It is a Monoid, bcoz there exist identity $e= 0$

Infact it is a group... Moreover an Abelian group.

Infact it is a group... Moreover an Abelian group.

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Is this monoid:

Addition modulo (take mode using m) on the set of Integers (Z m)={0,1,2,3,4,…..m-1}

i.e. For all a a (+ modulo using m) e = e (+ modulo using m) a =a

here, e is an identity element

Addition modulo (take mode using m) on the set of Integers (Z m)={0,1,2,3,4,…..m-1}

i.e. For all a a (+ modulo using m) e = e (+ modulo using m) a =a

here, e is an identity element

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@Nandkishor3939 I think idenitity element should belong to the given set, m is not in given set. Moreover identity element, if exists, is unique. here it is $0$

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yes, 0 is identity element

and inverse of any element w.r.t. 0

I mean 1 has inverse of 4

right?

that is why group

and inverse of any element w.r.t. 0

I mean 1 has inverse of 4

right?

that is why group