Question

Please show that in any monoid (semigroup with neutral element e) if some element a has inverse a^-1, this inverse is unique. This means no element can have more than one inverse. [Hint: Start from writing the definition of the inverse for element a. Consider an element a which has two inverses (a1)^-1 and (a2)^-1. Then think about the value of (a1)^-1a(a2)^-1]. Comment: This is about any monoid which has inverses for some elements, but not necessarily for all elements. However, this applies also to groups where every element has inverse. So, we have as a consequence in every group there is exactly one inverse for each element.

Answer 1

classmate Date Page above G has Introductor: Monold ( Semigroup) -(01, *) is onold 1 Gris closed under binary operation 2. Gis associative under t 3. G has an identity element é. If G has three proporties wor.t * then (G *) is monoid. Monoid G becomes group if every element in Inverse, Suppose some element of Monord G has inverse provet. It is unique. Proof? To prove Inverse of an element a Egillis unique Let a be an element of G and let e be the identity element of G. Then ate= exa= a. Suppose a has two inversee a and ão then atal=e=ā, * a atal=e=ã *a . a = axe (by the definition of al x (xal) identity element) ie=a az -la *a) *ay by the associative P =e*a ( by the def- of a) prove a - a به = ة Hence No element in G has more than one inverse,