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

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.

The following group is finitely generated but not finitely presentable: $$ G=\langle a, b, t; tab^iat^{-1}=ba^ib, i\in\mathbb{Z}\rangle $$ 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. 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.

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).