# Does there exist a group that is both a free product and a direct product of nontrivial groups?

Do there exist such nontrivial groups $$A$$, $$B$$, $$C$$ and $$D$$, such that $$A \times B \cong C \ast D$$?

I failed to construct any examples, so I decided to try to prove they do not exist by contradiction.

If such groups exist, then $$C$$ and $$D$$ are disjoint subgroups of $$A \times B$$. Suppose $$w \in F[x, y]\ \{e\}$$, where $$F[x, y]$$ is the free group with generators $$x$$ and $$y$$. Suppose $$(a_c, b_c) \in C$$, $$(a_d, b_d) \in D$$ and $$h: F[x, y] \rightarrow A \times B$$ is a homomorphism, that maps $$x$$ to $$(a_c, b_c)$$ and $$y$$ to $$(a_d, b_d)$$. Then, by definition of free product $$h(w) \neq e$$. Thus, either $$\pi_A(h(w)) \neq e$$ or $$\pi_B(h(w)) \neq e$$, where $$\pi_A$$ and $$\pi_B$$ are projections on $$A \times B$$ onto $$A$$ and $$B$$ respectively. Thus every group word is not an identity either for $$A$$ or for $$B$$ and that results in $$\{A, B\}$$ generating the variety of all groups. And here I am stuck, failing to determine anything else.

Or do the examples actually exist?

• – Michael Burr Feb 12 at 20:51

Let me just replicate YCor's comment in the link that was given in the comments, with more details :

in a free product $$C\ast D$$ with $$C,D$$ nontrivial, the intersection of any two nontrivial normal subgroups is nontrivial.

Indeed, let $$H,K$$ be two nontrivial normal subgroups of $$C*D$$. I'll assume for simplicity that $$|C|,|D|$$ are large enough so that for each $$x,y$$ there is a nontrivial $$z$$ with $$z^{-1}\neq x, z\neq y$$. For instance $$|C|, |D|\geq 4$$ is good enough (take $$x,y$$, then there are at most two nontrivial $$z$$ such that $$z=x$$ or $$z^{-1}=y$$ : $$x$$ and $$y^{-1}$$; so if $$|G|\geq 4$$, any nontrivial element different from $$x$$ and $$y^{-1}$$ works)

Let me say that $$c_1d_1\dots c_nd_n$$ is the reduced form of an element of $$C*D$$ if $$c_i\in C, d_j\in D$$, and the only $$c_i,d_j$$ allowed to be $$1$$ are $$c_1$$ and $$d_n$$. Clearly, if $$x=c_1d_1\dots c_nd_n$$ is the reduced form of $$x$$, and $$n\geq 2$$, then $$x\neq 1$$ in $$C*D$$ ("clearly" here is to be understood as : it's a classical property of free products), moreover, if $$n=1$$ this is $$1$$ if and only if $$c_1=d_1=1$$.

Now let $$x= c_1d_1\dots c_rd_r \in H, y=c'_1d'_1\dots c'_sd'_s \in K$$ be nontrivial elements, with obvious notations, written in reduced form. The point will be that $$[x,y]\in H\cap K$$ (this is obvious by normality), and that, up to changing $$x,y$$ a bit, this can't be the trivial element.

Now up to conjugation by some element of $$C$$, one may assume $$d_r = 1$$ and $$c_1\neq 1$$ (this can be done by the hypothesis on $$|C|$$ - and $$c_r \neq 1$$, but that follows from the reduced form); and up to conjugation by some element of $$D$$, $$c'_1 = 1$$ , $$d_s'\neq 1$$ (using the cardinality hypothesis on $$D$$ - and $$d_1' \neq 1$$, but again this follows from the reduced form).

So with these hypotheses $$[x,y] = \color{red}{c_1d_1\dots c_r}\color{blue}{d_1'\dots c_s'd_s'}\color{red}{c_r^{-1}\dots d_1^{-1}c_1^{-1}}\color{blue}{d_s'^{-1}c_s'^{-1}\dots d_1'^{-1}},$$ which is written in reduced form, and is thus $$\neq 1$$. Therefore $$[x,y]\in H\cap K\setminus\{1\}$$.

Apply this to your supposed $$A\times \{1\}, \{1\}\times B$$ to get a contradiction.

I don't know if there's an easier argument, or a not-too-complicated argument that encapsulates the low cardinality cases, but I guess for these you have to go "by hand" somehow; or perhaps you can adapt this argument to these cases by working a bit more. In any case I didn't want to bother with these cases, and this argument works in most cases and is pretty painless so in any case it's interesting to share

Let $$f:A\times B\rightarrow C*D$$ be an isomorphism where $$A,B,C,D$$ are non trivial groups. Let $$a_0\in A$$ not trivial, for every $$b\in B$$, $$f(a_0)$$ commutes with $$f(b)$$.

Now we look the proof of the The Corollary 4.1.6 p. 187 of Magnus Karrass and Solitar.

The first part of this proof asserts that if $$f(a_0)$$ is contained in the conjugated of a free factor, that is $$f(a_0)$$ is in $$gCg^{-1}$$ or is in $$gDg^{-1}$$ then so is $$f(b)$$ for every $$b\in B$$. Without restricting the generality, we suppose that $$f(a_0)$$ and so $$f(B)$$ are contained in $$gCg^{-1}$$. Let $$b_0$$ a non trivial element of $$B$$ since $$B$$ is contained in the free conjugated factor $$gCg^{-1}$$ the same argument shows that $$f(A)$$ is contained in $$gCg^{-1}$$ this implies that $$C*D$$ is contained in a free conjugated factor. Contradiction.

The second part of the proof shows that if $$f(a_0)$$ is not contained in a free conjugated factor, then there exists $$u_c$$ such $$f(a_0)$$ and $$f(b)$$ are a power of $$u_c$$.

You can express $$f(a_0)$$ uniquely as a reduced sequence (Theorem 4.1.) this implies that there exists a unique element $$u$$ with minimal length such that $$f(a_0)$$ is a power of $$u$$ and if $$f(a_0)$$ is a power of $$v$$, $$v$$ is a power of $$u$$. This impllies that every element of $$f(B)$$ are power of $$u$$ and $$f(B)$$ and $$B$$ are cyclic. A similar argument shows that $$f(A)$$ and $$A$$ are cyclic, we deduce that $$A\times B$$ is commutative. Contradicition since $$C*D$$ is not commutative.

Reference.

Combinatorial Group theory.

Magnus, Karrass and Solitar.

To be a free product $$C\ast D$$ implies the existence of an action on oriented tree such that $$C$$ is the stabilizer of some vertex $$v_0$$ and such that edge stabilizers are trivial (so that nontrivial element fix at most one vertex).

Let $$(a,b)$$ be a nontrivial element of $$C$$; we can suppose that $$a\neq 1$$ up to switch $$A$$ and $$B$$. Then in the Bass-Serre tree, $$(a,b)$$ fixes a unique vertex $$v_0$$, and hence this vertex is also fixed by the centralizer of $$(a,b)$$, and hence $$(a,1)$$ fixes $$v_0$$. Applying this to $$(a,1)$$ shows that $$B$$ fixes the vertex $$v_0$$. Choose $$1\neq b'\in B$$. It fixes the unique vertex $$v_0$$ and hence its centralizer fixes $$v_0$$, so $$A$$ fixes $$v_0$$. Finally $$G=A\times B$$ fixes $$v_0$$, so $$D=1$$.