# Can I recover a group by its homomorphisms?

There is finitely generated group $G$ which I don't know. For every finite group $H$ I can think of, I know the number of homomorphisms $G \to H$ up to conjugation. (By this I mean that two homomorphisms $\phi_1$ and $\phi_2$ are being considered equivalent if there is a $h \in H$ such that $\phi_1(g)h = h\phi_2(g)$ for all $g \in G$.)

Given these numbers, do I have enough information to recover $G$?

Edit: The question is motivated from physics. A flat $H$-connection on a manifold $M$ is a homomorphism $\pi_1(M) \to H$ and a gauge transformation is a conjugation. So I'm interested whether I can recover the fundamental group by counting equivalence classes of connections, for arbitrary finite gauge groups.

Edit 2: It would be interesting as well if we count the number of homomorphisms without taking the equivalence by conjugation into account.

• As anomaly's answer shows, you should restrict attention to groups which can be detected based on their finite quotients. These are the residually finite groups (en.wikipedia.org/wiki/Residually_finite_group). It's known, for example, that the fundamental groups of closed manifolds of dimension at most $3$ are residually finite. – Qiaochu Yuan Feb 23 '15 at 20:19
• In addition to @QiaochuYuan's comment, recent work of Bridson and Wilton essentially covers this topic. Specifically, this paper proves that, given two finitely presented, residually finite groups, it is undecidable whether their profinite completions are isomorphic. The "profinite completion" of a group is an object which encodes all its finite quotients. See also this related paper, which proves this it is undecidable whether a finitely presented group has a finite quotient (equivalently, has trivial profinite completion) – user1729 Feb 23 '15 at 21:24
• Well, no, since you might need infinitely many $H$es, but it still suggests that you won't make much progress without further assumptions. – Qiaochu Yuan Feb 23 '15 at 21:48
• Alright, so I should be asking "How much information about $\pi_1$ can I recover from counting $H$-connections?" or so. Possibly the Morita equivalence class or so..? – Turion Feb 23 '15 at 21:53
• @Turion: any finitely presented group is the fundamental group of some smooth closed $4$-manifold, and it's known that there are finitely presented groups that are not residually finite. – Qiaochu Yuan Mar 12 '15 at 21:43

No. If $G$ is infinite and simple, then any homomorphism $f:G \to H$ for $H$ finite must be trivial; otherwise, $\ker f$ would be a proper normal subgroup of $G$. Such (finitely-generated) groups exist, though their construction is nontrivial; look at the Thompson groups, for example, or one of Higman's examples.

• Very beautiful answer, +1. Can one say anything if $G$ is finite? – Alex Wertheim Feb 23 '15 at 14:44
• Thanks. That's a much harder question, and I don't have any ideas offhand. My instinct would be to say that the answer is no, by analogy with the case in representation theory where two groups can have the same character table; but that's nowhere near an actual proof. – anomaly Feb 23 '15 at 14:46
• @AlexWertheim: For $G$ finite the answer should be yes. If you can determine for every finite group $H$ whether it is a quotient of $G$, then the biggest quotient will be $G$ itself. I think you can get the number of surjective homomorphisms $G \to H$ modulo conjugation by induction: the number of all homomorphisms is given and every non-surjective homomorphism has a proper subgroup $U$ of $H$ as image, so you know their number by induction (you'll have to correct this number considering $N_H(U)$ and the conjugates $U^h$). – j.p. Feb 25 '15 at 13:30
• @j.p.: What about maps that are conjugate in some $H > K$ but not in $K$ itself? – anomaly Feb 25 '15 at 15:20
• @Turion: I hope to have some time for it at the weekend. – j.p. Feb 27 '15 at 16:35

For $G$ finite the answer is yes.

It is clearly enough to know which finite groups $H$ are quotients of $G$, as $G$ will the biggest of them. For this we determine the number of surjective homomorphisms $G \to H$ by induction: For $H=1$ there is exactly one. So we may assume that we know the number of surjective homomorphisms $G \to U$ for all proper subgroups $U$ of $H$.

By assumption we are given the number $n_H$ of homomorphisms $\varphi:G \to H$ up to conjugation by elements $h$ of $H$ where $\varphi^h(g) = (\varphi(g))^h$. Let $U = \varphi(G)$ be the image of $\varphi$ in $H$. For $h\in C_H(U)$ centralizing $U\le H$ we have $\varphi^h = \varphi$, and vice versa the equality implies that $h$ centralizes $U$. So counting modulo conjugation is the same as weighing each homomorphism with one over the index of the centralizer of its image. (For $h\in N_H(U)$ normalizing $U\le H$ we know $\varphi^h(G)=\varphi(G)$. For general $h\in H$ we get that that the image of $G$ in $H$ under $\varphi^h$ is a subgroup of $H$ conjugated to $U$: $\varphi^h(G) = U^h$.)

Writing the number $n_H$ of homomorphisms (up to conjugacy) as $$n_H = \sum_{\varphi:G\to H} \frac{|C_H(\mathrm{Im}(\varphi))|}{|H|} = \sum_{U\le H}\sum_{{\varphi:G\to U} \mbox{ surjective}} \frac{|C_H(U)|}{|H|}$$ the left side $n_H$ is known by assumption. By induction all but one term (for $U=H$) of the outer sum on the right side can be calculated, which enables us to determine the missing term $x = \sum_{{\varphi:G\to H} \mbox{ surjective}} \frac{|C_H(H)|}{|H|}$. The number of surjective homomorphisms $G\to H$ is then $x$ times the index of the center of $H$.

The proof can easily be adapted for the case without conjugation mentioned in edit 2.

• Instead of $n_G$ you mean $n_H$ in the big formula? On what exactly do you do induction? Is there some sort of lattice of groups and their subgroups? – Turion Mar 2 '15 at 14:25
• @Turion: Yes, thanks, this should be $n_G$ (also in the line below the formula). Induction is on the order of the group $H$. – j.p. Mar 2 '15 at 14:31
• Shouldn't it be $n_H$? – Turion Mar 11 '15 at 17:53
• @Turion: Do you have a clue about the answer to your question for finitely generated residually finite groups? – j.p. Mar 11 '15 at 18:49
• Given that your proof feels a bit like magic, no I don't :I – Turion Mar 11 '15 at 20:52