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 counterexample ?

  • 1
    $\begingroup$ Does the condition hold for fixed $H$ or for all $H$? $\endgroup$ Nov 7, 2011 at 16:32
  • 16
    $\begingroup$ @QiaochuYuan: One would presume the question is asking about a fixed $H$ - if it held for all $H$ then $G$ would necessarily be trivial (as taking $H=G$...). Although, interestingly, if one takes $C_2$ in the Hilbert Hotel example (so, $G\cong \bigoplus\limits_{i=1}^\infty C_2$) then $G\cong G/H$ for all finitely generated $H$. $\endgroup$
    – user1729
    Nov 7, 2011 at 16:46

10 Answers 10


Look at $G=\bigoplus\limits_{i=1}^\infty\ \mathbb Z$ and the subgroup $H=\mathbb Z\ \oplus\ \bigoplus\limits_{i=2}^\infty\ 0$.

  • 15
    $\begingroup$ aka as Hilbert's hotel... $\endgroup$
    – lhf
    Nov 7, 2011 at 16:23
  • 11
    $\begingroup$ @lhf: Surely Hilbert's Hotel is $G=\bigoplus\limits_{i=1}^\infty\ C_2$, as rooms are either occupied or not occupied...? (Alternatively - $C_3$ instead of $C_2$, adding the other option of "needing cleaned".) $\endgroup$
    – user1729
    Nov 7, 2011 at 16:38
  • $\begingroup$ @user1729, yes you're right. $\endgroup$
    – lhf
    Nov 7, 2011 at 16:41
  • 12
    $\begingroup$ @user1729 - To some of us, cleanliness falls along a continuum. :P $\endgroup$
    – user5137
    Nov 7, 2011 at 16:52
  • $\begingroup$ Also, Substituting $\mathbb{Z}$ for any other group still works! $\endgroup$ Mar 8, 2017 at 1:37

If $G\cong G/H$ implies $H$ is trivial then $G$ is called Hopfian. Otherwise, $G$ is called (imaginatively!) non-Hopfian. Non-Hopfian-ness is a truly nasty property!

A related property is that of Residual finiteness. A group is residually finite if for any two elements $g, h\in G$ there exists some homomorphism, $\phi$, to a finite group $H$, $\phi:G\rightarrow H$ such that $\phi(g)\mathrel{\neq_H} \phi(h)$. Equivalently, noting that $\phi(gh^{-1})=1$, $G$ is Residually finite if for any $g\in G$ there exists some $\phi:G\rightarrow H$, $H$ finite, such that $\phi(g)\mathrel{\neq_H} 1$. Note that a finite group is clearly residually finite, and the proof that a finite group is Hopfian is low-level.

It is interesting to note that if $G$ is finitely generated and non-Hopfian then $G$ is not Residually finite. A proof of this can be found in many/most graduate-level texts which cover infinite groups (for example, D. J. S. Robinson's book A Course in the Theory of Groups), or see this Math.SE answer. As has been pointed out in the comments below, this only holds if $G$ is finitely generated. For example, the free group on countably many generators is residually finite but is non-Hopfian.

As Joseph Cooper pointed out in his answer, certain Baumslag-Solitar groups are non-Hopfian, such as $$BS(2, 3)\cong \langle a, b;b^{-1}a^2b=a^3\rangle$$ (to see this, take the map $a\mapsto a^2, b\mapsto b$ and play around with it for a bit - it has non-trivial kernel, but the proof is non-trivial...you can find a proof in Magnus, Karrass and Solitar's book Combinatorial Group Theory, in their section on one-relator groups, or in this Math.SE answer). Indeed, there exists a classification of the Baumslag-Solitar groups with respect to their Hopficity and Residual-finiteness. Specifically,

The group $BS(m, n)=\langle a, b; b^{-1}a^mba^{n}\rangle$ for $m, n\in\mathbb{Z}$ is,

  • Residually finite if and only if $|m|=|n|$ or $|m|=1$ or $|n|=1$,

  • Hopfian if and only if it is Residually finite or the set of prime divisors of $m$ is equal to the set of prime divisors of $n$.

So, for example, taking $m$ and $n$ to be coprime and each of absolute value greater than $1$ will yield a non-Hopfian group. There is a paper of Meskin which generalises this to groups with with relation where the relation is of the form $uv^mu^{-1}v^n$ where $u, v$ are words in the generators. Indeed, his statement about residual-finiteness is identical to the case of the Baumslag-Solitar groups ($G$ is residually finite if and only if $m$ and $n$ are equal to each other or one in absolute value). Edit: I believe that Meskin's result on residual finiteness was the original result. That is, he proved the general case where the relation has the form $uv^mu^{-1}v^n$ and as a corollary got the first proof of the Baumslag-Solitar case.

Also, the fact that some Baumslag-Solitar groups are not Hopfian is surprising, as the Baumslag-Solitar groups are the $HNN$-extensions of $\mathbb{Z}$, and one would think that an extension of $\mathbb{Z}$ would be very nice indeed! However, that is clearly not the case...

(Note that Chris Leary asked only a few days ago what the history of Hopfian groups was - see here.)

  • $\begingroup$ It's not the case that residually finite groups are Hopfian. The free group on countably many generators is a counterexample. It is true that finitely generated residually finite groups are Hopfian. $\endgroup$ Nov 7, 2011 at 16:23
  • 2
    $\begingroup$ Sorry, I mentally put "finitely generated" before everything I read or write, and (unwittingly and unreasonably) expect everyone else to do the same! Duly Edited. $\endgroup$
    – user1729
    Nov 7, 2011 at 16:34
  • $\begingroup$ +1 Very interesting. Hope you don't mind my latex edit. $\endgroup$
    – Rasmus
    Nov 7, 2011 at 17:44

Probably the easiest example I can think of is $S^1$ under the doubling map (or more generally, multiplication by any positive integer).

More concretely, let $S^1=\{e^{i2\pi\theta}\in\mathbb{C}\mid \theta\in[0,1)\}$ and with group operation being multiplication inherited from $\mathbb{C}$ which is the same as addition of angles modulo $2\pi$. Then let $\rho_n\colon S^1\rightarrow S^1$ for $n\geq 2$ be given by $\rho_n(z)=z^n$. It's pretty easy to see that $\rho_n$ is a surjective group homomorphism with kernel isomorphic to $\mathbb{Z}/n\mathbb{Z}$ and so we have $$S^1/\ker\rho_n\cong S^1.$$

  • 1
    $\begingroup$ This is nice because it’s so geometric. $\endgroup$
    – Lubin
    Sep 11, 2013 at 21:09
  • $\begingroup$ This is beautiful $\endgroup$ Aug 30, 2015 at 20:08
  • $\begingroup$ This is a cute example. $\endgroup$
    – Boka Peer
    Feb 24, 2020 at 7:23

For a finitely generated non-abelian example, see the Baumslag-Solitar groups, in particular $\mathrm{BS}(2,3)\cong \langle a,b; b^{-1}a^2b=b^3\rangle$.


Sticking with an abelian theme, you could let $G=\mathbb{Z}(p^{\infty})$ and $H=\mathbb{Z}(p^{k})$ for a positive integer $k$.


The question is equivalent, through the first isomorphism theorem, to asking: does a surjective group morphism $G\to G$ necessarily have a trivial kernel? The answer (for infinite groups) is "no".

One can take $\def\Z{\Bbb Z}G=\Bbb Q/\Z$, and for some integer $n>1$ take the morphism of multiplication by$~n$; its kernel is the nontrivial subgroup $H=(\frac1n\Z)/\Z\cong\Z/n \Z$.

This example can also be realised taking for $G$ the group of all roots of unity and the morphism of taking $n$-th powers; its kernel is the subgroup $H$ of $n$-th roots of unity. The map $\Bbb Q\to\Bbb C$ with $t\mapsto\exp(2\pi\mathbf it)$ induces an isomorphism from the first realisation to this one.

To simplify, one could take for $n$ a prime number$~p$ and consider only the $p^i$-th roots of unity for some $i\in\Bbb N$. This gives the Prüfer $p$-group.

  • 3
    $\begingroup$ +1 Ah this is a nice countable subgroup of the example in my answer. $\endgroup$
    – Dan Rust
    Sep 11, 2013 at 21:43

A different kind of example giving a kind of realization of the Frobenius kernels. Let $p$ be a prime, $F=F_p=GF(p)$, let $x$ be an unknown, and let $R$ be the set of finite $F$-linear combinations of $x^q$, where $q$ is a non-negative rational number with the property that $p^\ell\cdot q$ is an integer for some natural number $\ell$. Then $R$ is a ring w.r.t. the natural operations. Let $A=R/I$ where $I$ is the ideal of $R$ generated by $x^p$. Then $\phi:A\rightarrow A, a\mapsto a^p,$ is a surjective endomorphism of $F$-algebras with kernel $J=$ the ideal generated by the coset $x+I$. Let $n\ge1$ be an integer. By functoriality of the affine group scheme $GL_n$ the mapping $\phi$ gives (acting entrywise) a surjective homomorphism of groups $\phi:GL_n(A)\rightarrow GL_n(A)$. So, with $G=GL_n(A)$ and $H=\ker\phi$, we get the desired isomorphism $G/H\simeq G$. The subgroup $H$ consists of those matrices that are congruent to the identity matrix modulo $J$.


I find it surprising nobody pointed out that the OP's conjecture is true, given a common conventional reading of the question.

When people ask a question about $G \to G/H$, very frequently they don't mean to say just any old map, but they mean to refer specifically to the projection map $x \to x H$ sending an element of $G$ to its equivalence class modulo $H$.

Other examples would be $G \to G \times H$ would be interpreted as the map $x \to (x,1)$, or $G \times H \to H$ would be the map $(x,y) \to y$.

Similarly, when people write an isomorphism $A \cong B$ in a situation where $A \to B$ would be given said conventional meaning, they don't mean $A \cong B$ to say merely that $A$ and $B$ are isomorphic, but that the specific map $A \to B$ chosen by convention is an isomorphism.

Given these conventional readings, it is true that if $G \cong G/H$, then $H$ is trivial. Making everything explicit, we have a theorem

If the projection map $x \mapsto xH$ from $G$ to $G/H$ is an isomorphism, then $H$ is indeed the trivial subgroup of $H$.

  • 6
    $\begingroup$ There is a difference between the kind of usual sloppiness of formulation you describe and this question: it is indeed not uncommon to understate a conclusion or a description of a known situation by saying just "isomorphic" while the isomorphism could be specified. But here you are reading more into a hypothesis than is actually there, and this is just not allowed. This question is perfectly analogous to "if $G$ is isomorphic to a subgroup $H$, does it follow that $H=G$?", where it would be obviously wrong to assume that isomorphism by inclusion is meant. $\endgroup$ Feb 14, 2014 at 6:19

Proving Baumslag-Solitar groups is a counter-example is not at all tough, so I will write this proof for $BS(m,n)$

Choose any two integers $m$ and $n$ such that $(m,n)=1$ and $m,n \neq -1,0,1$.

Let $G=BS(m,n)=\langle a,b\ |\ b^{-1}a^mb=a^n \rangle$.
Consider $\phi: G \to G$ where $\phi(a)=a^m$ and $\phi(b)=b$, it is easily checked that $\phi$ is well defined.

Claim-1 $\phi$ is surjective.

We see that $a^m,b \in \text{Im}(G)$, so $a^n\in \text{Im}(G)$ also. Now as $(m,n)=1$, $k_1m+k_2n=1$ for some $k_1,k_2\in \Bbb{Z}$, and as $\text{Im}(G)$ is a subgroup, $a=a^1\in \text{Im}(G)$, thus $\phi$ is an epimorphism.

Claim-2 Ker$(\phi)$ is non-trivial.

$[b^{-1}ab,a]=b^{-1}abab^{-1}a^{-1}ba^{-1}\neq 1$ by Britton's lemma, but $\phi([b^{-1}ab,a])=\phi([b^{-1}a^mb,a^m])=[b^n,b^m]=1$, so kernel is non-trivial. $\hspace{4cm}\blacksquare$


consider the group $\mathbb C^*$ of non-zero complex numbers under multiplication and define $f: \mathbb C^* \to \mathbb C^*$ as $f(z)=z^2 , \forall z \in \mathbb C^*$ ; note that $f$ is a homomorphism , it is surjective as if $0 \ne z \in \mathbb C$ , then $0 \ne \sqrt z \in \mathbb C$ and $f(\sqrt z)=z$ ; moreover $\ker f=\{z \in \mathbb C^* : z^2=1\}=\{-1,1\}$ is non-trivial and so by 1st Isomorphism theorem , $\mathbb C^*/\ker f \cong \mathbb C^* $

  • $\begingroup$ You could note that this a minor modification of this already existing answer, just replacing the circle $S^1$ by the $\mathbb C^{\times} = S^1 \times \mathbb R_{> 0} \cong S^1 \times \mathbb R.$ $\endgroup$
    – tracing
    Feb 8, 2015 at 14:37

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.