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We need to find all ring homomorphisms from $\mathbb Z_{20} \to \mathbb Z_{30} $ ; I read its solution somewhere which states that : $R : \mathbb Z_{20} \to \mathbb Z_{30}$ defined by $R(x) = ax$ , $a$ belongs to $\mathbb Z_{30}$ is a ring homomorphism if :

$1) \ a^2 = a$ and ,

$2) |a| \ \Big| \ 20 , \ \ \ |a| \ \Big| \ 30$

$1)$ is acceptable , but i couldn't understand $2)$.. why order of $a$ should divide both $20$ and $30$ ?

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2 Answers 2

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The order of $a$ obviously divides $30$. It must also divide $20$, because $$ 20a=R(20)=R(0)=0 $$

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I will state here one theorem and one property of group Homomorphism which will allow you to see why (2) hold true.

$\textbf{Lagrange's Theorem}$

If $G$ is a finite group and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$.

Proof: It's proof is based on cosets of $H$ in $G$ and can be found in any standard textbook.

$\textbf{Property of Group Homomorphism}$

If $|g|$ is finite, then $|\phi(g)|$ divides $|g|$.($\phi$ is some homomorphism)

Proof: $g^n=e_{G} \Rightarrow (\phi(g))^{n}=e_{\bar{G}} \Rightarrow |\phi(g)|$ divides $|g|$.

Now, back to the problem, since $\phi(1)=a$ and Ring form an abelian group under addition. So, form a cyclic subgroup of $<a>$ whose order must divide order of $Z_{30}$ and $|e_{z_{20}}|=20$ in $Z_{20}$.

Hence, $ |a| \ \Big| \ 20 , \ \ \ |a| \ \Big| \ 30$

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