Why is a symmetric group a group If the identity axiom for groups is the following $ae = ea = a$ ($e$ the identity element) and the operation for a symmetric group is function composition then how does a symmetric group pass this test? 
Specifically suppose that we have two bijections of set $S = \{1,2,3\}$, $a = (2,1,3)$ and identity $e=(1,2,3)$, then the identity axiom is only working for function composition when the identity is the left function ($ea \ne ae$). 
Can anyone explain if I understand this correctly?
 A: Short answer:
$(1,2,3)$ is not the identity. The identity element of the symmetric group maps each $x$ to itself.
Slightly longer answer:
Following a variation of logic, we can easily prove that $\mathbb R$ is not a group for addition. For example, take two elements of $\mathbb R$, $a=10$ and identity $e=498123921$. Then the identity axiom is not working, because $a+e\neq a$.
Longer answer:
The mistake you made is that you did not correctly and completely write out the identity axiom for groups.
In particular, the axioms states the following:

If $G$ is a group, then there exists some $e\in G$ such that for every $a\in G$, we have $ae=ea=a$.

The axiom does not claim that you can choose what the identity of the group is. It states that the identity of a group is an element that must satisfy a particular condition.
It is fairly easy to show that the permutation $e$ for which $e(1)=1, e(2)=2$ and $e(3)=3$ satisfies the condition and is therefore the identity of $S_3$.
A: If you mean $e = (1,2,3)$ to mean the function that maps $(1,2,3)$ to $(1,2,3)$ and $a = (2,1,3)$ to mean the function that maps $(1,2,3)$ to $(2,1,3)$, then it is certainly true that $ae = ea = a$. Just try it on each element. $ae$ maps $(1,2,3) \rightarrow (1,2,3) \rightarrow (2,1,3)$ and $ea$ maps $(1,2,3)\rightarrow (2,1,3)\rightarrow (2,1,3)$.
If you are instead using cycle notation, then the identity element is not $e = (1\;2\;3)$, but $e = (1)(2)(3)$. The same argument as the previous paragraph shows this is the identity element.
A: With the notation you appear to be using, this is what happens:-
If you do $a$ then $e$
1 goes to 2 then to 2
2 goes to 1 then to 1
3 goes to 3 then to 3
If you do $e$ then $a$
1 goes to 1 then to 2
2 goes to 2 then to 1
3 goes to 3 then to 3
I hope that helps you to clarify your thoughts.
