Coproduct of $C_2$ and $C_3$ in $\mathsf{Grp}$ I have been self-studying from Aluffi's Algebra: Chapter 0. I am looking at Chapter II section 3, exercise 3.8. Here it is:

As far as I can tell, I need to do the following: 


*

*Describe the projections $\pi_2 : C_2 \to G$ and $\pi_3 : C_3 \to G$.

*Show that the universal property for coproducts is satisfied: that is, for any group $A$ and any choice of group homomorphisms $\varphi_2 : C_2 \to A$ and $\varphi_3 : C_3 \to A$, there exists a unique group homomorphism $\sigma : G \to A$. 


Giving a more concrete description of the cyclic groups, take $C_2 = \{0, 1\}$ and $C_3 = \{0, 1, 2\}$. Then the projections are $\pi_2(k) = x^k$ where $k=0,1$, and $\pi_3(l) = y^l$ where $l=0,1,2$. I think that this takes care of (1).
Beyond this, it is not clear to me what I should do. Any help would be appreciated!
 A: You can look at $G$ as a group that has strings like $xyxy^2xy$ or $y^2xyxy$ as elements. 
Then $\phi:G\to A$ on e.g. $xyxy^2xy$ must be prescribed by:$$xyxy^2xy\mapsto\phi_2(x)\phi_3(y)\phi_2(x)\phi_3(y)^2\phi_2(x)\phi_3(y)$$
This $\phi$ must be shown to be a group homomorphism with $\phi\circ\pi_2=\phi_2$ and $\phi\circ\pi_3=\phi_3$, and must be shown to be unique in satisfying this.
A: A map out of a group is just given by a map on the generators such that the relations are satisfied. Thus a map from $G$ to $H$ is uniquely determined by two elements of $H$, one which squares and one which cubes to the identity. This is exactly the description of a pair of maps from $C_2$ and $C_3$ to $H$, and the correspondence is induced by maps from $C_2$ and $C_3$ to $H$ taking a generator to a generator, as you say.
A: I don't know if you are satisfied with the above two brilliant and suggestive answers given by @Kevin Arlin and @drhab. So I will elaborate on  @drhab's answer, using your notations.
You can look at $G$ as a group that has strings like $xyxy^2xy$, $y^2xyxy$, or $xyxy^2$ as elements. Take $xyxy^2$ as an example.

*

*The condition "$\sigma:G\to A$ is a group homomorphism" forces that
$$xyxy^2 \mapsto\sigma(x)\sigma(y)\sigma(x)\sigma(y)^2$$


*Besides, the condition "the diagram commutes such that $\sigma \circ \pi_2 = \varphi_2$ and $\sigma \circ \pi_3 = \varphi_3$" and the fact that $\pi_2(1)=x$ and $\pi_3(1)=y$ forces the above map becomes
$$xyxy^2 \mapsto\sigma(x)\sigma(y)\sigma(x)\sigma(y)^2 = \sigma \circ \pi_2(1)\cdot  \sigma \circ \pi_3(1) \cdot \sigma \circ \pi_2(1)  \cdot (\sigma \circ \pi_3(1))^2 = \varphi_2(1)\varphi_3(1)\varphi_2(1)\varphi_3(1)^2$$
In fact, the map $\sigma: G\to A$ is completely prescribed by the above 2 "forces". So the map $\sigma: G\to A$ is the unique group homomorphism.
The point behind the proof is just what @Kevin Arlin said in his answer: the group homomorphism $\sigma: G\to$ any set $A$ is completely determined by how the homomorphism act on the group generators (Meanwhile, it must satisfy all relations. See this post). Your functions $\pi_2$ and $\pi_3$ are "injections", they identify $[1]$ in $C_2$ with $x$ in $G$, $[1]$ in $C_3$ with $y$ in $G$. After this identification, $\varphi_2$ and $\varphi_3$ actually specify how $\sigma$ should act on generators $\{x,y\}$ in $G$. So $\sigma$ has no choice, as has been proved above.
