# Criterion for isomorphism of two groups given by generators and relations

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?

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 $H := \langle u_x : x \in X \rangle \subseteq U$.

• Why not just say the generators of the first group have to satisfy "exactly" the relations of the second group. As it is you've only required them to allow all those relations and possibly more. Just specify there can be no others. – Gregory Grant Feb 11 '16 at 11:23
• @GregoryGrant What you mean by "exactly", could you make this more formal? As I see it when someone says a group satisfies exactly $R$ over $X$ (i.e. just those relations, or no other nontrivial ones) that is just some other wording for $\langle X \mid R \rangle$, but this is not a handy criterion, as it is precisely the statement for which I want a more handsome criterion. – StefanH Feb 11 '16 at 11:41
• Perhaps you need to require a bijection on the generators that induces a bijection on the relations. I don't think there's a short-cut, the criteria you stated above clearly is not sufficient. You could have each group isomorphic to a subgroup of the other, yet the two groups are not isomorphic to each other. – Gregory Grant Feb 11 '16 at 11:50
• For example a free group of rank two contains a free group of countably infinite rank. – Gregory Grant Feb 11 '16 at 11:54
• By bijection on relations you mean the normal subgroups generated by the relations in the free group to be isomorphic? Regarding your last sentence, the homomorphisms are onto, and not just isomorphic to subgroups, guess this is stronger. Do you have an example at hand? – StefanH Feb 11 '16 at 11:57

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 $Q_8$.
• Thanks! But that was not what I was looking for specifically. I would be more interested in a proof showing that two groups given by generators and (different) relations are isomorphic, and these groups are preferable infinite! The last specifically because then the criterion that they both map surjectively onto each other is not sufficient to conclude that they are isomorphic (which would suffice in the finite case). In your proof for example you also exploit finiteness by using $|G| = |Q_8|$ and epimorphism $\Rightarrow G \cong Q_8$. – StefanH Feb 13 '16 at 22:13
• Btw I guess you mean $N = \{a,4,a^2b^{-1},abab^{-1}\}$ and $a^i b^jN$ (not $b^i$). But thanks for the effort, so one upvote! – StefanH Feb 13 '16 at 22:15