdot |\text{Ker} (\varphi)| $. In particular, this is the trivial homomorphism: $ \varphi(a) = \bar{0}, \quad \forall a \in \mathbb{Z}_{15} $

I don't know how could I find the others homomorphisms. :/

francolino
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2 Answers2

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First, we need to find $\phi \left ( \overline{1} \right )$. Suppose that $\phi \left ( \overline{1} \right )=\overline{a}$, where $a \in \left \{0, 1, 2, 3, 4, 5 \right \}$. We have $$\overline{0}_6=\phi \left ( \overline{0}_{15}\right )=\phi \left ( \overline{15} \right )=\phi \left ( 15.\overline{1}\right )=15\phi \left (\overline{1}\right )=15\overline{a}$$ (Note that $15.\overline{1}$ means $\overline{1}+\overline{1}+...+\overline{1}$ (15 times) ) Thus we get $15a$ divides by $6$. So $a$ can be $0,2,4$. Therefore, we have three homomorphisms.

Nal
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  • Why does $6$ divide $15a$? Can you write down the propriety that you used? – francolino Jul 02 '15 at 00:51
  • A homomorphism must map identity to identity. Since $15a = 0$ (mod $15$), it follows that $15\phi(a)$ maps to a multiple of $6$ (only multiples of $6$ are $0$ (mod $6$)). – David Wheeler Jul 02 '15 at 01:13
  • I don't get it. Why do you mention $ 15\phi(a) $? He didn't use it. Is it related to $ 15a $? I don't see why "it follows" that '$ 15\phi(a) $ maps to a multiple of $ 6 $'? – francolino Jul 02 '15 at 02:19
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    We get $15\overline{a}=\overline{15a}=\overline{0}_6$, and hence, 6 divides 15a – Nal Jul 02 '15 at 02:26
  • I finally get it. Thank you a lot. :) – francolino Jul 02 '15 at 02:37
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More generally, it is easy to prove, for any $m,n$ that $$\operatorname{Hom}_\mathbf Z(\mathbf Z/m\mathbf Z,\mathbf Z/n\mathbf Z)\simeq \mathbf Z/\gcd(m,n)\mathbf Z.$$

Indeed,$\;\operatorname{Hom}_\mathbf Z(\mathbf Z/m\mathbf Z,\mathbf Z/n\mathbf Z)\simeq \operatorname{Ann}_{\mathbf Z/n\mathbf Z}(m\mathbf Z)=\bigl\{x+n\mathbf Z\mid mx\in n\mathbf Z\bigr\}.$

Write $m=\gcd(m,n)m_1$, $n=\gcd(m,n)n_1$, where $m_1$ and $n_1$ are coprime. ‘$mx$ is divisible par $n$’ means that ‘$m_1x$ is divisible by $n_1$’ and, since $m_1$ and $n_1$ are coprime, it means ‘$x$ is divible by $n_1$’.

Thus $$\operatorname{Hom}_\mathbf Z(\mathbf Z/m\mathbf Z,\mathbf Z/n\mathbf Z)\simeq \operatorname{Ann}_{\mathbf Z/n\mathbf Z}(m\mathbf Z)=n_1\mathbf Z/n\mathbf Z\simeq \mathbf Z/\gcd(m,n)\mathbf Z.$$

Bernard
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