What does the homomorphism from $U(10) \rightarrow \mathbb{Z}_4$ mapping look like?

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Given a homomorphism function $f : U(10) \rightarrow \mathbb{Z}_4$ where $f([3]_{10}) = [3]_4$, find $f([1]_{10}), f([7]_{10})$, and $f([9])_{10}$.

I found that $U(10)$ can be generated by $[3]_{10}$. Then by homomorphism properties $f(a^n) = f(a)^n$.

So, I this is what I tried to get the $f([1]_{10})$:

$f([1]_{10}) = f(([3]_{10})^4) = f([3]_{10})^4 = ([3]_4)^4$

However, I'm not sure how to continue it as it is element in $\mathbb{Z}_4$ which the binary operation is addition, or is it just like the usual $3^4$ mod 4?

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This is a standard calculation that I think you would benefit from doing yourself; having said that, you have made an elementary mistake in what is otherwise the right idea. I will, thus, try to correct it here.


Let $(G, \circ), (H, \ast)$ be groups with sets $G,H$ and binary operations $\circ, \ast$, respectively.

We call a map $\varphi: G\to H$ a homomorphism if for all $a,b\in G$,

$$\boxed{\varphi(a\circ b)=\varphi(a)\ast\varphi(b).}$$

That is, homomorphisms respect operations.


So $$f([3]_{10}^4)=4f([3]_{10}).$$