2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Siddharth Venkatesh Apr 26 '14 at 5:50
  • $\begingroup$ @schzan Why are these groups called Hopfian? What did Hopf do with them? I couldn't find any references. $\endgroup$ – user39598 Aug 8 '17 at 0:57
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Very nice argument, +1. $\endgroup$ – Seirios Apr 26 '14 at 8:55
  • $\begingroup$ @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 $G\cong G/H$) $\endgroup$ – schzan Apr 27 '14 at 2:02
  • $\begingroup$ Well, G is not in general free, so that does not help you. $\endgroup$ – Mariano Suárez-Álvarez Apr 27 '14 at 2:11
  • $\begingroup$ 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) $\endgroup$ – schzan Apr 28 '14 at 20:36
  • $\begingroup$ @shzan, I did not use the fact that $G$ is free, because it isn't! $\endgroup$ – Mariano Suárez-Álvarez Apr 28 '14 at 22:45
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.