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Show that if $G_1$ and $G_2$ are two finite groups and the number of distinct group homomorphisms from $G_1$ to $H$ is equal to that of $G_2$ to $H$ for every finite group $H$, then that $G_1$ and $G_2$ are isomorphic .

It's a problem in IMS 2014 .

Actually, I have tried to show $|G_1|= |G_2|$ , primarily, and if there exists an onto homomorphism from $G_1$ onto $G_2$, then our assertion will be done. But, I can't approach properly! Here, I have found an answer on Aops

But, I want a different approach .

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  • $\begingroup$ What machinery do you have at hand? If you've read Artin's book or Dummit's, how many chapters have you read? $\endgroup$
    – Mikhail D
    Jul 27, 2019 at 16:44
  • $\begingroup$ @Mikhail D, Sir, I have read upto group homomorphisms, group actions, and some portions of finite abelian groups . $\endgroup$ Jul 27, 2019 at 16:56
  • $\begingroup$ After some contemplatiin, I arrived too at what boils down to the proof you reference - now I can't unthink it :( -- Why do you desire a different approach? Is there something you on't like with this one? $\endgroup$ May 17, 2020 at 14:59
  • $\begingroup$ @Hagen von Eitzen,Sir, actually, I thought the proof may use isomorphism theorems, or some theorems from Sylow's, but it mainly uses number theory, so I thought if it can be done on another way. $\endgroup$ May 18, 2020 at 4:12

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