When are two presentations of groups are isomorphic? In this post it is said:
[…] find a set of generators of the first group that satisfies the relations of the second group […]
But I doubt that, as this just shows that the first group is an epimorhpic image of the second group. For example consider
$$
\langle x \mid x^5 = 1 \rangle
$$
then all the generators of this group also satisfy $x^{10} = 1$, but we do not have an isomorphism between $\langle x \mid x^5 = 1 \rangle$ and $\langle x \mid x^{10} = 1 \rangle$. So a more precise way might be
Find a set of generators of the first group that satisfies the relations of the second group and vice versa.
More formally, if $G = \langle X \mid R \rangle$ and $H = \langle Y \mid Q \rangle$ for some generating sets $X, Y$ and relations $R, Q$. If we can find $Y \subseteq G$ such that $G = \langle Y \rangle$ and $Y$ satisfies the relations in $Q$, and vice versa we find in $H$ some $X \subseteq H$ such that $X$ satisfies the relations in $R$, can we conclude that $G \cong H$?
But all I see from that is the if it holds, we have an epimorphism $\varphi : G \twoheadrightarrow H$ and another epimorphism $\psi : H \twoheadrightarrow G$; see this post. If $H$ and $G$ would be finite, this would be sufficient to imply that $G$ and $H$ are isomorphic. But in general, I just see that $|H| = |G|$ as sets by the existence of surjective maps, but the bijective map does not have to be a homomorphism.
So is this the right criterion to show that two presentations are isomorphic, and if so how to prove it?
EDIT: Some thoughts by reading about the free groups.
i) The classic proof that two free groups over generating sets of the same cardinality are isomorphic goes by extending the bijection and its inverse between the two generating sets to an epimorphisms and note that the resulting epimorphisms are inverse to each other. But this does not seem to work for general groups given by generators and relations, as for example if $x$ is in the generating sets of one group and satisfies $x^2 = 1$, and it corresponds to a generating element of the second group which does not fulfills this relation, then the extended map is no homomorphism.
ii) The free group has the outstanding property that every epimorphism onto it splits. But this property is obviously not shared by arbitrary groups. Suppose we have this proprty, and that the epimorphisms are unique, then this is also not sufficient. If the above given epimorphism would be unique (which is it not in general, consider $\langle x,y \mid x^2, y^2 \rangle$, then for each group with generators $a,b$ of order two we could choose $x \mapsto a, y \mapsto b$ or the other way) and would split, i.e. we would have injective homomorphisms $\alpha : G \to H$ such that
$$
\alpha\psi = \mbox{id}_G
$$
and $\beta : H \to G$ such that $\beta\varphi = \mbox{id}_H$, but I do not see that this would imply $\beta = \psi$ and $\alpha = \varphi$.
iii) Also the projective property of the free group could not be generalised. Let $G = \langle X \mid R \rangle$ be a group given by generators and relations, and suppose we have some other groups $U, V$ with homomorphism $\alpha : G \to V$ and epimorphism $\beta : U \to V$. Then there exists some homomorphism $\varphi : G \to U$ such that $\varphi\beta = \alpha$.
Let $\beta(u_x) = \alpha(x)$ and set $\varphi(x) := u_x$. But it is not clear how to extend this to a homomorphism onto $H := \langle u_x : x \in X \rangle \subseteq U$.
Best Answer
In response to your request in the comments for an example of a proof of an isomorphism using generators and relations, I found the following proof in Hungerford's Algebra page 67. I'm not sure if this example will satisfy what you had in mind, if not let me know I can try to dig deeper.
Let $G$ be the group defined by generators $a,b$ and relations $a^4=e$, $a^2b^{-2}=e$ and $abab^{-1}=e$. Since $Q_8$, the quaternion group of order $8$, is generated by elements $a,b$ satisfying these relations (Exercise 4.14), there is an epimorphism $\phi:G\to Q_8$ by Theorem 9.5. Hence $|G|\geq|Q_8|=8$. Let $F$ be the free group on $\{a,b\}$ and $N$ the normal subgroup generated by $\{a^4,a^2b^{-2}\}$. It is not difficult to show that every element of $F/N$ is of the form $a^ib^iN$ with $0\leq i\leq 3$ and $j=0,1$, whence $|G|=|F/N|\leq8$. Therefore $|G|=8$ and $\phi$ is an isomorphism. Thus the group defined by the given generators and relations is isomorphic to $Q_8$.