>>    ∃ i ∈ S: ∀ x ∈ S, i # x = x # i = x.
>
> Hey, not only you added identity element, but you seems to imply also
> (but not really) that its also commutative:
>
> ∀ x ∈ S, ∀ y ∈ S, x # y = y # x.
>
> So please be more careful for my poor brain and write instead:
>
> ∃ i ∈ S: ∀ x ∈ S, i # x = x and x # i = x.
>
> There is commutativity for the identity element, but it is not needed to
> be commutative.

A monoid is /defined as/ having a two-sided identity.

Note that it is possible to have more than one left-identity, OR to have 
more than one right-identity. But it is impossible to have a 
left-identity and a distinct right-identity.

Suppose that

   ∃ i ∈ S: ∀ x ∈ S, i # x = x
   ∃ j ∈ S: ∀ x ∈ S, x # j = x

are the left and right identities of #. Now consider the term i#j. By 
the first equation, i#j=j. By the second equation i#j=i. Thus we have 
that i = j.

> Your definition of inverse element once again seems to imply general
> commutativity, it is not needed.

Again, a group is /defined as/ having two-sided inverses.

If you wish to split hairs, you don't actually need to /insist/ that the 
inverses by double-sided; the associativity of # and the existence of a 
two-sided inverse i already necessarily implies this fact:

Consider the term

   x # y # x

where y is the left-inverse of x. By that assumption, we have

   (x # y) # x = i # x

and by the definition of i as the two-sided identity, we have

   (x # y) # x = i # x = x

By the assumption of associativity, we have that

   (x # y) # x = x # (y # x)

It immediately follows that y # x must be the identity, and hence that y 
is ALSO the right-inverse of x. QED.

> (I like to think about the Quaternion group (non abelian group of order
> 8): 1&  -1 are their own inverse, but inverse of i,j,k are same negative
> (-i, -j, -k))

OK, so 1 is the identity element, -i is the inverse of i, -j is the 
inverse of j, and -k is the inverse of k.

> Yes, -i.i = i.-i = 1
> but i.j = -j.i = k

That is true - but irrelevant. You just proved that the group is 
non-Abelian. We know that. We're discussing the /inverses/ though, and 
if you look, we have:

   i . -i = 1 = -i . i
   j . -j = 1 = -j . j
   k . -k = 1 = -k . k

In other words, -i is the TWO-SIDED inverse of i, and so on and so forth 
- as I originally asserted.

Every term commutes with the identity and with its own inverse. It might 
not commute with anything else, but it does commute with these two 
special elements.

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