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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 $([0]_4, [2]_4, [1]_4)$ to map on, and $9$ possibilities. But answer isn't $9$. Where my logic failed me?

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  • $\begingroup$ 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...? $\endgroup$ – DonAntonio Apr 26 '14 at 12:24
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    $\begingroup$ What is the problem with $[3]_4$? It has also order 4 in $\mathbb{Z}_4$. $\endgroup$ – Josué Tonelli-Cueto Apr 26 '14 at 12:29
  • $\begingroup$ That may be my mistake. I didn't checked all the elements. Would it be 16 possibilities? $\endgroup$ – Dark Archon Apr 26 '14 at 12:33
  • $\begingroup$ @DarkArchon Yes, that's the answer. $\endgroup$ – Josué Tonelli-Cueto Apr 27 '14 at 9:56
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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.

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  • $\begingroup$ Why 8? Shouldn't it be 16: $ \beta([(1,0)])=([0]_4, [1]_4, [2]_4, [3]_4)$, same for $\beta([(1,0)])$? $\endgroup$ – Dark Archon Apr 26 '14 at 12:44
  • $\begingroup$ @DarkArchon Ooops. Of course. $\endgroup$ – Fredrik Meyer Apr 26 '14 at 13:01
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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.

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