Is the coproduct of any family of groups a free group

Question

In Lang’s Algebra, he constructs the coproduct of a family of groups in a similar way to the construction of the free group of a set $S$, $F(S)$. He later constructs the free product of $n$ groups and shows that if they are cyclic, then the free product is isomorphic to the free group on $n$ generators. I have also read elsewhere that the free product(of finitely many groups) is the coproduct of those groups. In light of the similar construction between the coproduct in the category of groups and $F(S)$, I am wondering if the coproduct of an arbitrary family {$G_i$} is also a free group. If you look in Lang, it seems that the coproduct of groups may be the free group generated by the set $S$ which has the same cardinality as the disjoint union of the groups {$G_i$}.

My apologies if this is trivial, I just can’t quite grasp whether or not it is true.

Answer 1

What Lang shows is that if the groups are infinite cyclic then their free product is a free group. But that's because the infinite cyclic group is itself free.

The free product of an arbitrary family of groups can be defined similarly to how the free product of a finite family is defined. But in general it will not be free.

Recall that a subgroup of a free group is necessarily free. Thus we have:

Theorem. Let $\{G_i\}_{i\in I}$ be a family of groups. The free product $\mathop{*}\limits_{i\in I}G_i$ is a free group if and only if each $G_i$ is free.

Proof. Since the free product contains a copy of each $G_i$, if the free product is free, then each $G_i$ is free.

Conversely, if $G_i$ is free on $X_i$, then the free product is free on $X = \mathop{\amalg}\limits_{i\in I} X_i$, the disjoint union of the $X_i$: given any set map $f\colon X\to K$ from $X$ to (the underlying set of) a group $K$, this map corresponds to a family of maps $f_i\colon X_i\to K$, and each $X_i$ induces a morphism $F_i\colon G_i\to K$ with $F_i|_{X_i}=f$. The universal property of the free product then induces a morphism $F\colon\mathop{*}\limits_{i\in I}G_i\to K$ with $F\ circ\iota_j= F_j$ for each $j$ (where $\iota_j$ is the embedding of $G_j$ into the free product. Thus, $F\circ\iota_j|_{X_j} = f_j$, and thus the restriction of $F$ to (the image of) $X$ is equal to $f$. Thus, the free product has the relevant universal property and hence is the free group on $X$. $\Box$

Answer 2

You are missing a vital assumption here: the coproduct of a family of infinite cyclic groups is free. There is, of course, only one infinite cyclic group: $\mathbb Z$. For example, the infinite dihedral group $\mathbb Z/2 * \mathbb Z/2$ is not free. I'll prove this using the fact that the free product of a family of groups is their coproduct, but as Arturo Magidin pointed out in the comments, you can simply use torsion.

By universal property, we have an isomorphism $\mathrm{Hom}(\mathbb Z/2 * \mathbb Z/2, \mathbb Z/3) \cong \mathrm{Hom}(\mathbb Z/2, \mathbb Z/3) \times \mathrm{Hom}(\mathbb Z/2, \mathbb Z/3)$. The only map $\mathbb Z/2 \longrightarrow \mathbb Z/3$ is trivial, so the only map $\mathbb Z/2 * \mathbb Z/2 \longrightarrow \mathbb Z/3$ is trivial as well. However, we have $\mathrm{Hom}(F(S), \mathbb Z/3) \cong 3^S$, which is a singleton iff $S = \emptyset$ iff $F(S)$ is the trivial group. But $\mathbb Z/2 * \mathbb Z/2$ is not trivial, so it cannot be a free group on any set $S$.

One large benefit of the approach Arturo mentions in the comments is that the essence can be phrased very concisely by saying "look at torsion." I arrived at this proof by a similarly concise thought process. I thought to myself "the objects have different universal properties." Thinking like this is valid due to the Yoneda lemma, and I think it is a worthwhile perspective to consider in algebra.