# Is the coproduct of any family of groups a free group

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.

• page number/section/chapter? Aug 19 '21 at 1:25
• @frogeyedpeas The existence of coproducts is on page 70 and 71 in section 12 of chapter 1. Aug 19 '21 at 1:26
• No, the coproduct of a family of groups need not be a free group. It is only a free group if the groups $G_i$ are free. For example, if a $G_i$ has torsion, then the coproduct contains torsion, and hence cannot be free. Aug 19 '21 at 1:52
• But it is true that the free product can be defined for an arbitrary family, and is the coproduct of that family. But the free product of groups need not be a free group. Aug 19 '21 at 1:53
• Can I just point out that the question in the subject line is completely ambiguous. Does "any family" mean some family or all families? Aug 19 '21 at 7:00

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 $$X$$. $$\Box$$

• Thank you for your answer! I am also wondering why Lang requires that the set-theoretic intersection of the groups $A$ and $B$ is {1} in the construction of the free product. He also mentions that the same is true of $A$ and $B$ taken as subgroups of $A*B$, so perhaps that’s the reason. Aug 19 '21 at 19:42
• @AdamFrench: Lang wants to make $A$ and $B$ literal subsets of the free product. In order to do that, you need the underlying sets to intersect trivially. Usually you construct the free product by first "disjointizing" $A$ and $B$, taking the set-theoretic disjoint union $(A\setminus\{1\})\amalg (B\setminus\{1\})\amalg \{1\}$; but if you look at the set theoretic construction, this amounts to constructing isomorphic copies of the set. Usually, you just need a set $X$, and embeddings $\iota_A\colon A\to X$, $\iota_B\colon B\to X$ of underlying sets that intersect at the image of $1$. cont) Aug 19 '21 at 19:47
• @AdamFrench: then you say you "identify" $A$ and $B$ with their images under the $\iota$. If you look at the definition of the coproduct he gives earlier, all you need is embeddings, not literal subgroups. Aug 19 '21 at 19:48

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.