TITLE: Explain why $U_{44} \cong (\mathbb{Z}_{10} \oplus \mathbb{Z}_2) $.
QUESTION [6 upvotes]: Explain why $\newcommand{\Z}{\mathbb{Z}} U_{44} \cong (\Z_{10} \oplus \Z_2) $ 
I know that $\Z_{20} \cong (\Z_{10} \oplus \Z_2)$, so if I can show $U_{44}   \cong \Z_{20}$, then I can conclude that $U_{44} \cong (\Z_{10} \oplus \Z_2) $ since isomorphism is transitive.
In order to show $U_{44} \cong \Z_{20}$, I need to show that the function $ \varphi: U_{44} \to \Z_{20}$ is bijective and structure preserved.  I know that $U_{44} \text{ and } \Z_{20}$ have order $20$, so the function can be bijective, but it's not enough to show $\varphi: U_{44} \to \Z_{20} $ is bijective.
I wonder if anyone can show me how to show $ \varphi: U_{44} \to \Z_{20} $ is bijective and structure preserving.

REPLY [7 votes]: Be careful: $$\quad \mathbb Z_{10} \times \mathbb Z_2 \not\cong \mathbb Z_{20}; \quad\text{since}\;\gcd(2, 10) = 2\neq 1$$
Use the facts that 

$U(44)= U(4\cdot 11)\cong U(4)\times U(11)$, since $4$ and $11$ are relatively prime.
$U(4)\cong \mathbb Z_2$ and $U(11)\cong \mathbb Z_{10}$. 

To conclude that $$U(44) \cong \mathbb Z_{2}\times \mathbb Z_{10}$$