# Collections in direct products and freeness

I am looking for references about the following type of questions:

Let $G$ and $H$ be two groups, let $(g_i:i\in I)\subset G$ and $(h_i:i\in I)\subset H$ be collections of group elements, and let $$F=\big\langle (g_i,h_i):i\in I\big\rangle\subset G\times H$$ be the group generated by every pair $(g_i,h_i)$ in the direct product $G\times H$. What are sufficient/necessary conditions for the groups $\big\langle(g_i,h_i)\big\rangle$ to be free from each other in $G\times H$, or, in other words, for $F$ to be isomorphic the free product of the groups generated by the $(g_i,h_i)$, that is, $$F\cong *_{i\in I}\big\langle(g_i,h_i)\big\rangle.$$

I have done some research on Google and MAthSciNet with keywords like "free product", "free", "direct product", but most of the papers I can find are about subgroups of direct product of free groups (for example https://people.maths.ox.ac.uk/bridson/papers/BHow05/BH2.pdf), or subgroups of free products of groups (such as http://www.ams.org/journals/tran/1970-150-01/S0002-9947-1970-0260879-9/S0002-9947-1970-0260879-9.pdf), which is not exactly what I'm looking for.

• Question referred to in arxiv.org/abs/1605.00288 by Benoit Collins and Pierre Yves Gaudreau Lamarre.
– YCor
May 10, 2019 at 20:08

To begin with: a necessary and sufficient condition for $(g_i,h_i)_{i\in I}$ to be free is that either $(g_i)_{i\in I}$ or $(h_i)_{i\in I}$ is free.

It is obviously sufficient. Let us check it's necessary; the case when $I$ is a singleton being straightforward (to say that a singleton family is free means that its single element has infinite order), assume it has at least 2 elements. If none of the two families is free, there exist nontrivial group words (in the free group over $I$), say $w,w'$ such that $w((g_i)_{i\in I})=1$ and $w'((h_i)_{i\in I})=1$. Since $I$ has at least 2 elements, there exists a conjugate of $w'$ (say, by some generator) that does not commute with $w$. Then possibly replacing $w'$ with this conjugate and then considering $w''=[w,w']$, we have $w''((g_i,h_i)_{i\in I})=1$ and $w''\neq 1$.

Now I answer the question, which regards freeness of arbitrary cyclic groups (recall that, by definition, a family $(H_i)$ of subgroups of a group $G$ is free if the canonical homomorphism from the free product of the $H_i$ into $G$ is injective).

Define $\gamma_i=(g_i,h_i)$; assume that no $\gamma_i$ equals 1 and that $I$ has at least 2 elements. For $g$ in a group, define $o(g)$ to be the order of $g$ (0 if $g$ is not torsion).

A necessary and sufficient condition for $(\langle(g_i,h_i)\rangle)_{i\in I}$ to be free is that either $o(h_i)$ divides $o(g_i)$ for all $i$ and $(\langle g_i\rangle)$ is free, or vice versa.

It is obviously sufficient. Otherwise, first assume that the divisibility property fails. Hence there exist indices $i,j$ and positive integers $k,\ell$ such that $g_i^k=1$, $h_i^k\neq 1$, $h_j^\ell=1$, $g_j^\ell\neq 1$. Then $\gamma_i^k\neq 1$, $\gamma_j^\ell\neq 1$, but $[\gamma_i^k,\gamma_j^\ell]\neq 1$, hence $(\langle\gamma_i\rangle)_{i\in I}$ is not free.

Otherwise we have, up to switching: $o(h_i)$ divides $o(g_i)$ for all $i$ (which forces $g_i\neq 1$ for all $i$ since $\gamma_i\neq 1$ was assumed), but the condition fails, which means that there exists nontrivial words $w,w'$ in the free product of cyclic groups ${\large\ast}_i\mathbf{Z}/o(g_i)\mathbf{Z}$ such that $w((g_i)_{i\in I})=1$ and $w'((h_i)_{i\in I})=1$. Indeed, define $w''$ to be the commutator of $w$ and $w'$, where $w'$ is possibly replaced with a conjugate if it accidentally () commutes with $w$; then $w''((\gamma_i)_{i\in I})\neq 1$ and it follows that $(\langle\gamma_i\rangle)_{i\in I}$ is not free. For () to work I need the fact that the centralizer of any nontrivial conjugacy class is trivial, a fact that holds in an arbitrary nontrivial free product with the unique exception of the infinite dihedral group.

(*) this fails when $I$ has only 2 elements $i,j$ and both $g_i,g_j$ have order 2 (hence $h_i^2=h_j^2=1$ as well by the divisibility assumption); then a stupid argument works: if both $(g_1,g_2)$ and $(h_1,h_2)$ are non-free, then both $\{g_1,g_2\}$ and $\{h_1,h_2\}$ generate finite groups, hence so does $\{\gamma_1,\gamma_2\}$ and hence $(\gamma_1,\gamma_2)$ is not free.

• PS: alternatively, if the nontrivial elements $w$ and $w'$ commute, then they have a nontrivial common power $w''$ (general fact in free groups) and that's enough, so no need to treat separately the case when $I$ is a singleton.
– YCor
Apr 30, 2015 at 17:43
• I'm not convinced about sufficiency. Suppose that $g_1$ and $h_2$ have order 2, and that $g_2$ and $h_1$ are of infinite order. Then, let $G=\langle g_1\rangle*\langle g_2\rangle$, and $H=\langle h_1\rangle*\langle h_2\rangle$. Then, by definition, $g_1$ is free from $g_2$, and $h_1$ is free from $h_2$. Yet, $(g_1,h_1)^2(g_2,h_2)^2(g_1,h_1)^{-2}(g_2,h_2)^{-2}=(e,h_1^2)(g_2^2,e)(e,h_1^{-2})(g_2^{-2},e)=(e,e)$, and so $(g_1,h_1)$ and $(g_2,h_2)$ are not free.
– user34424
Apr 30, 2015 at 17:51
• You're right, I misread your definition of free, because it is non-standard. You should say instead $(\langle g_i\rangle)$ is free. So $(g_i)$ is free in the standard terminology if and only if all $g_i$ have infinite order and $(\langle g_i\rangle)_{i\in I}$ is free.
– YCor
Apr 30, 2015 at 18:15
• Ah, I did not know of the standard definition. Sorry about the confusion!
– user34424
Apr 30, 2015 at 18:18
• I edited to answer the question in general.
– YCor
Apr 30, 2015 at 19:11