How to determine whether an isomorphism $\varphi: {U_{12}} \to U_5$ exists? I have 2 groups $U_5$ and $U_{12}$ , .. 
$U_5 = \{1,2,3,4\}, U_{12} = \{1,5,7,11\}$. 
I have to determine whether an isomorphism $\varphi: {U_{12}} \to U_5$ exists.
I started with the "$yes$" case: there is an isomorphism.
So I searched an isomorphism $\varphi$ , but I didn't found. So I guess there is no an isomorphism $\varphi$  . 
How can I prove it? or at least explain? please help.
 A: I don't know how much algebra you know but notice that $U_5$ is cyclic, with generator, say,  $[2]$. Indeed for any prime $p$, $u_p$ is going to be cyclic of order $p-1$.
Now look at $U_{12}$, is it cyclic? A rather tedious check tells you that $1^2,5^2,7^2,11^2$ all equal $1$ mod $12$ and so this group is not cyclic. In fact from this you can derive that $U_{12}$ is isomorphic to $C_2 \times C_2$. Now as isomorphic groups are either both cyclic or both not cyclic (as isomorphisms preserve the order of elements), these two groups are not isomorphic.
A: Note that $x^2\equiv 1\pmod {12}$ for all elements of $U_{12}$ whereas the corresponding property does not hold in $U_5$.
A: In $U_{12}$ everybpdy is his own inverse. This is not the case in $U_5$.
Along the same lines, $2$ is a generator of $U_5$, while $U_{12}$ has no generator. Since one group is cyclic and the other not, they cannot be isomorphic.
The same fact, restated: $U_{5}$ has an element of order $4$, while all elements of $U_{12}$ have order $1$ or $2$.
Remark: If you looked for an isomorphism, testing all viable possibilities, and nothing worked, a complete record of the testing is a proof of non-isomorphism.
However, a more cxommon way of showing non-isomorphism of two groups $A$ and $B$ is by using structural properties.
If you have a property $P$ that must be preserved by isomorphism, and $A$ has property $P$, but $B$ doesn't, then you know $A$ and $B$ cannot be isomorphic.
