Explain why $U_{44} \cong (\mathbb{Z}_{10} \oplus \mathbb{Z}_2) $.

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