Cayley's Theorem - Questions on Proof Blueprint [Fraleigh p. 82 theorem 8.16] Not a duplicate of this exquisite answer, the numbers in which I abide by here. Not querying the proof, hence please don't discourse on it. Proof blueprint:

Steps 1-2 in words. Left multiplication of every element in the group by a fixed element in the group constitutes a permutation of those elements. 
Steps 1-2 in math. We must prove $\lambda_x(g) = xg$ for all $g, x \in G$ is a bijection $: G \to G$.
  Or see Fraleigh p. 87 exercise 8.52: $p_x(g) = gx$ operates too.
Steps 3-4 in words. As the fixed elements used as multipliers $\color{brown}{\text{range over all elements in the group}}$, we get a family of permutations, one for each multiplier. The idea is to show that this family forms a group with the same fundamental structure as the original group.
Steps 3-4 in math. We must prove $\theta(x) = \lambda_x$ for all $x \in G$ is an injective homomorphism $: G \to S_G$. Or see Fraleigh p. 87 exercise 8.52: $u(x) = p_{x^{=1}}$ operates too.

Question (1.) Why left-multiply the elements in $g$ by $x$ in $\lambda_x(g) = xg$? To induce a permutation? 
(2.) Why does $\lambda_x(g) = xg$ only need to be a bijection? Why not an isomorphism?
(3.) I condone $\phi \to Im(\phi)$ is a surjection because image = codomain, but still don't understand why Fraleigh doesn't prove  $\theta$  is surjective? Related to this?   
(4.) Why do we need $\theta(x) = \lambda_x$? It feels redundant. $\lambda_x(g) = xg$ is true for all $x \in G$, hence doesn't $\lambda_x(g) = xg$ $\color{brown}{\text{range over all elements in the group}}$ already ?  My course doesn't cover Qiaochu Yuan's comment on currying on top of exquisite answer.
 A: Cayley's Theorem says that every group is isomorphic to a permutation group, that is, a subgroup of $S_n$ for some $n$.
1)  Yes.  We need to realize the elements of $G$ as permutations.  This is an example of the left regular representation.  That's why the permutation is denoted by $\lambda_x$: lambda is the Greek equivalent of "L", which stands for "left."  (Also, that almost certainly a $\rho_x$ for the right regular representation, not a $p_x$, since rho is the equivalent of "R", which stands for "right.")
2)  As I said above, we need to realize each element of $G$ as a permutation.  And a permutation is simply an bijection $G \to G$, not necessarily an isomorphism.
3)  In fact $\theta : G \to S_n$ is not surjective in general.  It only maps $G$ onto $S_n$ if $G$ happens to be isomorphic to $S_n$.  In general, $G$ is only isomorphic to a subgroup of $S_n$.  Of course $\theta$ maps onto its image---that's tautologically true.
4)  I think you're getting the maps mixed up.  Given $x \in G$, we want to find a permutation that is "the same" as $x$.  To do so, we examine how $x$ acts on $G$ by left multiplication by defining the map
\begin{align*}
\lambda_x : G &\to G\\
g & \mapsto xg \, .
\end{align*}
Just to emphasize, for each $x$, $\lambda_x$ is a map from $G$ to itself.  (In fact, $\lambda_x$ is a bijection since its inverse is just $\lambda_{x^{-1}}$.)  We have not yet involved a symmetric group.  Finally, we define
\begin{align*}
\theta : G &\to S_G\\
x &\mapsto \lambda_x
\end{align*}
where $S_G$ is the symmetric group on $G$, i.e., the set of all bijections of $G$ with the binary operation of functional composition.  That is, to each element $x \in G$ we associate the permutation of $G$ that is simply left multiplication by $x$.
Once we prove that $\theta$ is an injective homomorphism, then $\theta$ is an isomorphism onto its image, which is a subgroup of $S_G$.  This shows that $G$ is isomorphic to a subgroup of a permutation group.
Does that answer all your questions?
