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

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 $D=1$.