# Number of homomorphisms from $Z_4 \times Z_4$ to $Z_4$

So, I have to count number of homomorphisms like: $\beta:(\mathbb{Z}_4,+)\times (\mathbb{Z}_4,+) \to (\mathbb{Z}_4,+) .$ I count order of the $(\mathbb{Z}_4,+)$, it is $4$. So the elements $(1,0)$ and $(0, 1)$ should be mapped on elements of order $1,2$ or $4$ - divisors of $4$. That gives me elements $(_4, _4, _4)$ to map on, and $9$ possibilities. But answer isn't $9$. Where my logic failed me?

• An idea: any non-trivial (group) homomorphism has kernel of order $\;4\;$ or $\;8\;$ ,so why won't you count subgroups of that direct product with these orders...? – DonAntonio Apr 26 '14 at 12:24
• What is the problem with $_4$? It has also order 4 in $\mathbb{Z}_4$. – Josué Tonelli-Cueto Apr 26 '14 at 12:29
• That may be my mistake. I didn't checked all the elements. Would it be 16 possibilities? – Dark Archon Apr 26 '14 at 12:33
• @DarkArchon Yes, that's the answer. – Josué Tonelli-Cueto Apr 27 '14 at 9:56

In general you have $\hom(A \times B, C) \approx \hom(A ,C) \times \hom (B,C)$ for abelian groups $A,B,C$: So you only need to figure out what the homomorphisms $\hom(\mathbb Z_4, \mathbb Z_4)$ are, but these are given by multiplication, so they are all given by elements of $\mathbb Z_4$.
In conclusion, there are $16 =4 \cdot 4$ in total.
• Why 8? Shouldn't it be 16: $\beta([(1,0)])=(_4, _4, _4, _4)$, same for $\beta([(1,0)])$? – Dark Archon Apr 26 '14 at 12:44
Since all elements of $\mathbb Z_4$ have order dividing $4$, you can map the two generators of $\mathbb Z_4\times\mathbb Z_4$ anywhere you like.