Isomorphism between $U_{10}$ , $U_8$ , $U_5$

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 ?

Solution 1:

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.

Solution 2:

In a cyclic group, there is at most one element of order $2$: because a cyclic group of order $n$ has exactly one subgroup of order $d$ for each $d$ that divides $n$; if $2|n$, then there is exactly one subgroup of order $2$. But distinct elements of order $2$ generate distinct subgroups of order $2$.

Since $-1$ is always of order $2$ in $U_n$, we already know that $7$ is of order $2$ in $U_8$. To verify that $U_8$ is not cyclic, it is enough to note that $3^2\equiv 1\pmod{8}$. Thus, $U_8$ has at least two elements of order $2$, and so cannot be cyclic.

As an aside: it is incorrect to talk about the generator of a cyclic group, unless the group has order $2$; every cyclic group of order $n\gt 2$ has many generators; in fact, it has $\varphi(n)$ generators. So $3$ is a generator of $U_{10}$, but not the only one ($7$ also generates); $2$ is a generator of $U_5$, but it is not the only one ($3$ also generates).

Solution 3:

In $U_8$, $3^2=1$ and $5^2=1$, also $5\cdot 3= 7$.

So we can see that $$U_8= \langle 3\rangle\times \langle 5\rangle$$ that is isomorphic to $\mathbb Z_2\times \mathbb Z_2$.

So $U_8$ is not isomorphic with others... as $U_5$ and $U_{10}$ are cyclic group of same order. $U_5$ and $U_{10}$ are isomorphic by the last argument.