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

Solution 1:

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.

Solution 2:

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.