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I tried many examples , but i can't find any counterexample .
But I guess there are many counter examples , and specific sorts of groups or subgroups have this property (e.g abelian groups or normal subgroups).
Thus I have two question:

  1. Is there any counter example of group $G$ and its subgroup $H$ s.t there is no surjective homomorphism from $G$ to $H$ ?
    1. If exists some counterexamples , which sort of groups or subgroups have this property?

I would also appreciate any reference .

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    $\begingroup$ For an example consider the quaternion group of order $8$ which has a cyclic subgroup of order $4$ but no such quotient. $\endgroup$ – Tobias Kildetoft Mar 10 '16 at 13:36
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    $\begingroup$ If $G$ is a finite simple group, then any non-trivial morphism from $G$ must be injective, so it has no surjective morphism to any group that has a strictly lower cardinal (in particular, any strict subgroup). $\endgroup$ – Captain Lama Mar 10 '16 at 13:40
  • $\begingroup$ Do you really just want any old surjective homomorphism $G \rightarrow H$, or do you want this homomorphism to be a retraction of the inclusion $H \hookrightarrow G$? $\endgroup$ – goblin Mar 11 '16 at 5:33
  • $\begingroup$ Old surjective! seems sufficient . $\endgroup$ – user217174 Mar 11 '16 at 5:37
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Just a convincing example to show that this property is very far from being natural : take the good old symmetric group $S_n$ (let's take $n\geqslant 5$ to be safe). Then any group of order $n$ is a subgroup of $S_n$, but there is no surjective morphism from $S_n$ to any of them. This is because the only surjective morphisms from $S_n$ to any group are the identity, the signature and the trivial morphism.

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    $\begingroup$ I'm not really sure, I'd have to think more about it. My first intuition is that $G$ must be nilpotent. Then its p-Sylows must also have this property. But I'm not really sure about what that implies for a $p$-group, since the quaternion example shows that it is not automatic. $\endgroup$ – Captain Lama Mar 10 '16 at 14:13
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    $\begingroup$ Well, just to state the obvious there, it is at least true for finite abelian groups, since it is clearly true for cyclic groups. $\endgroup$ – Captain Lama Mar 10 '16 at 14:21
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    $\begingroup$ The required property can be rephrased as: for every subgroup $H$ of $G$, there exists a quotient of $G$ isomorphic to $H$. Finite abelian groups are well-known to have this property. (Although I am not sure I agree it follows immediately from the fact that cyclic groups have this property, as it is not preserved under direct products: SmallGroup(16,4) has the property but not its direct square.) $\endgroup$ – verret Mar 11 '16 at 2:16
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    $\begingroup$ I also think that a group $G$ having this property must be nilpotent. It's true if $G$ is soluble: let $P$ be a Sylow $p$-subgroup of $G$. Since $G$ is soluble, $P$ has a complement in $G$, namely $H$, the Hall $p'$-subgroup of $G$. So $G$ must have a quotient isomorphic to $H$, but the only way this can happen is if $P$ is normal. This shows that all Sylow subgroups of $G$ are normal hence it is nilpotent. As you say, the question then reduces to $p$-groups. It looks like nonabelian groups with this property are rather rare: up to order 32, there are only three: (16,4), (27,3), (32,23). $\endgroup$ – verret Mar 11 '16 at 2:18
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    $\begingroup$ For $p$ odd, the groups of order $p^3$ and exponent $p$ provide an infinite family of non-abelian examples. $\endgroup$ – verret Mar 11 '16 at 2:59
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Take $G = F(\{x_1,\dots, x_n\})$, the free group on $n$ generators. The commutator subgroup $G' = [G,G]$ is a free group of infinite rank and thus $G$ cannot surject onto $G'$, as it simply doesn't have enough generators (it has finite rank).

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