Suppose we have two groups $G,H$, and the premise is that every homomorphism from $G$ to $H$ is trivial.
This certainly can happen when either of $G,H$ is trivial, but is it possible that neither of $G,H$ are trivial?
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3What about $\Bbb Z/2\Bbb Z$ and $\Bbb Z/3\Bbb Z$? – J. W. Tanner Mar 14 '21 at 21:42
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1Yes. Take $C_p$ and $C_q$ where $p$ and $q$ are distinct primes. – Sofía Marlasca Aparicio Mar 14 '21 at 21:42
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1More generally, if $G$ and $H$ have finite coprime exponents, there will not be any nontrivial homomorphisms $G \to H$. – diracdeltafunk Mar 15 '21 at 01:14
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First, remember that there are $(m,n)$ number of homomorphisms between $\mathbb{Z}_n$ and $\mathbb{Z}_m$, see Find the number of homomorphisms between $\mathbb{Z}_m$ and $\mathbb{Z}_n$. Hence if $n$ amd $m$ are coprime, there is only one homomorphism. So it is not necessary for both groups to be trivial.
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There are other examples. If $G=\mathbb{Q}$ and $H=\mathbb{Z}$ are the additive group of rational numbers and the additive group of integers respectively, then every homomorphism from $G$ to $H$ is trivial.
kabenyuk
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By the First homomorphism theorem, this also happens if $G$ is simple and $|H|< |G|$. For example, $A_5\to S_4$.