# Finitely generated abelian group is Hopfian.

Let $G$ be a finitely generated abelian group and $H$ be its proper subgroup. Can we prove the result using short exact sequence, $$1\to H\to G \to G/H\to 1.$$ Any other method is also welcome.

• You should be able to use the classification theorem for finitely generated abelian groups. Any finitely generated abelian group is of the form $G = \mathbb{Z}^{r} + H$, where $H$ is a finite group that is a direct sum of cyclic groups. – Siddharth Venkatesh Apr 26 '14 at 5:50
• @schzan Why are these groups called Hopfian? What did Hopf do with them? I couldn't find any references. – user39598 Aug 8 '17 at 0:57

Let $G$ be a f.g. abelian group and $f:G\to G$ a surjection and $K_n$ be the kernel of $f^n$. We can consider the short exact sequence $$0\to \mathbb Q\otimes K_n\to\mathbb Q\otimes G\xrightarrow{1\otimes f^n}\mathbb Q\otimes G\to 0$$ of finite dimensional rational vector spaces. Since the map $\mathbb Q\otimes G\xrightarrow{1\otimes f^n}\mathbb Q\otimes G$ is a surjective endomorphism of a finite dimensional vector space it is injective, and $\mathbb Q\otimes K_n=0$. This implies that $K_n$ is a torsion group.

Now the sequence $(K_n)_{n\geq1}$ is increasing, and all its elements are contained in the torsion subgroup of $G$, which is finite. It follows that the sequence stabilizes: there is an $m$ such that $K_n=K_m$ for all $n\geq m$.

Suppose $g\in G$ is such that $f(g)=0$. There is an $h\in G$ such that $f^m(h)=g$, and then $f^{m+1}(h)=0$ so that $h\in K_{m+1}=K_m$ and we see that $g=f^m(h)=0$. The map $f$ is thus injective.

• Very nice argument, +1. – Seirios Apr 26 '14 at 8:55
• @Mariano, thakns for the answer. I also found that, Every exact sequence of finitely generated free abelian groups is split. Thus, we obtain $$G=H\bigoplus G/H,$$ which is possible only when $H={1}$. Is this correct?(where I assumed G to be non Hopfian, and used $G\cong G/H$) – schzan Apr 27 '14 at 2:02
• Well, G is not in general free, so that does not help you. – Mariano Suárez-Álvarez Apr 27 '14 at 2:11
• ohh..could you please tell me where the statement uses the fact that it is free(which aspect of free groups was required)? I am sorry if this comes as a surprise but I would really appreciate if you could explain what is the difference between finitely generated abelian group and finitely generated free abelian group.(suggested links are also welcome) – schzan Apr 28 '14 at 20:36
• @shzan, I did not use the fact that $G$ is free, because it isn't! – Mariano Suárez-Álvarez Apr 28 '14 at 22:45

Let $G$ be a finitely generated abelian group of rank $r$ and $\varphi : G \twoheadrightarrow G$ be an epimorphism.

• Let $g_1,\dots,g_r \in G$ be $r$ $\mathbb{Z}$-independent elements. Because $\varphi$ is surjective, there exists $h_i \in G$ such that $\varphi(h_i)=g_i$. Because $\varphi$ is $\mathbb{Z}$-linear, $h_1,\dots,h_r$ are $\mathbb{Z}$-independent.

• Let $g \in G$ such that $\varphi(g) \in \mathrm{Tor}(G)$. The family $\{h_1,\dots, h_r,g\}$ is $\mathbb{Z}$-dependent, so there exist $n,n_1,\dots,n_r$ such that $ng= n_1h_1+ \cdots +n_rh_r$; moreover, we may suppose $n \neq 0$. Applying $\varphi$ to the previous equality and multiplying by a well-chosen $k$, we obtain $$0=kn \varphi(g)= kn_1g_1 + \cdots + kn_rg_r,$$ hence $n_1= \cdots = n_r=0$. Therefore, $ng=0$ and $g \in \mathrm{Tor}(G)$.

• We deduce that $\mathrm{ker}(\varphi)= \mathrm{ker}(\tilde{\varphi})$ where $\tilde{\varphi}$ denotes the restriction of $\varphi$ to $\mathrm{Tor}(G)$. Moreover, $\tilde{\varphi}$ defines an epimorphism $\mathrm{Tor}(G) \twoheadrightarrow \mathrm{Tor}(G)$. But $\mathrm{Tor}(G)$ is a finite group, so $\tilde{\varphi}$ is an isomorphism. We conclude that $$\mathrm{ker}(\varphi)= \mathrm{ker}(\tilde{\varphi})= \{0\},$$ that is $\varphi$ is an isomorphism.

Maybe this answer is not needed but I find it to be a quick way to prove that fg abelian groups are hopfian.

Since $$\mathbb{Z}$$ is a Noetherian ring and abelian groups are $$\mathbb{Z} -$$modules it follows that fg abelian groups are Noetherian modules.

There is a well known proposition that says that an epimorphism $$f:M\to M$$ (where $$M$$ is a Noetherian module) is an isomorphism.