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If $A$ is a square matrix and there exists a square matrix $B$ such that $AB =1$, than it is known that $BA=1$. This property is proved with some properties from linear algebra. Although I've never seen it be proved just by structures of matrix multiplication, I couldn't find a counterexample of a set with structures of matrix multiplication but left inverse doesn't imply right inverse.

To be more specific, let $X$ be a set and binary operation $\cdot$ is defined on $X$. If $\cdot$ is associative and $X$ has left and right identity(which will be the same), than does $A \cdot B = 1$ for some $A, B\in X$ implies $B \cdot A = 1$?

If not, what other properties of matrix multiplication should we add to this structure of $(X,\cdot)$ in order to get the property?

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  • $\begingroup$ Left and right identity will not necessily be the same. $\endgroup$ – amsmath Oct 25 '19 at 2:47
  • $\begingroup$ It is not true in general; see math.stackexchange.com/questions/70777/… $\endgroup$ – Elliot G Oct 25 '19 at 2:50
  • $\begingroup$ @amsmath No, I mean I'd like to know the operation such that if there exists an element which has a left inverse then the element also has a right inverse(Then the two inverses must be same.). $\endgroup$ – coxehj4142 Oct 25 '19 at 2:52
  • $\begingroup$ @coxehj4142 Then why do you write something about left and right identities? $\endgroup$ – amsmath Oct 25 '19 at 2:54
  • $\begingroup$ It is not enough; a monoid may have elements with one-sided inverses but no inverse on the other side. If $A$ is infinite, the set of all functions $f\colon A\to A$ is a monoid under composition, and if $f$ is surjective but not injective then there exists $g$ such that $f\circ g=\mathrm{id}_A$, but there is no element such that $h\circ f=\mathrm{id}_A$. It really is something special about matrices and the way they act, not about the monoid structure. $\endgroup$ – Arturo Magidin Oct 25 '19 at 2:56
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Take $X = \{f:\mathbb R\to\mathbb R\}$ equipped with composition $\circ$. Can you think of a function that is surjective but not injective?

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If you just consider one operation, that is, if you work with monoids, the answer is no. A counterexample is the bicyclic monoid, which is the quotient of the free monoid on two generators $a$ and $b$ under the relation $ab = 1$.

However, let me answer your last question:

What other properties of matrix multiplication should we add to this structure of $(X,\cdot)$ in order to get the property?

It turns out that the property still holds if you work on a commutative semiring, which is, roughly speaking, a ring without subtraction. For a proof, see [1].

[1] Reutenauer, Christophe; Straubing, Howard. Inversion of matrices over a commutative semiring. J. Algebra 88 (1984), no. 2, 350--360.

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