Presence of Identity Element and abelian groups I.N.Herstein in Topics of Algebra defines a group as a set having a special element $i$ such that:
$a,i\in A(S) $ which satisfies $i\cdot a = a\cdot i = a$ 
In this way, this follows commutativity holds true for identity element, then doesnt it run contradictory to the non-necessary condition that $a \cdot b \neq b \cdot a$ ?
In similar vein is the set of integers under subtraction a group? {my reasoning is no it isnt, because $a-0 = a \neq 0-a$} 
Soham
 A: The commutativity of the identity in a group is part of its definition, even in a non-commutative group.  And you are correct about subtraction not being a valid group operator: it's not even associative.
A: Non-commutative denotes the negation of the commutative law $$\rm\:\color{#C00} \lnot\ [\color{#C00}\forall\,x,y\!:\ xy \color{#C00}= yx]\, \equiv\, \color{#C00}\exists\,x,y\!:\ xy\color{#C00}\ne yx\:$$  In words, commutativity fails to be true universally (for all elements), precisely when there exists at least one counterexample. It doesn't mean that commutativity always fails.
A: The non-necessary condition is $\forall a,b \in A, a \cdot b \not= b \cdot a$. This doesn't rule out that it may be true for some $a, b \in A$. In fact, it could be true for all elements in the group. Then we call it an Abelian group, which is still a group, nonetheless.
And you are correct, the integers (or rationals or real numbers) with subtraction does not form a group. You gave one reason. You could also check associativity.
A: The "unnecessary condition" is that $a\cdot b= b \cdot a $ for all $a,b$. We don't need this fact to prove that the identity commutes. Let's assume there are two: an left identity $i$ and a right identity $j$. 
We find that $i\cdot j=i=j$, since we can use the definition of either identity. So they're actually the same. In the immortal words of highlander, "there can be only one". So the same identity element has to work for both sides if it's going to work at all.
