I put this question on Stackexchange :


<https://math.stackexchange.com/questions/1659760/when-are-groups-subgroups-of-a-same-group>


but it got no answer, so I post it here.


Let $\mathcal{G}$ be a nonempty set of groups. When is it true that there exists a group $B$ ($B$ as "big") such that every group in $\mathcal{G}$ is a subgroup of $B$ ? I mean "a subgroup" in the strict sense, not "isomorphic to a subgroup".


Clearly, the following conditions are necessary for the existence of such a group $B$ (where $x \ \underset{G}{\star} \ y$ denotes the product of $x$ by $y$ in $G$) :


1° the groups in $\mathcal{G}$ have all the same identity;


2° if $G$ and $H$ are groups in $\mathcal{G}$, if $x$ and $y$ are in $G \cap H$, then $x \ \underset{G}{\star} \ y = x \ \underset{H}{\star} \ y$;


3° if $G$ and $H$ are groups in $\mathcal{G}$, if $x$ is in $G \cap H$, then $x$ has the same inverse in $G$ and in $H$;


4° if $G, H, K, L$ are in $\mathcal{G}$, if both expressions $x \underset{G}{\star} (y \ \underset{H}{\star} \ z)$ and $(x \ \underset{K}{\star} \ y) \underset{L}{\star} \ z$ make sense, these expression are equal.


(In fact, 2° is implied by the conjunction of 1° and 4°.)


I wonder if the conjunction of these conditions is sufficient for the existence of a group $B$ such that every group member of $\mathcal{G}$ is a subgroup of $B$.


Generally, if $\mathcal{G}$ is a (let us say nonempty) set of groups, here is a necessary and sufficient condition for the existence of a group $B$ such that every group member of $\mathcal{G}$ is a subgroup of $B$.


Let $V$ denote the union (in the usual sense, not the "disjoint union") of the groups members of $\mathcal{G}$. Let $Mo(V)$ denote the free monoid on $V$. Let $R\_{\mathcal{G}}$ denote the "smallest" equivalence relation in $Mo(V)$ compatible with the monoid law of $Mo(V)$ and such that


5° for every $G$ in $\mathcal{G}$, the unity of $G$ is congruent modulo $R\_{\mathcal{G}}$ with the identity (empty word) of $Mo(V)$;


6° for every $G$ in $\mathcal{G}$, for all $x$ and $y$ in $G$, the element (of length two) $(x, y)$ of $Mo(V)$ is congruent modulo $R\_{\mathcal{G}}$ with the element (of length 1) $(x \ \underset{G}{\star} \ y)$.


Let $\mathcal{S}(\mathcal{G})$ (as "sum") denote the quotient monoid $Mo(V)/R\_{\mathcal{G}}$ and $\varphi$ denote the canonical monoïd morphism from $Mo(V)$ onto this quotient. Then $\mathcal{S}(\mathcal{G})$ is a group and, if I'm not wrong, the two following conditions are equivalent :


7° there exists a group $B$ such that every group in $\mathcal{G}$ is a subgroup of $B$;


8° for all $G,H$ in $\mathcal{G}$, for all $x$ in $G$, for all $y$ in $H$, the equality $\varphi(x') = \varphi(y')$, where $x'$ and $y'$ denote the images of $x$ and $y$ by the canonical injection from $V$ into $Mo(V)$, implies $x = y$ (in other words, the restriction of $\varphi$ to the set of the words of length 1 is injective.


Here is the proof that 7° implies 8°. Assume 7°. If $(x\_{1}, \ldots , x\_{r})$ and $(y\_{1}, \ldots , y\_{s})$ are elements of $Mo(V)$ congruent modulo $R\_{\mathcal{G}}$, then $x\_{1} \ldots x\_{r} = y\_{1} \ldots y\_{s}$ in $B$. Taking $r = s = 1$, we obtain 8°. In order to prove that 8° implies 7°, one can construct $B$ isomorphic to $\mathcal{S}(\mathcal{G})$.


My question is : are the (equivalent) conditions 7° and 8° equivalent to the conjonction of 1°, 3° and 4° ? It is perhaps not difficult to solve, but a hint to the literature (if any) could help me to save time. Thanks in advance.



When are groups subgroups of a same group?
Too long for a comment.


You are going to need more, as Derek Holt suggests. This should be clear if you look at Hanna Neumann's work on *Generalized free products with amalgamated subgroups*, Amer. J. Math **70** (1948) 590-625 and **71** (1949) 491-540. 


The goal there is weaker than yours, but the conditions required are already a bit stronger, since they require looking at "triple intersections". Explicitly, Hanna Neumann considers a family $\mathcal{G}=\{G\_{\alpha}\}\_{\alpha\in A}$ of groups, for each pair of indices $\alpha\neq\beta$, subgroups $U\_{\alpha\beta}\leq G\_{\alpha}$ and $U\_{\beta\alpha}\leq G\_{\beta}$, together with isomorphisms $f\_{\alpha\beta}\colon U\_{\alpha\beta}\to U\_{\beta\alpha}$ such that $f\_{\alpha\beta}^{-1}=f\_{\beta\alpha}$. She looks for conditions under which there exists a group $\mathfrak{F}$ and monomorphisms $\mu\_{\alpha}\colon G\_{\alpha}\to \mathfrak{F}$ such that $\mu\_{\alpha}|\_{U\_{\alpha\beta}} = \mu\_{\beta}\circ f\_{\alpha\beta}$ for all $\alpha\neq \beta$. (I.e., only looking up to isomorphism).


Early on (Theorem 3.0 in the first part), she gives the following necessary condition:



> 
> For any three distinct indices $\alpha,\beta,\gamma$, the meets $$U\_{\alpha\beta\gamma}=U\_{\alpha\beta}\cap U\_{\alpha\gamma},\quad U\_{\beta\gamma\alpha}=U\_{\beta\alpha}\cap U\_{\beta\gamma},\qquad U\_{\gamma\alpha\beta}=U\_{\gamma\alpha}\cap U\_{\gamma\beta}$$
>  are isomorphic and that the maps $f\_{\alpha\beta}$, $f\_{\beta\gamma}$, $f\_{\gamma\alpha}$ provide the isomorphism between them which, moreover, is such that for every element $u\_{\alpha\beta\gamma}$ of $U\_{\alpha\beta\gamma}$, $$f\_{\beta\gamma}(f\_{\alpha\beta}(u\_{\alpha\beta\gamma})) = f\_{\alpha\gamma}(u\_{\alpha\beta\gamma}).$$
> 
> 
> 


(She also shows that it is enough to consider the subgroup of $G\_{\alpha}$ generated by all the $U\_{\alpha\beta}$; I suspect that the same will be true for your problem). 


However, the conditions are not sufficient. Here is her example:


Let $G\_1=\langle a\_1\rangle\times\langle b\_1\rangle$, where $a\_1$ and $b\_1$ generate infinite cyclic groups. $U\_{12}=\langle a\_1\rangle$ and $U\_{13}=\langle a\_2\rangle$. 


Let $G\_2 = \langle a\_2,c\_2\mid c\_2a\_2=a\_2c\_2^3\rangle$, with $U\_{21}=\langle a\_2\rangle$, $U\_{23}=\langle c\_2\rangle$.


Let $G\_3 = \langle b\_3,c\_3\mid (b\_3c\_3)^2=1\rangle$, with $U\_{31}=\langle b\_3\rangle$, $U\_{32}=\langle c\_3\rangle$. 


The triple intersections are all trivial, so the condition is met. 


The specified isomorphisms identify $a\_1$ with $a\_2$ ( call the resulting element $a$), $b\_2$ with $b\_3$ (call it $b$), and $c\_1$ with $c\_3$ (call it $c$). No overgroup exists containing isomorphic copies of $G\_1$, $G\_2$, and $G\_3$ intersecting in the required subgroups. From $(bc)^2 = 1$ we get $a^{-1}(bc)^2a = (bc^3)^2 = 1$, but this relation between $b$ and $c$ does not follow from $(bc)^2 = 1$. 


I haven't thought about whether you can translate the example from "isomorphic" to "equal"; though I expect you can. In any case, your set-up must certainly satisfy Neumann's necessary condition, for otherwise not even the corresponding abstract group exists. And that condition does not follow from your 1--4, unless I am very mistaken.




---


**Added.** It seems, from comments, that your requirement that we have equality of sets rather than the more common "specificied subgroups identified via specified isomorphisms" stems from a (imho misguided) comment to your question in math.stackexchange, where the poster suggested that the direct product of a family of groups is an example of a group containing (isomorphic copies of) the original groups. 


I think you took the wrong lesson from that comment (and that commenter missed the point). The "right" way to do what you want is to make the data a triple, $(\{G\_{\alpha}\}\_{\alpha\in A}, \{U\_{\alpha\beta}\mid \alpha,\beta\in A, \alpha\neq \beta\}, \{f\_{\alpha\beta}\mid \alpha,\beta\in A\alpha\neq \beta\})$ where $G\_{\alpha}$ is a group, $U\_{\alpha\beta}$ is a subgroup of $G\_{\alpha}$, and $f\_{\alpha\beta}\colon U\_{\alpha\beta}\to U\_{\beta\alpha}$ is an isomorphism (you may also require $f\_{\alpha\beta}^{-1}=f\_{\beta\alpha}$). The question is then when does there exist a group $B$, and group embeddings $g\_{\alpha}\colon G\_{\alpha}\to B$, such that $g\_{\alpha}(u) = g\_{\beta}(f\_{\alpha\beta}(u))$ for all $\alpha\neq \beta$ (for weak amalgams); or with the additional condition that the intersection $f\_{\alpha}(G\_{\alpha})\cap f\_{\beta}(G\_{\beta})$ in $B$ be exactly $f\_{\alpha}(U\_{\alpha\beta})=f\_{\beta}(U\_{\beta\alpha})$ (for strong amalgams).


Phrased this way, you have **exactly** the subject of Hanna Neumann's work mentioned above. 


