I'm trying to decide which groups are isomorphic to one another .
$$\begin{align*}
U_5 &= \{1,2,3,4\}\\
U_8 &= \{1,3,5,7\}\\
U_{10} &= \{1,3,7,9\}
\end{align*}$$
I've checked and verified that $U_{10}$ and $U_5$ are cyclic , by finding the creator generator of each group: The creator generator of $U_5$ is $2$ and the creator generator of $U_{10}$ is $3$).
The question is, how can I verify that $U_8$ is not cyclic ? Is there a fast way to determine that, without checking if one of the elements creates generates all other elements in the group ?
Best Answer
First of all in algebra people tend not to call them "creators", we tend to say generators if I am correct. Secondly, to check a group is not cyclic you would have to verify that none of its members generates the entire group. In your case it can be done extremely easy by inspection: Note that $$\langle 3\rangle =\langle 3,3^2=1\rangle$$ and the same hold for $5$ and $7$.
To create an isomorphism just map the generator of one of your groups to the generator of the other. That is, $\varphi:U_5\rightarrow U_{10}$ will be defined as $\varphi(2)=3$, everything else will be determined from here:$$\varphi(4)=\varphi(2\cdot 2)=\varphi(2)\varphi(2)=3\cdot 3=9$$$$\varphi(3)=\varphi(2\cdot 4)=\varphi(2)\varphi(4)=3\cdot 9=2$$$$\varphi(1)=\varphi(4\cdot 4)=\varphi(4)\varphi(4)=9\cdot 9=1$$ Note the last line was not even necessary since in a homomorphism identities are mapped to identities.
Hope this helps.