Question

Show that${\mathit{U}}_5$ is isomorphic to ${\mathit{U}}_{\mathbf{10}}$.

Short answer

It has been proved that $U_5$is isomorphic to$U_{10}$

Step-by-step solution

Observe U5 and U10

See that$U_5=\left\{1,2,3,4\right\}$ is a cyclic group of order 4 generated by 2.

And,$U_{10}=\left\{1,3,7,9\right\}$ is a cyclic group with order 4 generated by 3.

Define a map from U5 to U10 

Define a map $\phi :U_5\to U_{10}$such that $\phi \left(2^k\right)=3^k$for some integer k.

This is a well-defined map.

We can be seen that,

$$\begin{gathered} \phi \left(1\right)=1 \\ \phi \left(2\right)=3 \\ \phi \left(3\right)=7 \\ \phi \left(4\right)=9 \end{gathered}$$

Hence, this a bijection map.

Prove that the map is isomorphic

Since the map is already a bijection, only homomorphism needs to be checked.

Let $k,m\in \mathrm{\mathbb{Z}}$, then

$\begin{array}{rcl}\phi \left(2^k{2}^m\right)& =& \phi \left(2^{k+m}\right)\\ & =& 3^{k+m}\\ & =& 3^k{3}^m\end{array}$

Therefore, the map is homomorphic.

Hence,$\phi$ is an isomorphism.

Conclusion

Hence, we conclude that$U_5$ is isomorphic to$U_{10}$