Identity element as product of transposition. My lecture notes indicates that the identity element of a symmetric group $S_{n}$ is 
$\left (  \right )=\left ( 12 \right )\left ( 12 \right )$
Just to state:

The identity element $b \in G$ where G is a group is the element b such that 
  $\forall a,b \in G$ 
  $a*b=b*a=a$

In a permutation group, however, we speak of elements as being permutations. 
And of course, by a certain theorem, any permutation of a finite set can be written as a cycle or as a product of disjoint cycles 
giving 
$\left ( a_{1}a_{2}\cdot \cdot \cdot a_{m} \right )\left ( b_{1}b_{2}\cdot \cdot \cdot b_{k} \right )\cdot \cdot \cdot $
In short, I am perhaps confused with how $\left (  \right )=\left ( 12 \right )\left ( 12 \right )$. In part because the notation looks strange.
 A: Some reminders about permutations:
A permutation is by definition a bijective function from a set to itself.  The set of all permutations of $\{1,2,3,\dots,n\}$ (along with the operation of function composition) is referred to as the symmetric group $S_n$.
For example with $n=4$ we could have as an example of a permutation: $f=\{(1,2),(2,1),(3,3),(4,4)\}$, in other words $f(1)=2, f(2)=1, f(3)=3, f(4)=4$.
To simplify notation, we can refer to this in two-line format:  $f=\begin{pmatrix} 1&2&3&4\\f(1)&f(2)&f(3)&f(4)\end{pmatrix} = \begin{pmatrix}1&2&3&4\\2&1&3&4\end{pmatrix}$
One can also prefer to write this in cyclic notation as well by "following the bouncing ball," keeping track of where one element gets mapped under repeated applications of the permutation until arriving back where it started.  In the above example, $f=(2~1)$
The identity permutation is the identity function $e(x)=x$.  In the context of $S_4$, that would be $e=\{(1,1),(2,2),(3,3),(4,4)\}$ or equivalently as $\begin{pmatrix}1&2&3&4\\1&2&3&4\end{pmatrix}$ or equivalently as $(1)$ or $(~)$ depending on your preferred notation.
That the identity permutation is indeed the identity for the group $S_n$ is immediate from how it is defined since:
$(e\circ f)(x)=e(f(x))=f(x)$ for all $x$, so $e\circ f = f$.  Also $(f\circ e)(x)=f(e(x))=f(x)$ for all $x$, so $f\circ e = f$.
That the identity permutation can be written as the product of two equal transpositions follows from the fact that transpositions are self inverses.
Using the above example of $f=(2~1)$ again, we have:
$(f\circ f)(x)=f(f(x))=\begin{cases} f(f(1))&\text{if}~x=1\\ f(f(2))&\text{if}~x=2\\ f(f(x))&\text{for all other}~x\end{cases}=\begin{cases} f(2)&\text{if}~x=1\\ f(1)&\text{if}~x=2\\ f(x)&\text{for all other}~x\end{cases}=\begin{cases} 1&\text{if}~x=1\\ 2&\text{if}~x=2\\ x&\text{for all other}~x\end{cases}=x$ for all $x$.
Thus $f\circ f=(2~1)(2~1)=e$
