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I'm trying to understand proof of Cayley's theorem from Fraleigh's "A First Course In Abstract Algebra" p.82. I have read this post, but there are still a few things that I don't understand.

Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$.

Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$.

To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$.

1). When we prove that $\phi$ is one to one why we only consider the case when argument of $\lambda_x$ is $e$? Why don't we consider all cases? Of course there is nothing special, for example: $\lambda_x (a) = \lambda_y (a)$, so $xa=ya$ and again $x=y$, but still...

2). Can someone prove that $\phi$ is onto? Before the theorem there is a lemma:

Let $G$ and $G'$ be groups and let $\phi : G \rightarrow G'$ be a one-to-one function such that $\phi(xy) = \phi(x) \phi(y)$ for all $x, y \in G$. Then $\phi[G]$ is a subgroup of $G'$ and $\phi$ provides an isomorphism of $G$ with $\phi[G]$.

This lemma is being used in proof, but I don't really understand it. Is there any way to prove the theorem without using this lemma? I think we only need to prove that $\phi$ is onto.

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Let's examine Cayley's Theorem at a high level. The statement of the theorem is that every group is isomorphic to some subgroup of a Symmetry group ($\text{Sym}(G)$ in particular). The proof here actually tells you that when $G$ acts on itself by left multiplication, there exists an injective homomorphism $\phi : G \to \text{Sym}(G)$. The $\lambda_{x}$ functions are the permutations associated with $G$. That is, the action of $G$ on itself by left multiplication simply permutes the elements of $G$. Thinking of actions as permutations is a very helpful and useful mentality.

The map of $x \mapsto \lambda_{x}$ is injective by the cancellation property of groups. Suppose $\lambda_{x} = \lambda_{y}$. Then $xg = yg$ for any $g \in G$. If $g \neq e$, we simply multiply both sides by $g^{-1}$ to obtain that $x = y$. Considering $xe = ye$ makes life easier.

Note that $\phi$ is usually not surjective. The lemma you are quoting is the First Isomorphism Theorem in disguise. Do you see why $\phi[G] \leq G^{\prime}$? Apply the subgroup test. Is $e_{G^{\prime}} \in \phi[G]$? Is $\phi[G]$ closed under the operation of $G^{\prime}$? Is $\phi[G]$ closed under inverses? These properties all follow from the fact that $\phi$ is a homomorphism. Take some time to attempt this lemma. It's a good exercise, and the fact that $\phi[G] \leq G^{\prime}$ is very important.

In the proof of Cayley's Theorem, we use the fact that $G \cong \phi[G]$ to deduce that $G$ is indeed a subgroup of $\text{Sym}(G)$.

Note: Every group action exhibits a permutation representation. If we are careful in stating Cayley's Theorem, we can get this added result with really no extra work.

Theorem: Let $G$ be a group acting on the set $A$. Then there exists a homomorphism $\phi : G \to \text{Sym}(A)$.

Here, we map $\phi(x) = \lambda_{x}$ again, though this map need not be injective unless the action is faithful (this is precisely the definition of a faithful action). The First Isomorphism Theorem gives us that $G/\text{ker}(\phi) \cong \phi(G)$. When $\phi$ is injective, $\text{ker}(\phi) = \{e_{G}\}$.

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