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I need to prove the following:

Let $G$ be a group. Then it's simple if and only if there is only surjective homomorphism $G \to G'$ for $G' = \{ e \}$ or $G' \cong G$.

Not sure how to approach this problem. So far I only have proved that if $G$ is simple and a homomorphism is surjective, then so is $G'$, but I'm not sure this is relevant.

The question is not a duplicate, since it also asks how to prove ($G$ is simple, $\phi: G \to G'$ is a surjective homomorphism) $\Rightarrow$ ($G' = \{ e \}$ or $G' \cong G$).

user1729
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Jxt921
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    Kernels of group homomorphisms are normal, and normal subgroups are kernels of the associated quotient homomorphism. – Max Jun 30 '16 at 00:39
  • Hint: The kernel of a homomorphism is a normal subgroup. – James Jun 30 '16 at 00:40
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    This is probably not true as stated (though I don't know a counterexample off the top of my head). You need to say not just $G'\cong G$ but that in that case the surjective homomorphism in question is an isomorphism. – Eric Wofsey Jun 30 '16 at 00:58
  • @user1729 I don't think it is a duplicate, and I edited to explain why. – Jxt921 Jun 30 '16 at 19:20
  • @Jxt921 I disagree: I believe this is a duplicate, because the linked question gives a counter-example to your statement. One direction holds, yes, but you were not asking about one direction but both. I suggest you ask the other direction in a separate question. – user1729 Jul 05 '16 at 08:46

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This is not true. It holds for finite groups, but not in general.

The issue is that a surjective homomorphism $G\rightarrow G$ need not be an isomorphism. If there exists a non-injective surjective homomorphism $G\rightarrow G$ then $G$ is called non-Hopfian. See here for more details and explicit examples (easiest example: $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}\times\cdots$).

The Prüfer group $\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ gives a couner-example to your specific question: it is not simple, but all homomorphic images are either isomorphic to itself or are trivial (see here).

If you restate your question to be about finite groups instead then it is a relatively straight-forward exercises on the first isomorphism theorem, as the comments and the other answer are getting at.

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user1729
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  • I see. However, the question in the link you gave says it hold for finitely generated groups and not just finite. – Jxt921 Jun 30 '16 at 19:12
  • And yes, if $\phi: G \to G'$ is a surjective homomorphism, then $G/ \ker \phi \cong G'$, but since $\ker \phi$ is normal and $G$ is simple, then $G' \cong G/ { e } \cong G$ or $G' \cong G/G \cong { e }$. – Jxt921 Jun 30 '16 at 19:15