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

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.

• Actually, \begin{align*} \Bbb{Z}/20\Bbb{Z}&\cong\Bbb{Z}/4\Bbb{Z}\times\Bbb{Z}/5\Bbb{Z}\\ &\not\cong\Bbb{Z}/10\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}. \end{align*} ($\gcd(10,2) = 2\neq 1$) – Stahl Apr 18 '13 at 0:22

## 4 Answers

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

• In my answer, $\times$ denotes what you might know as the external direct product: $\otimes$ or in the case of $\mathbb Z_n \times \mathbb Z_m$ it denotes what some call the external direct sum: i.e., $\mathbb Z_{10} \otimes \mathbb Z_2 \not\cong \mathbb Z_{20},$ etc...with the conclusion that $U_{44} \cong \mathbb Z_2 \otimes \mathbb Z_{10}$. I apologize for any confusion notation may be causing! – Namaste Apr 18 '13 at 1:39
• don't worry, I understand it perfectly, thank you very much for your answer. – Diane Vanderwaif Apr 18 '13 at 2:09

Hint: List the elements and compute their order. From this it should be quite clear.

In general, for $n=p_1^{e_1}\cdots p_k^{e_k}$, $U_n\cong(\mathbb{Z}/p_1^{e_1}\mathbb{Z})^\times\times\cdots\times(\mathbb{Z}/p_k^{e_k}\mathbb{Z})^\times$ by the Chinese Remainder Theorem.

Note that $U(44)= U(11*4)\cong {\mathbb Z_{10}} \oplus {\mathbb Z_2}$, since $\gcd(11,4)=1.$

In general, if $\gcd(a,b)=1$, then $U(ab)\cong {\mathbb Z_a}\oplus{\mathbb Z_b}$