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 $\{y\}$).
Best Answer
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.