Can somebody explain Cayley's theorem to me? I need some help understanding his theorem that every group is isomorphic to a group of permutations.
I understand what isomorphism means but I'm not very clear with the idea of a group of permutation. I mean, I understand a given set's elements can be permuted to get a permutation of the set. But, how can there be a group who's elements are permutations ? What would the operation be ?
My book says that in the younger days of abstract algebra, a group meant a group of permutation before its definition got expanded and that is the reason this theorem is so important.
As far as the proof goes, I understand what why the function (from any given group to a group of permutation) would be invective and surjective but not why it obeys $f(a).f(b) = f(a*b)$.
 A: The group operation in a "group of permutations" (meaning a subgroup of a symmetric group) is just composition of functions. If you have two permutations $f$ and $g$ of some set, the product $fg$ is just $f \circ g$. That is, you perform the permutation $g$, then perform the permutation $f$.
Cayley's Theorem is saying that all groups can be thought of in this way. Although a group is defined abstractly, this theorem says that it can be viewed as a set of permutations, where the group multiplication represents composing the permutations.
The proof of Cayley's Theorem is essentially to just look at a group $G$ acting on itself by multiplication. Then each $g \in G$ can be thought of as a permutation of $G$, where the permutation is just multiplying all the group elements by $G$. More technically, if $\mathrm{Sym}(G)$ is the set of all permutations of $G$ (as a set), then you get a map $\varphi \colon G \to \mathrm{Sym}(G)$ where $\varphi(g)$ is a permutation defined as follows: for each $h \in G$,
$$ \varphi(g)(h) = gh $$
This is a group homomorphism, so the image $\varphi(G)$ (which is isomorphic to $G$ itself, since $\varphi$ is injective) is a subgroup of $\mathrm{Sym}(G)$.
A: What does a symmetry mean but some transformation you can do behind somebodies back without them noticing the difference when they turn back?
If a put a bunch of blocks on a table in a line, then there is a group of symmetries where the elements are built swapping two of the blocks at a time. The composition of two sequences of swaps is just a longer sequence of swaps, i.e. concatenation. Clearly a sequence of swaps can be undone, and there is the empty sequence of swaps (do nothing), which is the identity element under this concatenation operation. So the swaps generate a group, the group of permutations.
This is the same as the group of all bijections from the set ${1, \ldots, n}\$ to itself, where the operation is composition of functions. (I am sweeping under the rug the true fact that any rearrangement of the blocks can be achieved by swapping only two at a time. But this should be intuitive... you only have two hands, after all.)
If your group has order $n$, this group is isomorphic to the set of all bijections from the SET $G$ to itself (because $G$ and $\{1, \ldots, n\}$ are isomorphic / bijective sets). But each element in $g$ induces in a natural way a bijection from $G$ to itself, namely the multiplication by $g$ on the left map: $x \to gx$. But the multiplication by $gh$ map is the same as the multiplying first by $h$, and then by $g$, which is the same as the group law in the group of bijections. It is straightforward to check that the only group element acting as the identity under translation is the identity (hint: you can recover $g$ by looking at the image of the identity under the multiply by $g$ on the left map).
A: We recall Cayley's Theorem and its proof.

Theorem: Let $G$ be a finite group of order $n$. Then $G$ is isomorphic to a subgroup of $S_n$.

Proof:
The group $G$ acts on $n$ objects, namely its elements, by the rule $g \cdot h = gh$. It is clear that this action is faithful, for if $g \cdot h = h$ for all $h$, then $g = e$ by group axioms. Thus the action gives rise to an injective homomorphism $\phi : G \to S_n$. By the First Isomorphism Theorem,
$$G = G/\ker\phi \simeq \operatorname{im}\phi \leq S_n,$$
as desired.
What is happening here? A group action, roughly speaking, is a way to represent the symmetries of the object. Group actions from a group $G$ to a (finite) set $A$ are in a one-to-one correspondence with homomorphisms from $G$ to $S_n$. This is because for every $g \in G$, we can define a map $\sigma_g : A \to A$ by
$$\phi_g(a) = g \cdot a.$$
It can be shown easily that $\phi_g$ is a bijection from $A$ to $A$, and so there is a map $G \to S_n$ given by
$$\phi(g) = \sigma_g.$$
It can be shown that this gives rise to every $S_n$. Conversely, every homomorphism from $G$ to $S_n$ represents a group action, which is exactly the action described above. 
In the proof of Cayley's theorem, we are choosing a set of $n$ objects to act on and extracting the homomorphism. This homomorphism happens to be injective, so the First Isomorphism Theorem shows that $G$ can be embedded within $S_n$, with elements "acting" as permutations.
