What is a right-invertible and left-invertible operation? im reading a book and it says:


*

*An operation O is right invertible or left invertible in the set K if for any two elements x and y of the set K there always exists an element z of K such that x=yOz or x=zOy.

*An operation O which is both right and left invertible is simply invertible in the class K.

*K is a group with respect to O if this K is closed under O and O is associative and invertible in K.
Wikipedia says:
"To qualify as a group the set and operation, (G, •), must satisfy four requirements:


*

*Closure:
For all a, b in G, the result of the operation, a•b, is also in G.

*Associativity:
For all a, b and c in G, (a•b)•c = a•(b • c).

*Identity element:
There exists an element e in G such that, for every element a in G, the equation e•a=a•e=a holds.

*Inverse element:
For each a in G, there exists an element b in G, such that a•b=b•a=e, where e is the identity element."
I don't really get it, the closure property and the associativity property are the same in both definitions of group, but how right invertible and left invertible are the same of "identity element" or "inverse element"?
 A: In your post there are two concepts that are strictly linked but not the same...
One is that invertible operation, while the other is that of inverse element of an algebraic structure.
For e.g. a group $G$, the second one refer to the operation "$\circ"$ defined on the structure, but in addition needs  the existence of a "neutral (or: identical) element" $e$:

"An element of the algebraic structure $G$ with a two-sided inverse in $G$ is called invertible in $G$." 


See some useful schema regarding the classification of algebraic structures:

Magma, Semigroup, Monoid, etc.

A: There is one small wrinkle - your book should also require $K$ to be nonempty, otherwise the whole thing is vacuously true for an empty set, which we still don't consider to be a group.
Claim: A group in Wikipedia sense is a group in your book's sense.
Proof: $K$ is nonempty as $e\in K$. Take two elements $x, y\in K$ and:


*

*Let $z=y^{-1}\circ x$, where $y^{-1}$ is the inverse in the Wikipedia sense. Then, $y\circ z=y\circ(y^{-1}\circ x)=(y\circ y^{-1})\circ x = e\circ x = x$.

*Similar, let $z=x\circ y^{-1}$ and it is easy to prove that $z\circ y=x$.


Claim: A group in your book's sense is a group in Wikipedia sense.
Proof: First, as $G$ is nonempty (see my first sentence above), take any $x\in G$ and, applying the existence for left and right inverses on $x$ and $x$ (again), you can find such $e_1, e_2\in G$ so that $x=e_1\cdot x=x\cdot e_2$.
Now, for any other $y\in G$, there is $z\in G$ such that $y=x\cdot z$, so we have: $e_1\cdot y=e_1\cdot (x\cdot z)=(e_1\cdot x)\cdot z=x\cdot z=y$. Similarly, $y\cdot e_2=y$ for all $y\in G$.
Furthermore, this also implies that $e_1=e_1\cdot e_2=e_2$, so those two elements are equal and so we have an element $e$ such that $e\cdot y=y\cdot e=y$ for all $y\in G$ - the identity element in Wikipedia sense.
The last bit is easy. For any $x\in G$, there is $y\in G$ such that $x\cdot y=e$ (apply right invertibility to $x$ and $e$), and similarly there is $z\in G$ such that $z\cdot x=e$ (left invertibility). In fact, then we can prove that $y=z$ because $y=e\cdot y=(z\cdot x)\cdot y=z\cdot(x\cdot y)=z\cdot e=z$, which finishes the proof.
