Is there difference between finitely presented groups and finitely generated groups? 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?
 A: 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).
A: Yes. For instance, $(\mathbb{Z},+)$ is finitelty presented (it is generated by $1$), but it is not finite.
On the other hand, every finite group is finitely presented.
A: 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.
A: 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.
A: 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.
