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Let $G$ be a group and suppose that $G$ is the set-theoretic union of a collection $\mathcal U$ of subgroups. I will call such a collection a "covering" of $G$.

Suppose that for each $L \in \mathcal U$, we are given a homomorphism $L \xrightarrow{\rho_L} K$ into a common group $K$, in such a way that $\rho_L|_{L \cap L'} = \rho_{L'}|_{L \cap L'}$ for any two $L, L' \in \mathcal U$. Then the collection $\{\rho_L\}_{L \in \mathcal U}$ glues set-theoretically to give a map $G \xrightarrow{\rho} K$, which may or may not be a group homomorphism.

We can measure the obstruction to $\rho$ being a homomorphism as follows: one has the complex of pointed sets

$$* \to \mathrm{Hom}(G, K) \to \prod_{H \in \mathcal U} \mathrm{Hom}(H, K) \rightrightarrows \prod_{(H,H') \in \mathcal U\times U} \mathrm{Hom}(H \cap H', K).$$

Let $H(\mathcal U, K)$ denote the middle cohomology of this complex, in the category of pointed sets. In words, $H(\mathcal U, K)$ consists of the set of $\mathcal U$-compatible collections of homomorphisms $L\to K$, in which we have collapsed to a point all of those compatible collections coming from actual homomorphisms $G \to K$. Thus, $\rho$ is a homomorphism if and only if the class of the cocycle $\{\rho_L\}$ in $H(\mathcal U, K)$ is the trivial class (the distinguished point of $H(\mathcal U, K)$).

Question For which groups $G$ do we have $H(\mathcal U, K) = *$ for every covering $\mathcal U$ and every group $K$?

It is easy to see that a cyclic group has this property, since we have $G \in \mathcal U$ for every covering $\mathcal U$ of $G$ (some $L \in \mathcal U$ must contain the generator, so $\rho$ is trivially a homomorphism). This is pretty much the only example that I know. Some fairly natural questions are:

If $G$ is a free group on finitely many generators, do we have $H(\mathcal U, K)=*$ for every $\mathcal U, K$? What if $G$ is free abelian on finitely many generators?

Remark: One could also ask these questions by requiring only that $H(\mathcal U, K)=*$ for every finite covering of $G$. If a group $G$ has the property that every finite covering contains $G$, as in the case where $G$ is cyclic, then $H(\mathcal U, K)=*$ always. One example is the Prüfer group $\mathbf Q_p/\mathbf Z_p$ (which however is still an inductive limit of cyclic groups)... We could also change the question (drastically!) by requiring that $K$ be abelian, so that the complex above becomes a complex of abelian groups, and taking cohomology there instead.

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If I understand your question correctly, and I'm not mistaken, the answer for a non-cyclic, finitely generated free group $F$ is negative.

Fix a prime $p$. Clearly $F$ has a non-cyclic group $E$ of order $p^2$ as an image. Suppose $E = \langle a, b \rangle$, and let $K = \langle g \rangle$ be a cyclic group of order $p$. Now $E$ is the set-theoretic union of $\langle a \rangle$ and the $\langle a^i b \rangle$, for $0 \le i < p$. Clearly two distinct such subgroups intersect trivially.

If you define $\rho_{\langle a \rangle} : a \mapsto 1$, $\rho_{\langle b \rangle} : b \mapsto 1$ and $\rho_{\langle a^i b \rangle} : a^i b \mapsto g$, then it is clear that you cannot put these compatible pieces together to make a homomorphism $E \to K$.

Taking counterimages in $F$, one sees that this fails for $F$ as well.

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    $\begingroup$ Dear Andreas: Excellent! Do you think there are any non-cyclic examples? $\endgroup$ – Bruno Joyal Jul 15 '14 at 4:10
  • $\begingroup$ @Bruno, thanks. It is difficult to say at this stage. Perhaps there's a way to deal with finite simple groups, $A_5$ is easy, but there might be some general argument. Then there might be a way to treat non-cyclic groups that have nontrivial abelian quotients, the remaining case being when these questions are only cyclic. Would you like to try and see whether we can write a paper together? $\endgroup$ – Andreas Caranti Jul 15 '14 at 9:16
  • $\begingroup$ I meant, finite abelian quotients. $\endgroup$ – Andreas Caranti Jul 15 '14 at 9:28

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