# How to find a bijection from the set $Hom(G_1\times G_2, G_3\times G_4 )$ to the cartesian product of four Hom sets?

Suppose that $G_1, G_2, G_3, G_4$ be groups and consider $Hom(A, B)$ as collection of group homomorphisms from the group $A$ to the group $B$. We know that in general it is not a group but a set.

I know that there is a bijection from the set $Hom(G_1\times G_2, G_3\times G_4 )$ to the set $Hom(G_1, G_3)\times Hom(G_1, G_4)\times Hom(G_2, G_3)\times Hom(G_2, G_4)$. I have come to know from my seniors that its easy to prove this result using module theory or some thing like which I have zero knowledge.

But I want to establish this using simple elementary set theoratic approach. I mean is it possible to construct any explicit bijection here using elementary group theory only ? I am confused if my question is correct or not and if correct, then how to complete it ?

• The result is not correct. It fails even for $\operatorname{Hom}(G_1\times G_2,G_3)$ if you take for example $G_1 = G_2 = C_2$ and $G_3 = S_3$. There are $4$ homs from each copy into $S_3$, but not $16$ homs from their product. – Tobias Kildetoft Sep 11 '15 at 11:33
• Are the groups commutative? – Bernard Sep 11 '15 at 11:34
• Ohh My god. This is not true in general ? I thought it is true for arbitrary groups. :-( – Anjan3 Sep 11 '15 at 11:35
• @Bernard no. I am just considering arbitrary groups. Is commutativeness important here sir ? – Anjan3 Sep 11 '15 at 11:36
• The problem with your question is that if you have commutative groups, you do have an isomorphism. However $G_1\times G_2$ should be the coproduct, which happens to be the same as the direct product in the case of commutative groups, but it is no more true for general groups. – Bernard Sep 11 '15 at 11:40

We do have $\text{Hom}(A, B \times C) \cong \text{Hom}(A, B) \times \text{Hom}(A, C)$ by the following isomorphism:
$$\phi: f \mapsto (\pi_1 \circ f, \pi_2 \circ f)$$
where $\pi_1 : B \times C \to B$ is the projection onto the first coordinate, and $\pi_2: B \times C \to C$ onto the second. You can check that this is an isomorphism.
Therefore, $$\text{Hom}(G_1 \times G_2, G_3 \times G_4) \cong \text{Hom}(G_1 \times G_2, G_3) \times \text{Hom}(G_1 \times G_2, G_4)$$
However, we can't necessarily do better: $\text{Hom}(G_1 \times G_2, G_3)$ is not necessarily $\text{Hom}(G_1, G_3) \times \text{Hom}(G_2, G_3)$. I see that Tobias Kildetoft's given the example $G_1 = G_2 = C_2$ and $G_3 = S_3$ in the comments.