Do algebraic structures exist where the identity element is not commutative?

Question

In studying abstract algebra I have noticed that even when commutativity is not assumed, it is still assumed over the identity element. I am thrown off by this.

Answer 1

There are indeed examples (and nontrivial examples at that) in which identity elements do not commute. In particular the first that comes to mind is for left and right inverses of linear transformations on infinite dimensional vector spaces. For example:



Consider the linear transformation that maps infinite sequences of real numbers (a1, a2, a3, ...) to the vector (a2, a3, a4, ...)



You can verify that this is in fact a linear operator however the left sided inverse of this operator does not exist (namely because given (a2, a3, a4...) it is impossible to recover the a1 that was lost when performing the operation).



However this operator certainly has a right inverse, namely (a1, a2, a3,...) maps to (k, a1, a2, a3,...)



You can verify that this is a right inverse.





Other specific examples are the square root function, and integration/differentiation. In both of these cases one direction has a left/right inverse and the other direction does not have any inverse, or does not have the same inverse. Thus the identity is not commutative.



In particular this commonly happens for transformations that are one to one but not onto, or vice versa. From this it follows that these require infinite dimensional vector spaces.

Answer 2

Everything you are likely to study, if it has an identity, will have the identity commute with everything. As has been mentioned, any exceptions would be weird things.



A good example is 2x2 matrices, where the identity is

1 0

0 1.



This works on both the left and the right, while in general matrix multiplication isn't commutative.



You might encounter problems where each element as a left identity and a right identity:

La = aR = a, but then you may end up having to prove that L = R.

Answer 3

It's not assumed. The identity element is BY DEFINITION commutative. I assume you're studying groups. There are structures that are degenerate of groups that have only a subset of the properties of groups.



Magmas are sets closed under a binary operation



Semigroups are sets closed under an associative binary operation



Monoids are sets closed under an associative binary operation which has an identity element



Quasigroups are sets closed under a binary operation with inverses, i.e. unique solutions to

ax = b

ya = b



A Loop is a Quasigroup with an identity





You'll probably never talk about these as they're weird and don't have good examples.





You can also have left and right identities, i.e. non-equal elements a and b such that ax = x, xb = x for all x in your set. This is no longer a group (not sure if it has a specific name). I think the easiest set to think about when this could occur would be non-square matrices, where multiplication is only well-defined for one direction.



Hopefully this helps ease the pain... I know Abstract Algebra can be painful.

Answer 4

the commutative regulation holds: A + B = B + A The Associative regulation holds: A + B + C = (A + B) + C = A + (B + C) the identity aspect? do you advise identity matrix? a multiplicative identity matrix is a unit matrix the position all the elements on the finest idagonal equivalent to a million ( and it should be a sq. matrix) e.g. (imagine the brackets) a million 0 0 0 a million 0 0 0 a million or a million 0 0 a million

Answer 5

Commutativity is not assumed over the identity element, it is taken from the definition. The identity element of a group (G, #) is e with a # e = e = e # a for all a ϵ G.



Since the identity element is by definition commutative, then logically it impossible for there to be an algebraic structure where it is not commutative.



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@ Shane



A supposed identity that only works in one direction is no identity at all.