Write $\mathbb{Z}_{20}^{\times}$ as a product of p-power cyclic groups.

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Can anyone let me know if my answer is okay?

Write $\mathbb{Z}_{20}^{\times}$ as a product of p-power cyclic groups.

I showed that in general $\mathbb{Z}_{n}^{\times}=U(n)\leq \mathbb{Z}_{n}$, and since such a group is abelian and finite, $U(n)$ is abelian and finite by inheritance.

Then I applied the fundamental theorem of abelian finite groups, where $20=2^{2}\cdot 5$. Therefore, a product would be $U(20)=\mathbb{Z}_{4}\times \mathbb{Z}_{5}$. Is this product correct?

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You are right in saying that $U(n)$ is abelian, but for the wrong reasons: note that $\mathbb{Z}_n$ is a group with respect to $+$ whereas $U(n)$ is a group with respect to the product. In any case, $\mathbb{Z}_n$ is a commutative ring and so its group of unities will be an abelian group.

Your observation about the structure theorem is also somewhat correct, but not fully. Understanding $|U(n)|$ will give you useful information. Note that, in general,

$$ |U(n)| = |\{k : \operatorname{lcd}(k,n) = 1\}| = \varphi(n) $$

with $\varphi$ being Euler's function. In particular, $U(20)$ does not have twenty elements, it has eight. Thus, it is a $2$-group and by the structure theorem on abelian groups, it is one of the following:

$$ \mathbb{Z}_8 \ ; \ \mathbb{Z}_4 \oplus \mathbb{Z}_2 \ ; \ \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2. $$

Now, note that for example $3^2 = 9 \not \equiv 1 \pmod{20}$ and so not every element in $U(20)$ has order two. That discards $(\mathbb{Z}_2)^3$ because all of its elements have order two. In the same fashion, you can check that all elements are annihilated by $4$, i.e. $g^4 = 1$ for any $g \in U(20)$. Thus, there are no elements of order $8$ and in consequence,

$$ U(20) \simeq \mathbb{Z}_4 \oplus \mathbb{Z}_2. $$

Another way of seeing this is that given $x \in U(20)$,

$$ x^4 \equiv 1 \pmod{20} $$

corresponds by the Chinese remainder theorem to a solution of

$$ \cases{x^4 \equiv 1 \pmod{4} \\ x^4 \equiv 1 \pmod{5}} $$

and this is always the case: you can check by hand the case modulo $4$, and the other is a consequence of Fermat's little theorem.