When are groups subgroups of a same group?

Question

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.

Answer 1

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.

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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.