The identity element in multiplication Let's get on with the question. Is it true that the multiplication is always commutative when we multiply an entity (number, matrix etc.) with the identity entity? I mean, the statement is true in the scope of numbers ($a×1=1×a=a$) and even in the scope of matrices , where multiplication is not commutative from general perspective.
If it is true, then what would be the proof of it? (an as general as possible proof). The smallest but strong starting point of it would be of great use. Thanks in advance!
 A: An identity need not commute with every entity. There are left identities and right identities. For example, with 2 elements $a$ and $b$, define $a \times b = b$ and $b \times a = a$. In this case both $a$ and $b$ are left identities, and there is no defined right identity. If $a$ and $b$ happen to equal each other, then we can call it a two-sided identity, which is what we usually deal with when we have identities--just not always.
A: Say there were two distinct identities, a left one and a right one
$$
e_L: \forall a (e_La = a) \\
e_R: \forall a (ae_R = a) 
$$
Then by the first of those statements, taking $a=e_R$, 
$$
e_Le_R=e_R
$$
And by the second of those statements, taking $a=e_L$, 
$$
e_Le_R=e_L
$$
So by transitivity of equality,
$$
e_L = e_R
$$
This is true for any set equipped with a multiplication operation, if that operation has a left-identity element and a right-identity element.
A: By definition an identity element $I$ under $\times$ is an element such that $I\times a=a \times I=a$ for all $a$. So yes it must commute with any element.
A one-sided identety however doesn't. A left identity element $I_L$ under $\times$ is defined such that $I_L\times a = a$ for all $a$, no further requirement is implied. And such an element doesn't have to commute, you can for example define $\times$ as $\xi\times\eta =\eta$ and then any $a$ is an identity, but if $a\ne b$ we have $a\times b = b\ne a = b\times a$.
However if you have both a right and left identity they are both the same since $I_L = I_L\times I_R = I_R$. Also they are unique since if $I'_L = I_R = I_L$. Also it is an identity since $I_L\times a = a = a\times I_R = a\times I_L$.
