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