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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 ?

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    $\begingroup$ Does the condition hold for fixed $H$ or for all $H$? $\endgroup$ Nov 7, 2011 at 16:32
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    $\begingroup$ @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

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

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    $\begingroup$ aka as Hilbert's hotel... $\endgroup$
    – lhf
    Nov 7, 2011 at 16:23
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    $\begingroup$ @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
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    $\begingroup$ @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
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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.)

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $$S^1/\ker\rho_n\cong S^1.$$

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

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

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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.

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

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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 $H$.

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    $\begingroup$ 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
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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. $\hspace{4cm}\blacksquare$

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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^* $

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  • $\begingroup$ 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

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