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Let $G$ and $H$ be finite groups.

If $|G|$ and $|H|$ are coprime, the only homomorphism $G\rightarrow H$ is the trivial homomorphism.

But the condition is not necessary. If $G=A_4$ and $H=C_2 \times C_2\times C_2\times C_2$, then the only homomorphism $G\rightarrow H$ is the trivial one, but $|G|=12$ and $|H|=16$ are not coprime.

This can be shown by using the special property $ord(x)=2\ ,\ ord(y)=3 \implies ord(xy)=3$ , holding in the group $A_4$. How can I generalize this example ?

Is there an easy criterion, when there is only one homomorphism $G\rightarrow H$ for finite groups $G$ and $H$ ?

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    $\begingroup$ Unless you're ready to accept a tautological criterion such as "No nonzero quotient of $G$ is isomorphic to a subgroup of $H$", I don't think you're going to get anything useful. $\endgroup$ – Najib Idrissi Feb 29 '16 at 9:54
  • $\begingroup$ @NajibIdrissi At least that means that if $G$ is simple and cannot be injected into $H$, then this is true. But it is still far from a full characterisation (both the example above, and $G = C_{6}, H = C_{35}$ fail to fulfill that criterion) $\endgroup$ – Arthur Feb 29 '16 at 9:56
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    $\begingroup$ Observe that in the example you mention, $\;A_4\;$ has only one non-trivial normal subgroup $\;V\;$ ' which is of order four, and thus $\;A_4/V\cong C_3\;$ , which clearly cannot be a subgroup of your $\;H\;$ , so perhaps a more general partial condition is: for any non-trivial $\;N\lhd G\;$, $\;|G/N| \nmid |H|\;$ $\endgroup$ – DonAntonio Feb 29 '16 at 9:57
  • $\begingroup$ @Joanpemo I would change that to "For any proper $N \lhd G$", since it might be that $G$ can be injected into $H$. For instance, the canonical inclusion $A_5 \subseteq A_6$ would vacuously fulfill your condition. $\endgroup$ – Arthur Feb 29 '16 at 10:00
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    $\begingroup$ There are probably smaller examples, but there are two nonisomorphic simple groups of order $20160$, namely $A_8$ and ${\rm PSL}(3,4)$. So in that case, there is only one homomorphism $G \to H$, but your condition on orders fails with $N=1$. $\endgroup$ – Derek Holt Feb 29 '16 at 16:43

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