Cayley's Theorem - Discourse on Fraleigh's proof or Alternative Easier proof I have hard time understanding Fraleigh's proof. Can someone either explain Fraleigh's or provide an alternative, perhaps easier proof?
Thanks so much.
 A: Let $n=|G|$.  Think about the elements of $S_n$ - these are permutations of the numbers $1$ to $n$.  Now let's arbitrarily number the elements of $G$ so that $G=\{g_1,g_2,\ldots, g_n\}$.  Now, we could talk about the permutations in $S_n$ being applied to these elements of $G$, correct?  For example, $(123)$ would permute $g_1$ and $g_2$ and $g_3$ accordingly.  We call this "letting $S_n$ act on $G$."
Cayley's theorem does exactly what we just did, only backwards.  He starts by labeling $G=\{g_1,g_2,\ldots, g_n\}$.  Then he picks an element $x\in G$ and multiplies everything in $G$ on the left by $x$.  Doing so rearranges the elements in $G$ in some way from their original order.  He then finds the associated permutation in $S_n$, which he calls $\theta_x$.

Example. Let $|G|=3$ and suppose you start with elements $g_1,g_2,g_3$ (in that order).  Then, suppose you multiply everything by some $x\in G$, and you end up with $g_2,g_3,g_1$ (in that order).  Then $x$ will be associated with the permutation $\theta_x=(123)$ in $S_3$.

Once he finds this with all the $x\in G$, he looks at the subset $\{\theta_x:x\in G\}\subseteq S_n$.  He then proves that that set is a subgroup, and in fact is isomorphic to $G$.
So, to sum everything up, Cayley's theorem simply shows that the permutations of a $G$'s elements induced by left multiplication by each element $x\in G$ form a group which itself isomorphic to $G$.  Read the above explanation a few times and I think you'll see.
A: It's basically all about that the maps $g\cdot:G\to G$ (sending $x\mapsto gx$) are distinct permutations, that is, $(g\cdot)\in Sym(G)$, moreover this assignment $G\to Sym(G)$, sending $g\mapsto (g\cdot)$ is an injective group homomorphism.
