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trying to think of any residually finite group which is not Hopf. Any help?

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1) $G = \bigoplus_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ is residually finite but not Hopfian. It is clearly not Hopfian, and any non-identity element $g$ must have $g_i \neq 0$ for at least one $i$. Then $\pi_i(G) \cong \mathbb{Z}/2\mathbb{Z}$ is a finite quotient in which the image of $g$ is nontrivial.

2) An infinitely generated free group is residually finite and not Hopfian. This is a basic example that any connoisseur of group theory should work out carefully for herself.

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Every finitely generated residually finite group is already Hopfian. So we need an infinitely generated group. An example is given here: http://groupprops.subwiki.org/wiki/Residually_finite_not_implies_Hopfian.

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  • $\begingroup$ The example given there doesn't seem to be fully correct: e.g. the additive group of a nontrivial $\mathbb{R}$-vector space is divisible, hence not residually finite. $\endgroup$ – Pete L. Clark Feb 10 '14 at 21:30
  • $\begingroup$ (It can be fixed by replacing "field" by "finite field" in the first sentence of the proof. This seems to be just a typo. Anyway, I added an answer just to be sure.) $\endgroup$ – Pete L. Clark Feb 10 '14 at 21:36
  • $\begingroup$ Yes, thank you ! $\endgroup$ – Dietrich Burde Feb 10 '14 at 21:37
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As Dietrich Burde has pointed out, you need to consider infinitely generated groups. This is because of the following result due to Mal'cev.

Theorem (Mal'cev): A finitely generated, residually finite group is Hopfian.

Proof: Suppose $G$ is residually finite and non-Hopfian, and we shall find a contradiction. As $G$ is non-Hopfian, there exists a map $\phi: G\rightarrow G$ such that $\ker\phi\neq 1$. Consider $g\in\ker\phi\setminus\{1\}$. Then as $G$ is residually finite there exists a subgroup $N_g\unlhd_fG$ such that $g\not\in N_g$, with associated map $\psi: G\rightarrow K=G/N_g$. As $G$ is finitely generated there are only finitely many, $n$ say, homomorphisms $G\rightarrow K=G/N_g$. We shall denote these homomorphisms by $\psi_1, \ldots, \psi_n$. Now, each of the maps $\phi\psi_i$ are distinct and so they constitute all the homomorphisms from $G$ to $K$. However, $\phi\psi_i(g)=_K1$ for all $i$ but $\psi(g)\neq 1$. This is a contradiction, as required.

Note that residual finiteness is a very strong property for finitely presented groups, but relaxing finite presentability to finite generation looses you some properties, and relaxing finite generation loses you even more. For example, a finitely presented, residually finite group has soluble word problem but there exists finitely generated, residually finite groups which are not even recursively presentable! Finite generation gives you Hopfian and also a residually finite automorphism group, but again relaxing this looses these conditions.

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