If we take the following endomorphism, $\phi:R[t] \to R[t]$ by $\sum_{i = 0}^n a_it^i \mapsto \sum_{i = 0}^{\lfloor n/2 \rfloor} a_{2i} t^i$, it is surjective but not injective. (It just removes odd coefficients and pushes everything down).

Is there a similar endomorphism from $\mathbb{Z} \to \mathbb{Z}$? If so, can you give an example. Otherwise what are the conditions required for a surjective endomorphism to exist that is not injective.

  • $\begingroup$ What would be the preimage of a generator of $\mathbb{Z}$ under such a homomorphism? $\endgroup$ – dls Feb 17 '12 at 3:17
  • $\begingroup$ The image of a noninjective homomorphism from $\mathbb Z$ to a group is finite. The only surjective homomorphisms $\mathbb Z\to\mathbb Z$ are $x\mapsto\pm x$. $\endgroup$ – Jonas Meyer Feb 17 '12 at 3:22
  • $\begingroup$ Awesome, I see now we can't have it for $\mathbb{Z}$. Is there a way to determine in general which groups you can find one and which groups you can't? $\endgroup$ – zrbecker Feb 17 '12 at 3:59

If I tell you the terminology for the property you're studying, it will become easy for you to look up examples and nonexamples. A group $G$ such that every surjective homomorphism $f: G \rightarrow G$ is an isomorphism is Hopfian. So for instance wikipedia gives an introduction to Hopfian groups. As a very rough rule of thumb, among the groups which are easy to write down, the finitely generated ones tend to be Hopfian and the non-finitely generated ones tend not to be.

Some examples (ordered roughly in order of difficulty):

$\bullet$ A simple group is Hopfian. (Already this does not completely obey the rule of thumb!)
$\bullet$ Any finite group is Hopfian.

This implies: for every cardinal number $\kappa$ there is a Hopfian group of cardinality $\kappa$. The eminent group theorist Gilbert Baumslag claimed in a 1962 paper that for every cardinal number $\kappa$ there is a Hopfian abelian group of cardinality $\kappa$. However in 1963 he announced that his proof was incorrect, and apparently the problem remains open.

$\bullet$ A finitely generated abelian group -- or better stated, any Noetherian $\mathbb{Z}$-module -- is Hopfian.
$\bullet$ The additive group $(\mathbb{Q},+)$ of the rational numbers is Hopfian. (Against the rule of thumb!)
$\bullet$ A free group is Hopfian iff it is finitely generated.
$\bullet$ For any nontrivial group $G$, $\bigoplus_{i=1}^{\infty} G$ is not Hopfian.
$\bullet$ In particular, the additive group $(\mathbb{R},+)$ of the real numbers is not Hopfian.
$\bullet$ A finitely generated residually finite group is Hopfian.
$\bullet$ The Baumslag-Solitar group $B(2,3) = \langle x,y \ | \ y x^2 y^{-1} = x^3 \rangle$ is not Hopfian.

One can similarly define Hopfian and co-Hopfian objects in an arbitrary category. For instance, Hopfian $R$-modules appear in the (brief) $\S 3.8.2$ of these notes, in which it is noted that for any commutative ring $R$, every finitely generated $R$-module is Hopfian. (As alluded to above, this is almost trivial for Noetherian $R$-modules, hence for finitely generated modules over a Noetherian ring.)

  • $\begingroup$ Thanks, this is exactly what I was looking for. $\endgroup$ – zrbecker Feb 17 '12 at 6:36
  • $\begingroup$ Do you have a reference for the real numbers is not Hopfian? $\endgroup$ – Sigur Aug 1 '15 at 11:55
  • $\begingroup$ @Sigur: This is because $(\mathbb{R},+)$ is torsionfree and divisible and thus a $\mathbb{Q}$-vector space. So -- since every vector space has a basis, by the Axiom of Choice -- it is isomorphic to the direct sum of copies of $(\mathbb{Q},+)$ indexed by a set of continuum cardinality. This makes the Hopfian property clear. (The literal answer to your question is no: I don't have a reference.) $\endgroup$ – Pete L. Clark Aug 1 '15 at 15:53
  • $\begingroup$ @PeteL.Clark, thanks for attention. Now it is more clear. I'll study it. $\endgroup$ – Sigur Aug 1 '15 at 16:44

This is the same as requiring that a group be isomorphic to a quotient of itself by a proper normal subgroup. For $\mathbb Z$ this is impossible, and one way to see it is to note that all of the proper quotients of $\mathbb Z$ are finite. In the case of $R[t]$, as a group it is isomorphic to $R\oplus R\oplus R\oplus \cdots$, and a simpler surjective but not injective homomorphism is $(a_0,a_1,a_2,\ldots)\mapsto (a_1,a_2,a_3,\ldots)$. In terms of polynomials, this is the map $p(t)\mapsto \frac{p(t)-p(0)}{t}$.


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