Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [hopfian]

For questions about the Hopf property in groups and its failure.

A group $G$ is Hopfian (or Hopf) if every surjective endomorphism $G\rightarrow G$ has trivial kernel. Otherwise, the group is called non-Hopfian (or non-Hopf).

Examples of non-Hopfian groups include:

  • the direct sum of infinitely many infinite cyclic groups $G=\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots$, and
  • Baumslag-Solitar groups $BS(m, n)=\langle b, s\mid bs^mb^{-1}=s^n\rangle$ where $m$ and $n$ are coprime and each of absolute value greater than $1$ (the classical example is $BS(2, 3)$, due to Baumslag and Solitar in 1962).

See also:

37 questions
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Does $G\cong G/H$ imply that $H$ is trivial?

Let $G$ be any group such that $$G\cong G/H$$ where $H$ is a normal subgroup of $G$. If $G$ is finite, then $H$ is the trivial subgroup $\{e\}$. Does the result still hold when $G$ is infinite ? In what kind of group could I search for a…
Klaus
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When can a pair of groups be embedded in each other?

This is a question I made up, but couldn't solve even after some days' thought. Also if any terminology is unclear or nonstandard, please complain. Given groups $G$ and $H$, we say that $G$ can be embedded in $H$ if there exists an injective…
Srivatsan
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6 answers

If $G$ is a group and $N$ is a nontrivial normal subgroup, can $G/N \cong G$?

I know $G/N$ is isomorphic to a proper subgroup of $G$ in this case, so the gut instinct I had was 'no'. But there are examples of groups that are isomorphic to proper subgroups, such as the integers being isomorphic to the even integers, so that…
NotAwake
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Sort-of-simple non-Hopfian groups

A finite simple group is one which has no homomorphic images apart from itself and the trivial group. However, the simple-groups tag does not include the condition "finite". My question is the following. Is the following true? Claim: A simple group…
user1729
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Is retract of a finitely generated Hopfian group Hopfian?

A subgroup $H$ of a group $G$ is called retract of $G$ if there exists homomorphism $r:G\longrightarrow H$ so that $r\circ i=id_H$, where $i:H\hookrightarrow‎‎ G$ denotes the inclusion map. Also, recall that a group ‎$‎G‎$‎ is Hopfian if every…
user481657
9
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3 answers

The Hopfian property for groups

Let $G$ be a group, which for my purposes would be abelian. To say that $G$ has the Hopf property is to say that every epimorphism of $G$ is an automorphism. Does anyone happen to recall the context in which Hopf first used this concept, and a…
Chris Leary
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The free group $F_3$ being a quotient of $F_2$

Every finitely generated free group is a subgroup of $F_2$, the free group on two generators. This is an elementary fact, as is the fact that $G$, finitely presented, is the quotient of $F(|S|)$ the free group on some set of generators $S$ for $G$.…
Andrew Marshall
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(Non-Hopfian) groups that only have quotients that are themselves or the trivial group.

A group is non-Hopfian provided it is isomorphic to a proper quotient. The classic, finitely presented, example of such a group is the Baumslag-Solitar group $$BS(2,3)= \langle x,t \mid t^{-1}x^2 t =x^3 \rangle \cong \langle x,t \mid t^{-1}x^2 t…
user29123
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3 answers

Homomorphism/map in both direction implies isomorphism/homeomorphism or not?

I was working on a homework, and my first attempt get me to a deadend, but I was eventually able to solve it using a different method. But the fail attempt make me curious, and I wonder if it could have been fixed at all. So I got 2 topological…
anonymous
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Groups $G$ for which every finitely generated $\mathbb{Z}G$-module is Hopfian

Let $\mathbb{Z}G$ be the group ring over a group $G$. It is a well-known fact that every finitely generated module over a commutative ring is Hopfian. Hence if $G$ is abelian, then every finitely generated $\mathbb{Z}G$-module is Hopfian. On the…
M.Ramana
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non-residually finite group

Let $G$ be the subgroup of $\text{Bij}(\mathbb{Z})$ generated by $\sigma : n \mapsto n+1$ and $\tau$ which switches $0$ and $1$. How can we prove that $G$ is not residually finite? Is it hopfian?
Seirios
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Can you always find a surjective endomorphism of groups such that it is not injective?

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…
zrbecker
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Showing $\langle a,p,q\mid p^{-1}ap=a^2, q^{-1}aq=a^2\rangle$ is non-hopfian (from first principles).

According to a search on Approach0, this question is new to MSE. Motivation for Study: I'm doing some light reading of some notes by Miller on combinatorial group theory. Hopfian groups have just been defined. Whilst I've seen the term used…
Shaun
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Noetherian implies Hopfian

Can someone provide me with a reference for the following 2 statements, or indicate a proof? A finitely generated nilpotent group is Noetherian A Noetherian group is Hopfian.
user114539
3
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3 answers

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.
schzan
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