5. Show that $$ U(5) $$ is isomorphic to $$ U(10) $$, but $$ U(12) $$ is not.

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1. **Determine the Units Groups**

- $$ U(5) = \{ [1]_5, [2]_5, [3]_5, [4]_5 \} $$ since $$5$$ is prime.
   - $$ U(10) = \{ [1]_{10}, [3]_{10}, [7]_{10}, [9]_{10} \} $$ because these are the numbers less than $$10$$ and relatively prime to $$10$$.
   - $$ U(12) = \{ [1]_{12}, [5]_{12}, [7]_{12}, [11]_{12} \} $$ as these are the numbers less than $$12$$ and coprime to $$12$$.

2. **Examine the Group Structure**

- **For $$ U(5) $$:**  
     Since $$5$$ is prime, $$ U(5) $$ is cyclic with order $$4$$. For example, the element $$[2]_5$$ is a generator since:
     $$
     [2]^1 = [2]_5,\quad [2]^2 = [4]_5,\quad [2]^3 = [3]_5,\quad [2]^4 = [1]_5.
     $$
     
   - **For $$ U(10) $$:**  
     We check if it is cyclic. Consider the element $$[3]_{10}$$:
     $$
     [3]^1 = [3]_{10},\quad [3]^2 = [9]_{10},\quad [3]^3 = [27]_{10} = [7]_{10},\quad [3]^4 = [81]_{10} = [1]_{10}.
     $$
     Hence, $$[3]_{10}$$ is of order $$4$$, so $$ U(10) $$ is cyclic and isomorphic to $$ U(5) $$.

- **For $$ U(12) $$:**  
     Let’s check the orders of non-identity elements:
     
     - $$[5]_{12}$$:  
       $$
       [5]^2 = [25]_{12} = [1]_{12}.
       $$
     - $$[7]_{12}$$:  
       $$
       [7]^2 = [49]_{12} = [1]_{12}.
       $$
     - $$[11]_{12}$$:  
       $$
       [11]^2 = [121]_{12} = [1]_{12}.
       $$
       
     Each non-identity element has order $$2$$ which implies that $$ U(12) $$ is isomorphic to the Klein four-group and is not cyclic.

3. **Establishing the Isomorphism Between $$ U(5) $$ and $$ U(10) $$**

Define a map 
   $$
   \phi : U(10) \to U(5) \quad \text{by} \quad \phi([a]_{10}) = [a]_5.
   $$
   - **Well-defined:**  
     If $$[a]_{10} = [b]_{10}$$, then $$a \equiv b \ (\bmod\ 10)$$. Consequently, $$a \equiv b \ (\bmod\ 5)$$ so that $$\phi([a]_{10}) = \phi([b]_{10})$$.
   
   - **Homomorphism:**  
     For any $$[a]_{10}, [b]_{10} \in U(10)$$,
     $$
     \phi([a]_{10}[b]_{10}) = \phi([ab]_{10}) = [ab]_5 = [a]_5 [b]_5 = \phi([a]_{10})\phi([b]_{10}).
     $$
   
   - **Bijectivity:**  
     Both $$ U(10) $$ and $$ U(5) $$ have $$4$$ elements. Hence, an onto homomorphism between groups of the same finite order is also one-to-one.
     
   Thus, $$\phi$$ is an isomorphism, showing that 
   $$
   U(5) \cong U(10).
   $$

4. **Showing $$ U(12) $$ is Not Isomorphic to $$ U(5) $$**

- $$ U(5) $$ is cyclic of order $$4$$; it has an element of order $$4$$ (for instance, $$[2]_5$$).
   - In contrast, every non-identity element in $$ U(12) $$ has order $$2$$. Therefore, $$ U(12) $$ is not cyclic.
     
   Since group isomorphisms preserve element orders, a cyclic group of order $$4$$ cannot be isomorphic to a non-cyclic group where every element has order dividing $$2$$. Thus, 
   $$
   U(12) \not\cong U(5) \quad \text{(and also not isomorphic to } U(10)\text{)}.
   $$

5. **Conclusion**

$$
   U(5) \cong U(10)\quad\text{(both are cyclic of order $$4$$)}, \quad \text{but} \quad U(12) \not\cong U(5)\quad \text{(since $$ U(12) $$ is not cyclic)}.
   $$
$$ U(5) $$ is isomorphic to $$ U(10) $$ because both are cyclic groups of order 4. However, $$ U(12) $$ is not isomorphic to $$ U(5) $$ because $$ U(12) $$ is not cyclic.

Show that is isomorphic to , but is not.

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