# Prove that the free group generated by two elements is a coproduct of the integers by itself in Grp

That's a problem from Algebra: Chapter $0$ by P. Aluffi, p.78, ex.5.6.

One needs to prove that the group $F(\{x,y\})$ is a coproduct $\mathbb Z*\mathbb Z$ of $\mathbb Z$ by itself in the category $\text{Grp}$.

The author gives a hint to a reader: ''With due care, the universal property for one turns into universal property for the other''.

He assumes that the reader is familiar with basics of the category theory (functors hasn't been introduced in the book yet) and definition of the free groups via universal property.

As of now, I don't have any more ideas. Just wrote out the universal property for coproducts and free group of $\{x,y\}$ and observed that $\mathbb Z$ itself is a free group of a set with the cardinality of 1 (hence of $\{x\}$ and $\{y\}$).

• Can you find an inclusion $\{x,y\}\to Z*Z$ such that any function $\{x,y\}\to G$ extends uniquely to a homomorphism $Z*Z\to G$ ? That would prove that $Z*Z$ is the free group on two generators. Commented Jul 5, 2015 at 17:36

The short answer is that the free group functor $F \colon \mathsf{Set} \to \mathsf{Grp}$ is left adjoint to the forgetful functor and so commutes with colimits, hence $$F(\{x,y\}) \simeq F(\{x\}\sqcup\{y\}) \simeq F(\{x\}) \ast F(\{y\}) \simeq \mathbb Z \ast \mathbb Z .$$

But (adjoint) functors are not supposed to be known at this stage of the book, so just show it directly: show that $F(\{x,y\})$ is indeed the vertex of an initial cocone over the diagram $$\mathbb Z \qquad \mathbb Z.$$ Denote $\iota_1 \colon \mathbb Z \to F(\{x,y\}), 1 \mapsto x$ and $\iota_2 \colon \mathbb Z \to F(\{x,y\}), 1 \mapsto y$. Now, for any diagram $$\mathbb Z \stackrel f \to G \stackrel g \leftarrow \mathbb Z,$$ one can define $h_{f,g} \colon F(\{x,y\}) \to G$ on the generators by $x \mapsto f(1), y\mapsto g(1)$, making the following diagram commute $$\begin{array}{ccccc} \mathbb Z & \stackrel {\iota_1} \to & F(\{x,y\}) & \stackrel{\iota_2} \leftarrow & \mathbb Z \\ & {}_f \searrow & \ \ \ \ \ \downarrow {}_{h_{f,g}} & \swarrow {}_g & \\ & & G & & \end{array}$$ Of course, such an $h_{f,g}$ is necessarily unique: it is determined by the image of $x$ and $y$, which are given by the commutative diagram.

Hence, we have our initial cocone: $$\mathbb Z \stackrel{\iota_1} \to F(\{x,y\}) \stackrel{\iota_2} \leftarrow \mathbb Z,$$ showing that $F(\{x,y\})$ is a coproduct of $\mathbb Z$ by itself in the category of groups.

• And in the end we have to check that such unique homomorphism $F(\{x,y\}$ exists. Let $w$ be an arbitrary element of $F(\{x,y\}$( not necessarily an image of an element of $Z$ ).If $h$ it exists than it is a homomorphism. $w = \prod_{i=1}^nz_i, z_i \in \{x,x^{-1},y,y^{-1}\}$. Then $h(w) = \prod_{i=1}^nf(z_i)$. We check that it is indeed a homomorphism. Is that right? Commented Jul 7, 2015 at 15:58
• There's a mistake in my last comment. $h(w) = \prod_{i=1}^nd(z_i)$, where $d(z_i) =$ \begin{cases} f(z_i), & \mbox{if } z_i \in \{x^{-1},x\} \\ g(z_i), & \mbox{if } z_i \in \{y^{-1},y\} \end{cases} Commented Jul 7, 2015 at 22:21
• I'm not sure I understand your comments. The morphism $h_{f,g}$ exists because I defined it in my answer. By definition, if $A$ is a set, a group morphism $FA \to G$ is exactly the same as a function from A to (the underlying set of) $G$. In categorical terms, we have a bijection $\mathsf{Grp}(FA,G) \simeq \mathsf{Set}(A,UG)$ natural in both $A$ and $G$ (where $U$ is the forgetful functor $\mathsf{Grp} \to \mathsf{Set}$).
– Pece
Commented Jul 8, 2015 at 5:46
• Yes, you defined it, but we need to check it is indeed a homomorphism. Commented Jul 8, 2015 at 14:04
• @Jxt921 I think what Pece is saying is that since we are defining $h_{f,g}$ in terms of $f$ and $g$, which are already group homomorphisms, then naturally $h_{f,g}$ will also be a group homomorphism. Commented May 28, 2019 at 0:09

Just be careful that the group law of $Z$ is $+$, while that of $G$ is usually a multiplication. Let us abbreviate the free group on two generators simply by $F$. Map $x$ to $(1,0)$ and $y$ to $(0,1)$. By the universal property of free groups, this extends uniquely to a group homorphism $\phi$ from $F$ to $Z * Z$.

On the other hand, you have $f_1: Z \to Z * Z$ and $f_2: Z \to Z * Z$ corresponding to the two copies of $Z$. Consider the group homomorphisms $g_1: F({x}) \to F$ and $g_2: F({y}) \to F$, defined by sending $x$ to $x$, and $y$ to $y$, respectively. Note that both $F({x})$ and $F({y})$ are isomorphic to Z (map $x^n$ to $n$, etc.). Then by the universal property of coproducts, $g_1$ and $g_2$ induce a unique group homomorphism $\psi$ from $Z * Z$ to $F$, such that the diagram commutes.

Claim: $\phi$ and $\psi$ are inverses of each other. Consider $\psi \circ \phi$ from $F$ to $F$. It maps $x$ and $y$ to $x$ and $y$ respectively, just like the identity homomorphism, so it must be the identity, by the uniqueness part of the universal property for free groups.

Similarly, check that $\phi \circ \psi$ from $Z * Z$ to $Z * Z$ is the identity, using the universal property of the coproduct. "Et voila!"

What is a coproduct? an initial element of the double coslice category. (Sorry or my bad terminology, from a previous answer i guess the right term is cocone). As we have to act in the category group, the goal of that coproduct is an arbitrary group G. So we need to prove that given two group morphisms α, β: ℤ → G there are i₁, i₂ : ℤ → F({x, y}) such that there is a unique group homomorphism ψ : F({x, y}) → G such that ψi₁ = α and ψi₂ = β.

What is the stuff we can make use of? As a free group, for any set function f : {x, y} → G, we have a unique ψ group homomorphism: F({x, y}) → G, such that f = ψj with j being defined in the usual way as j: {x, y} → F({x, y}) given by j(x) = x, j(y) = y where the x, y on the right side stand for letters in F({x, y}).

Similar is that both universal properties yield some unique group homomorphism ψ aiming at G. But the coproduct starts from two group homomorphisms a, β with domain ℤ and the free group from a single set function f from the set {x, y}. So a route looks like constructing a function f, that somehow emulates α and β.

First define i₁', i₂' : {1} → {x, y} with i₁'(1) = x and i₂'(1) = y.

The morphisms α, β define the set functions α', β' : {1} → G, by α'(1) = α(1) and β'(1) = β(1). Let's use these to a define f : {x, y} → G (also a set function) according to that, such that f(x) = α'(1) and f(y) = β'(1). We can use f to poke at the free group, returning us a nice, useful ψ: By the universal property of the free group F({x, y}) there exists a unique group homomorphism ψ : F({x, y}) → G such that ψj = f. Note that ψji₁' = fi₁' = a' and ψji₂' = fi₂' = β'.

Now looking at our original α, β we define i₁, i₂ : ℤ → F({x, y}) as
i₁(0) = i₂(0) = () (the empty word) and
i₁(n) = (ji₁'(1))ⁿ and
i₂(n) = (ji₂'(1))ⁿ.

For n = 0 we get ψi₁(0) = ψi₂(0) = eG because ψ as a group homomorphism must map its identity (the empty word) to G's identity. As also α and β were group homomorphisms they had to map 0 (ℤ's identity) to G's identity too and for 0 α = ψi₁ and β = ψi₂

For n ≠ 0 we have.

ψi₁(n) =
ψ(ji₁'(1))ⁿ =
by ψ being a group homomorphism
(ψji₁'(1))ⁿ =
(α'(1))ⁿ =
by equality in G
(α(1))ⁿ =
by α being a group homomorphism
α(n).

And likewise for β and ψi₂(n). As ψ was defined uniquely by its free group property it still is defined uniquely and (F({x, y}), i₁, i₂) work as a coproduct for ℤ in the category Grp.

As a comment, which I'm not very sure of: F({x, y}) can act as ℤ*ℤ here because we don't need commutativity in Grp, which the canonical ℤ*ℤ usually has in Ab.