Cayley's Theorem (Simple) I don't really like to ask questions where i don't understand whats going on at all, but i just can seem to wrap my head around Cayley's Theorem, we went over it in class and i also watched a YouTube lecture that had a proof of theorem in it. I can't quite wrap my head around it though.
A question in my textbook is as follows  
Apply Cayley's Theorem to the group $U_{12}$. Write an explicit group
isomorphism from this group to a specic set of permutations of $U_{12}$.
Where $U_{12}$ is $\mathbb Z_{12}$ with all the stuff that does not have a multiplicative inverse thrown out.
Can someone Translate what this says to English and then perhaps back to math in a simpler way perhaps?
 A: I'll do the whole exercise step-by-step as illustration, but for $U(8)$ instead.


*

*Find the elements of $U(8)=({\bf Z}/8{\bf Z})^\times$: $\{1,3,5,7\}$ (technically, their equivalence classes).

*For each element in $a\in U(8)$, check what left-translation* does to the group's elements: $$\begin{array}{|c|c|c|c|c|} \hline a & x & 1 & 3 & 5 & 7 \\ \hline 1 & 1\cdot x & 1 & 3 & 5 & 7 \\ \hline 3 & 3\cdot x & 3 & 1 & 7 & 5 \\ \hline 5 & 5\cdot x & 5 & 7 & 1 & 3 \\ \hline 7 & 7\cdot x & 7 & 5 & 3 & 1 \\ \hline \end{array}$$

*Let $\psi:U(8)\to {\rm Sym}(U(8))$ be our desired homomorphism from $U(8)$ into the group of set-theoretic permutations of the elements of $U(8)$. Then for each $a\in U(8)$, $\psi(a)\in {\rm Sym}$ is the permutation $x\mapsto ax$. Explicitly, for example, in two-line notation, $$\psi(5)=\begin{pmatrix} 1 & 3 & 5 & 7 \\ 5 & 7 & 1 & 3\end{pmatrix},$$ simply by reading off the second-to-last row in the table I made.


Of course, whatever notation and level of explicitness the author expects of you is your duty to find out yourself, but this is the idea.
*By left translation we mean the map $x\mapsto ax$ under the group operation. In an abelian group, left and right translation are in fact the same.
A: Consider the group $U_{12}$ which has $\phi(12)=4$ elements. Lets call them 
$x_1,x_2,x_3,x_4$. You know the four elements, you'll need to write them explicitly.
Now Cayley theorem says that each one of the four elements defines a permutation:
$$f_i(x_j)=x_ix_j \,.$$
The problem asks you to explicitly write out the four elements, and calculate the four functions...
A: A group is essentially a collection of invertible transformations you can do on something. Each element is a transformation, and the group law is composition. Cayley's theorem is basically what rigorously justifies this idea: groups can be seen as a collection of ways of shuffling objects around, which means they should act like a subgroup of some symmetric group (i.e. fundamentally all groups are just permuting things). So they want you to note a couple things:
-$U_{12}$ is isomorphic to some permutation group
-Each element of $U_{12}$ is then a way of shuffling objects around.
-this means that you can look at multiplication in $U_{12}$ as a way of shuffling its own elements around! So it should match up with a subgroup of the permutation group on $4$ elements.
As an example: for any $g\in U_{12}$, $1g=g$. So $1$ represents the identity permutation. For others you may need to be more explicit: $11\cdot 1=11,\ \ 11\cdot 5=7,\ \ 11\cdot 7=5,\ \ 11\cdot 11=1$. So $11$ interchanges the pairs $(1,11)$ and $(5,7)$. That can be viewed as an element in a permutation group as well. The idea is to find a specific subgroup of the permutation group that captures the behaviour of $U_{12}$ in this way, by matching each element of $U_{12}$ with a certain type of permutation.
Overall goal: match each element of $U_{12}$ to an element in the permutation group on four elements, in such a way that this permutation subgroup has the same group structure as $U_{12}$.
