Coproduct of $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/3\mathbb{Z}$ in $\textbf{Grp}$? Define a group $G$ with two generators $a$, $b$, subject only to the relations $a^2 = e_G$, $b^3 = e_G$. How do I see that $G$ is a coproduct of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ in $\textbf{Grp}$?
Thoughts on the problem so far: I probably want to show that it satisfies the correct universal property? I am having trouble constructing the necessary homomorphism explicitly using the two given homomorphisms (in the definition of coproduct). Can anyone help?
 A: In general, the coproduct of any kind of algebraic structures can be presented by the disjoint union of generators with the disjoint union of 'relations'.
In your case $\Bbb Z/2\Bbb Z\cong\langle a\mid a^2=e\rangle\,$ and $\ \Bbb Z/3\Bbb Z\cong\langle b\mid b^3=e\rangle$, so just take the union of the presentations.
Here, by a 'relation' in the general setting, we mean a pair $(\sigma,\tau)$ of elements of a free algebra $F_X$ (whose elements are just the formal terms composed of the given operations applied to the given 'generator elements' $\in X$).
Now a presentation can be defined as a pair $\langle X\mid R\rangle$ where $X$ is a set (of generators) and $R$ is a set of relations on the free algebra $F_X$. What it 'presents' is the quotient structure $F_X/(R)$ where $(R)$ is the smallest congruence containing $R$. 
Also note that, any structure $\mathfrak M$ is presented in a trivial way by $\langle M\mid \Delta_M\rangle$ with all of its elements and all valid relations.

A function $f:X\to Y$ will be a morphism of presentations $\langle X,R\rangle\to\langle Y,T\rangle\ $ iff for all $(\sigma,\tau)\in R$ we have $\ (f(\sigma),\,f(\tau))\in T$. 
Finally, the functor $\mathfrak M\mapsto \langle M\mid \Delta_M\rangle$ is right adjoint to $\langle X\mid R\rangle\mapsto F_X/(R)$, hence it preserves colimits.
A: If $G$ is a finitely-generated group, then any group homomorphism $\varphi: G \to H$ is determined entirely by its action on the generators of $G$, much like a linear map from a vector space $V$ is determined entirely by its action on a basis of $V$. This means two things.


*

*If we want to "build" a homomorphism $\varphi: G \to H$, we are done once we decide which elements of $H$ we want to map the generators of $G$ to.

*If we have a homomorphism $\varphi: G \to H$, then we can write it in terms of its action of the generators, even if we have not specified what the action is. 


For example, let $\mathbb{Z}/5\mathbb{Z} = \langle a : a^5 = e\rangle$ be the cyclic group of order $5$ (generated by $a$). Every element $g$ of $\mathbb{Z}/5\mathbb{Z}$ can be written as $g = a^k$ for some positive integer $k$, so every homomorphism $\varphi: \mathbb{Z}/5\mathbb{Z} \to H$ can be written as$$\varphi(g) = \varphi(a^k) = \varphi(a)^k.$$
Both $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are cyclic, so let us denote their generators by $g_2$ and $g_3$, respectively. Normally, we would use "$1$" as the generator, but it will become confusing unless we distinguish the $1$ in $\mathbb{Z}/2\mathbb{Z}$ from the $1$ in $\mathbb{Z}/3\mathbb{Z}$.
Let us build two homomorphisms by thinking about where we should send generators. 


*

*We have $\mathbb{Z}/2\mathbb{Z} = \langle g_2 : g_2^2 = e\rangle$, so any homomorphism $i: \mathbb{Z}/2\mathbb{Z} \to G$ is determined by the value of $i(g_2)$. Intuitively, which element of $G$ "corresponds" most naturally to $g_2$? 

*Likewise, we have $\mathbb{Z}/3\mathbb{Z} = \langle g_3 : g_3^3 = e\rangle$, so any homomorphism $j: \mathbb{Z}/3\mathbb{Z} \to G$ is determined by the value of $j(g_3)$. Intuitively, which element of $G$ "corresponds" most naturally to $g_3$?

