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A group is said to be finitely generated if it can be generated by a finite set of generators.

Question : Is there difference between finitely presented groups and finitely generated groups?

user1729
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    Baumslag showed that a wreath product $A \wr B$ is finitely presented iff $B$ is finite or $A$ is trivial; his article is available here: http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN266833020_0075&DMDID=DMDLOG_0011. However, I do not know whether there exist other (and simpler) proofs. – Seirios Oct 31 '13 at 17:13
  • I know his proof, but I was hoping there was some direct and simpler way too. – W4cc0 Oct 31 '13 at 17:16
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    I'd rather say that I don't know if it possible to write a more complicated proof than Baumslag's :) – YCor Nov 01 '13 at 17:11
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    It can be generated by a finite set of generators, but generation is something that need not stop. For example, even if you have one generator, then you can consider all words of all lengths in it and it's formal inverse, and that gives you a group which has infinitely many elements, but is only finitely generated. So a finitely generated group is not small in terms of size, but small in terms of what really matters : its set of generators. So there is a HUGE difference between finite groups and finitely generated groups. – Sarvesh Ravichandran Iyer Feb 21 '18 at 12:32
  • A finitely generated group has finitely many generators. A finitely presented group also has finitely many relationships between the generators. – Michael Burr Feb 21 '18 at 12:36
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    related – Michael Burr Feb 21 '18 at 12:38
  • related – Michael Burr Feb 21 '18 at 12:49

4 Answers4

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The following group is finitely generated but not finitely presentable: $$ G=\langle a, b, t\mid t[a^i, b^j]t^{-1}=[a^i, b^j], i, j\in\mathbb{Z}\rangle $$ It is clearly finitely generated. To see that it is not finitely presentable, note that it is an HNN-extension whose associated subgroup is free of infinite rank (the associated subgroup is in fact the derived subgroup $F(a, b)'$, which is not finitely generated). This means that the given presentation is aspherical*, and hence minimal. It is then "well known" that such a group $G$ cannot be finitely presented. One reason is as follows: suppose that $H$ is a finitely presentable group, and that $H$ has presentation $\langle \mathbb{x}; \mathbf{r}\rangle$ with $\mathbf{x}$ finite and $\mathbf{r}$ infinite. Then all but finitely many of the relators are redundant: there exists a subset $\mathbf{s}\subset \mathbf{r}$ such that $\mathbf{s}$ is finite and such that $\langle\langle\mathbf{s}\rangle\rangle=\langle\langle\mathbf{r}\rangle\rangle$. In our example, this cannot happen by asphericity/minimality. Hence, $G$ is not finitely presentable.

*Chiswell, I.M., D.J. Collins, and J.Huebschmann. Aspherical group presentations. Math. Z. 178.1 (1981): 1-36.

user1729
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  • Let $x_k=ab^ka$, for $k\in\mathbb Z$, then these $x_k$ are not free. For example $x_ix_j^{-1}x_kx_l^{-1}=1$ when $i-j+k-l=0$. So, at least these $x_k$ are not a basis of a subgroup. Now, how to prove that the associated group is of infinite rank and why is that an HNN extension? – Tommy W. Cai Nov 05 '21 at 08:24
  • You're right, it is not free with that basis (although it is free, as subgroups of free groups are always free). I believe it is finitely generated by ${a^2, aba}$ (so is 2-generated). I've edited in a correct subgroup, with justification. There is an underlying point though: the specific subgroup doesn't matter. So long as you know that $F(a, b)$ contains non-finitely generated subgroups, this construction works. – user1729 Nov 05 '21 at 10:00
  • (I don't understand your question "why is that an HNN extension?" The only thing which could fail is that the associated subgroups may not have been isomorphic. Is this what you were wondering? If so: they were isomorphic, as the automorphism $a\mapsto b$, $b\mapsto a$ of $F(a, b)$ induced an isomorphism between them.) – user1729 Nov 05 '21 at 10:03
  • Yeah, you argument that it's isomorphism so that it's HNN will probably work, although you should make sure that you are sending a basis to a basis, (the conjugation shouldn't send a set of free elements to a set which are not free, for example) which I believe your original construction should probably be fine on this point, but don't know how to prove. (Anyway, once you chose an infinitely generated subgroup with a known basis, this problem is solved.) – Tommy W. Cai Nov 05 '21 at 20:14
  • @Tommy The stable letter $t$ is being induced by an automorphism of the big group, so induces an isomorphism of subgroups. For example, if we pick an arbitrary set of words $w_1(a,b),\ldots, w_n(a,b)$, then the group $$\langle a,b,t\mid w_1(a,b)^t=w_1(b,a),\ldots, w_n(a,b)^t=w_n(b,a)\rangle$$ is an HNN-extension. (Regardless, in my corrected answer I've made the action of $t$ trivial, to avoid this subtlety.) – user1729 Nov 05 '21 at 21:00
  • Yes, you are right. – Tommy W. Cai Nov 06 '21 at 13:08
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I'm not sure how much group theory you're willing to assume. Does the following argument answer your question?

The given group $G$ has presentation $$ \langle a,b \mid [a^{-n}ba^n, b] = 1\text{ for }n\in\mathbb{N}\rangle. $$ Consider then the following sequence of groups $$ G_n \;=\; \langle a,b \mid [a^{-1}ba,b] = \cdots = [a^{-n}ba^n,b]=1\rangle $$ These groups fit into a natural sequence of quotient homomorphisms $$ G_1 \to G_2 \to G_3 \to \cdots $$ Then $G$ is finitely presented if and only if this sequence eventually stabilizes.

There are many different ways to show that this sequence does not stabilize. For example, each $G_n$ has a natural homomorphism to $\mathbb{Z}$ which sends $a$ to $1$ and $b$ to $0$. By Schreier's lemma, the kernel $K_n$ of this homomorphism is generated by the elements $b_i = a^{-i}ba^i$ and has presentation $$ K_n = \langle \ldots,b_{-1},b_0,b_1,b_2\ldots \mid [b_i,b_j]=1\text{ for }|i-j|\leq n\rangle. $$ The resulting group clearly depends on $n$. For example, if we fix a value of $m$ and add the relations $b_i = 1$ for $i \in \mathbb{Z}-\{0,m\}$, the resulting quotient of $K_n$ is abelian if $m\leq n$, and is a nonabelian free group of rank two if $m > n$.

Another alternative is to consider the quotient of $G_n$ obtained by adding the relations $a^m = b^2 = 1$. The resulting quotient is finite if and only if $m \leq 2n+1$, so all of the $G_n$'s must be different.

Jim Belk
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  • Thanks! However there are some things I don't know very well. How we say that $G$ has that presentation? And how do you apply the Schreier's Lemma? – W4cc0 Oct 31 '13 at 17:55
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    The presentation for $G$ was obtained as follows. There's an obvious infinite presentation for $\mathbb{Z}^\infty$. Since $G = \mathbb{Z}^\infty \rtimes \mathbb{Z}$, we get a resulting presentation for $G$, namely that all conjugates of $b$ by powers of $a$ commute. You can then reduce to the given presentation using Tietze transformations. – Jim Belk Nov 01 '13 at 00:28
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    By "Schreier's Lemma", I was referring to the usual algorithm (which doesn't really have a name) for finding a presentation for a subgroup when you know a presentation for the whole group as well as representatives for the cosets. You can actually avoid using it in this case by arguing that the semidirect product of $K_n$ with $\mathbb{Z}$ (where the generator for $\mathbb{Z}$ acts by mapping $b_i$ to $b_{i+1}$) is isomorphic to $G_n$. – Jim Belk Nov 01 '13 at 00:36
  • Perfect! Thank you! :) – W4cc0 Nov 01 '13 at 10:42
  • Are there natural numbers $n$ and $m$ so that $G_n$ and $G_m$ are isomorphic? – Mojtaba Aug 28 '17 at 06:21
  • @W4cc0 : Could you please help me – Mojtaba Aug 29 '17 at 10:06
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    @Mojtaba Presumably they are all non-isomorphic, though it isn't necessary to prove this to resolve the current question. (It's enough to prove that the natural quotient maps $G_n\to G_{n+1}$ are not isomorphisms.) If you want a proof that they are non-isomorphic you should post it as a new question. – Jim Belk Aug 29 '17 at 13:57
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Rips' construction gives lots of examples of finitely generated groups which are not finitely presentable. Rips* proved the following result.

Theorem. For every finitely presented group $Q$ there exists a hyperbolic group $H$ and a finitely generated, normal subgroup $N$ of $H$ such that $H/N\cong Q$.

Given a finite presentation of $Q$, Rips explicitly constructs the group $H$. This result is usually referred to as Rips' construction. It turns out that in Rips' construction the subgroup $N$ is finitely presentable if and only if the image group $Q$ is finite (see Exercise II.5.47, p227, of Bridson and Haefliger, Metic spaces of non-positive curvature - one direction is obvious, while the other direction is highly non-trivial).

* E. Rips, Subgroups of small Cancellation Groups, Bulletin of the London Mathematical Society, Volume 14, Issue 1, 1 January 1982, pp45–47, doi link.

user1729
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Here is a sketch of proof:

  • $\langle a,t \mid [t^nat^{-n},t^{m}at^{-m} ]=1, \ n,m \in \mathbb{Z} \rangle$ is a presentation of $L:= \mathbb{Z} \wr \mathbb{Z}$.

  • If $L$ is finitely presented, there exists a finite set $S \subset \mathbb{Z}$ such that $$\langle a,t \mid [t^nat^{-n},t^{m}at^{-m}]=1, \ n,m \in S \rangle$$ is isomorphic to $L$, so it is sufficient to show that $$L_k= \langle a,t \mid [t^nat^{-n},t^{m}at^{-m}]=1, -k \leq n,m \leq k \rangle$$ is not isomorphic to $L$.

  • Setting $a_n=t^nat^{-n}$, we get $$L_k= \langle a_{-k}, \dots, a_k,t \mid [a_n,a_{m}]=1, a_{n+1}=ta_nt^{-1}, -k \leq n,m \leq k \rangle.$$

  • Notice that $L_k$ is a HNN extension and use Britton lemma to show that the subgroup $\langle a_0,t \rangle$ is isomorphic to the free product $\mathbb{Z} \ast \mathbb{Z}$.

  • Therefore, $L_k$ is not solvable so it cannot be isomorphic to $L$ (which is solvable).

Seirios
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