Questions & Answers

Question

Answers

A. A finite group

B. Only a semi-group

C. Only a monoid

D. An infinite group

Answer
Verified

Let us first understand what a group, monoid and semigroup is:

Group theory, in modern algebra, the study of groups, which are systems consisting of a set of elements and a binary operation that can be applied to two elements of the set, which together satisfy certain axioms. If we have a group X with binary operation *, then it must follow certain axioms. The axioms are:

The group must be closed under given binary operation that is $a * b \in X$ $\forall a,b \in X$.

It must obey associative law that is $a*(b*c) = (a*b)*c$ $\forall a,b,c \in X$.

It must contain an identity that is $a*e = e*a = a$ $\forall a \in X$.

It must contain an inverse that is $a*b = b*a = e$ $\forall a,b \in X$. Here, b is the inverse of a.

If a set and binary operation follows only property (1) and (2) , then we have a semigroup and if it follows (1), (2) and (3), then we have a monoid.

Consider the set of Integers and binary operation ‘+’.

Let $m,n \in {\mathbb{Z}^ + }$, then $m + n \in {\mathbb{Z}^ + }$ because sum of two positive integers will always result in a positive integer. Hence, it is closed under ‘+’.

Let $m,n,p \in {\mathbb{Z}^ + }$, then $m + (n + p) = m + n + p = (m + n) + p$. Hence, it is associative.

Let $m \in {\mathbb{Z}^ + }$, then $m + e = e + m = m$, but that is only possible if $e = 0$ but $0 \notin {\mathbb{Z}^ + }$. So, there is no identity.

We don’t even need to check the fourth property because even the third is not satisfied here.

Hence, only 1) and 2) are satisfied and according to the definition of a semi-group, we have a semigroup here.

Hence, it is a semi-group.

The students might get the idea that if third is followed, then we do not need to check fourth. But, we might end up making a mistake to call it a monoid but it is a group actually.

Students might make the mistake of considering the third property to be followed as well by forgetting to check if $e \in X$ or not. So, keep in mind to check that.