For $G =\mathbb Z_{10}^{\times} = \{[1], [3], [7], [9]\}$, use cycle notation to write out the permutation determined by each element $a\in G$ 
For $G =\mathbb Z_{10}^{\times} = \{[1], [3], [7], [9]\}$, use cycle notation to write out the permutation determined by each element $a\in G$ as in the proof of Cayley's Theorem.

Frankly, I don't even understand the question. What is "the permutation defined by each element?" I looked up some proofs of Cayley's theorem and it seemed irrelevant. And I am very tired... I would appreciate some guidance before I am chasing my tail here? :) thanks in advance.
 A: I will show you one possibility: the left-regular representation of $G$ over itself.
To each element $g \in G$ we will assign a bijection $G \to G$. This bijection will be, then, a permutation of the elements of $G$. I will call each such bijection $L_g$ to indicate its dependence on $g$, and it will permute the set $G = \{[1],[3],[7],[9]\}$.
Specifically, we set $L_g(x) = gx$ that is, we "left-multiply" by $g$. Here is how this plays out in the case of $g = [3]$:
$L_{[3]}([1]) = [3][1] = [3]\\L_{[3]}([3]) = [3][3] = [9]\\L_{[3]}([7]) = [3][7] = [1]\\L_{[3]}([9]) = [3][9] = [7].$
Evidently, then, this is the permutation $([1]\to[3]\to[9]\to[7])$ (and, of course, $[7]$ back to $[1]$), a $4$-cycle.
Moreover, the mapping $g \mapsto L_g$ which takes $G$ into (but not onto) $\text{Sym}(G)$ (considering the latter $G$ as merely a set), is actually a group homomorphism; that is, for $g,h \in G$:
$L_{gh} = L_g \circ L_h$ (since $(gh)x = g(hx)$ by the associative law, for any $x \in G$).
If you see that $g \mapsto L_g$ is also injective (because $gx = hx \implies g = h$, for any $x \in G$, so if $L_g = L_h$ then $g = h$ must be the case), you can see then we have a "copy" of $G$ hidden inside the group of permutations of four things (a beast usually referred to as $S_4$ since the "names" of the four things isn't that important for group theory purposes).
