TITLE: Can someone explain Cayley's Theorem step by step?
QUESTION [53 upvotes]: This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:

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, $\lambda_{xy} = \lambda_x \lambda_y$.

REPLY [6 votes]: For each $x\in G$, define function $\lambda_x: G \to G$ such that $\lambda_x(g)=x\cdot g$ ($\cdot$ here means the group operation),  for all $g\in G$. Since $G$ is a group, it is closed under the group operation. So, $\lambda_x$ maps $G$ to $G$, which means that $\lambda_x$ is well-defined. 
We also know that $x^{-1}\in G$. Thus, $\forall c\in G$, $x^{-1}\cdot c\in G.$ By our definition, $\lambda_x(x^{-1}\cdot c)=x\cdot (x^{-1}\cdot c)=(x\cdot x^{-1})\cdot c=c$. This means that for each $c\in G$, there exists $u\in G$ (with $u=x^{-1}\cdot c$) such that $\lambda_x(u)=c$, i.e. $\lambda_x$ is surjective. 
Then we show that $\lambda_x$ is injective. 
For $a,b\in G$, suppose $\lambda_x(a)=\lambda_x(b)$, i.e. $x\cdot a = x\cdot b$. Then we have $x^{-1}\cdot x\cdot a=x^{-1}\cdot x\cdot b \implies a=b$. Thus, $\lambda_x$ is injective.
Now that we've shown that $\lambda_x$ is bijective, it is a permutation of $G$. Thus, $\lambda_x\in S_G$. We now define $\phi: G\to S_G$ by defining $\phi(x)=\lambda_x$ for all $x\in G$.
What we want to show next is that $\phi$ is injective. Once this is proven, we know that $\phi: G\to \phi(G)$ is bijective. To show that $\phi$ is injective, suppose $x,y\in G$ and $\phi(x)=\phi(y)$. Then, $\lambda_x=\lambda_y$, meaning that for all $c\in G$, $\lambda_x(c)=\lambda_y(c)$, i.e. $x\cdot c = y\cdot c.$ Let $c=e$. Then, $x\cdot e=y\cdot e$, so $x=y$. Thus, $\phi$ is injective.
For $\lambda_x,\lambda_y\in \phi(G)$, $\lambda_x\cdot\lambda_y\in \phi(G)$ because $\lambda_x\cdot\lambda_y(c)=x\cdot y\cdot c=(x\cdot y)\cdot c=\lambda_{x\cdot y} c$. Also, observe that $\lambda_{-x}\in \phi(G)$. Thus, $\phi(G)$ is a subgroup of $S_G$, 
Lastly, we only need to show that $G$ is isomorphic to $\phi(G)$. This requires $\phi(x\cdot y)=\phi(x)\cdot \phi(y)$. 
Note that $\phi(x\cdot y)=\lambda_{xy}$ and $\phi(x)\cdot\phi(y)=\lambda_x\cdot\lambda_y.$ By our definition, for any $c\in G$, we have $\lambda_{x\cdot y}(c)=(x\cdot y)\cdot c=x\cdot (y\cdot c)=\lambda_x(\lambda_y(c))=\lambda_x\cdot \lambda_y(c)$. Thus, $\phi(x\cdot y)=\phi(x)\cdot \phi(y)$.