# An example of a residually finite group which is not Hopf

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

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.

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.

• 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. – Pete L. Clark Feb 10 '14 at 21:30
• (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.) – Pete L. Clark Feb 10 '14 at 21:36
• Yes, thank you ! – Dietrich Burde Feb 10 '14 at 21:37

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.