Questions tagged [hopfian-groups]

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

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How do you prove that $BS(m,n)$ is hopfian when m and n have the same prime factors?

In SOME TWO-GENERATOR ONE-RELATOR NON-HOPFIAN GROUPS , the authors say that if $\pi(m) = \pi(n)$ (where $\pi(k)$ is the set of prime factors of $k \in \mathbb{N}$), then every surjective morphism $\...
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2answers
121 views

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 ...
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If $G$ is a group and $\mathrm{End}(G) \cong \mathbb{Z},$ why is $G$ Hopfian?

This is supposed to be well-known and easy to establish. If $G$ is Hopfian, then any surjective endomorphism $f:G \rightarrow G$ must be an automorphism, hence a unit in the endomorphism ring, hence ...
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1answer
100 views

Is this surjective homomorphism necessarily injective? (Free Product of Cyclic groups)

Let $G=\langle g_1,\dots,g_m\mid g_1^{k_1}=\dots=g_m^{k_m}=1\rangle$. (Where $k_1,\dots,k_m$ are integers.) Let $\psi:G\to G$ be a surjective homomorphism. Is $\psi$ necessarily injective? In other ...
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1answer
123 views

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 ...
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A non co-hopfian simple group?

Is there a simple group (all normal subgroups are the group itself or the trivial group) that is non co-hopfian (there exists an injective group endomorphism that is not surjective) ?
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1answer
84 views

Is a finitely generated module over a Hopfian ring Hopfian?

An $R$-module $M$ is called Hopfian if every $R$-epimorphism ‎$‎f :M‎\longrightarrow ‎M$‎ is an automorphism. Suppose that $R$ is a Hopfian ring and $M$ is a finitely generated $R$-module. Is $M$...
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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, ...
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1answer
155 views

Is a free group of finite rank Hopfian?

Recall that a group ‎$‎G‎$‎ is Hopfian if every epimorphism ‎$‎f :G‎\longrightarrow ‎G‎‎$‎ is an automorphism (equivalently, ‎$‎N=1‎$‎ is the only normal subgroup for which ‎$‎G/N\cong G‎$‎). Is a ...
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To prove that a hopfian group is not free.

In the book "presentation of groups by johnson" in page $36$ they are trying to obtain a group that is locally free but not free (Example $2$).They have proved that the group $U=\bigcup_{n \in \mathbb{...
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Groups with a dense chain of (isomorphic) subgroups

Are there any interesting non-Abelian examples of groups which are equal to the union of a dense chain of normal subgroups, with each subgroup isomorphic to the original group. That is, a group $G$ ...
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1answer
226 views

(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^...
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2answers
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Does $\mathbb{Z}* \mathbb{Z} /N = \mathbb{Z} * \mathbb{Z} \implies N$ is trivial?

As the title suggests, my question is does $\mathbb{Z}* \mathbb{Z} /N = \mathbb{Z} * \mathbb{Z}$ imply that $N$ is trivial? Here, $\mathbb{Z} * \mathbb{Z}$ is the free product of $\mathbb{Z}$ and $\...
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2answers
260 views

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.
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2answers
300 views

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.
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2answers
346 views

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 ...
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3answers
429 views

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?
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6answers
657 views

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

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 ...
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3answers
677 views

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 ...
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2answers
234 views

Must a surjection $F_2\to F_2$ also be an injection?

In other words, are there words $w_1$ and $w_2$ in the free group $F_2$ such that $F_2=\langle w_1, w_2\rangle$ but which are not a free basis for $F_2$? I'm sure I'm missing a simple argument or a ...
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2answers
664 views

Let $G$ be a finite group and $H\triangleleft G$ a normal subgroup. Prove that $|G/H| =|G|$ if, and only if, $H = \{e\}$.

The group $G$ is a finite group, a group with finite number of elements, and $H\triangleleft G $a normal subgroup. How can we prove that the index $|G/H|=|G|$ iff $H=\{e\}$, the identity element?
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1answer
667 views

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?
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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 ...
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10answers
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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 ...
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3answers
352 views

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 ...
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1answer
772 views

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 ...
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2answers
261 views

$\langle X|\emptyset\rangle\ncong\langle X|R\rangle$ for finitely presented groups (exercise in Massey)

Let $:F_n$ denote the free group of rank $n$. How can I solve the exercise 7.6.3.(b), page 234, in Massey's Algebraic Topology? I'm guessing there's been made a mistake and (b) actually reads "If $G$ ...
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338 views

Epimorphism from GL(2,Z) to GL(2,Z)

Is there an epimorphism $f\colon \mathrm{GL}(2,\mathbb{Z})\to \mathrm{GL}(2,\mathbb{Z})$ which is not injective? Here, $\mathrm{GL}(2,\mathbb{Z})$ is the group of invertible $2\times 2$ matrices with ...
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2answers
786 views

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 ...