Let $G_{1},G_{2},\dots, G_{n}$ be any finite collection of finite groups. I need to prove that there exists a finite group $\mathbf{G}$ such that each $\mathbf{G_{i}}$ is a homomorphic image of $\mathbf{G}$.
Since this problem is in the section we're doing on direct products, I let $G = G_{1} \times G_{2} \times \cdots \times G_{n}$. By a definition we were given of direct product, $G$ is a group with componentwise multiplication as the group operation; i.e., for $x_{i}, \, y_{i} \in G_{i}$, $(x_{1},x_{2}, \cdots, x_{n})\cdot (y_{1},y_{2},\cdots, y_{n}) = (x_{1}y_{1}, x_{2}y_{2},\cdots, x_{n}y_{n}) \in G$.
Now, I know that $\forall x_{i} \in G_{i}$, we have the projection epimorphism $\pi_{i}(x_{1},x_{2},\cdots, x_{i}, \cdots, x_{n}) = x_{i}$, and so we are able to map any element of $G$ to an element of one of the $G_{i}$, but how do we map the entire group $G$ to the entire group $G_{i}$?
Am I able to use this projection epimorphism, and if so, how do I "sup it up" so that it can take care of whole groups and not just elements of groups?
If not, how do I prove this?
Thank you.