# Prove that inv(G) is a group. G is a monoid

• grimster
In summary, G is a monoid and Inv(G) is a group. A monoid is a set with an associative law of composition and a unit element. Since Inv(G) contains all elements that have an inverse in G, it is a group. To prove this, it must be shown that for each element x in Inv(G), its inverse is also in G. This can be done by considering the element y in G that satisfies xy=yx=1. As y satisfies the definition of Inv(G), it must also be in Inv(G). Additionally, the existence of an identity element e in Inv(G) and the associativity of the law of composition further supports the proof. Although the presented proof may seem sloppy, it can
grimster
G is a monoid. Inv(G) = {a E G, exists b so that a b = b a = 1}

prove that Inv(G) is a group. it's pretty obvious that inv(G) is a group. a monoid is a set with a law of composition which is associative and has a unit element. so inv(G) is clearly a group, because for all a in inv(G) there is an inverse element. but how do i prove this?

i guess i have to show that for each x in inv(G), the inverse of x is also in G. but how do i do that?

Last edited:
Given x in inv(G), you know there is a y in G so that xy=yx=1.
Let's look at y. Guess what? yx=xy=1. But then y is in inv(G) by definition...
1 is in inv(G), * is associative... did I forget something?

this is what i have so far. e is the identity element.

there is an element c such that c a = e.
a=ea=cba=cbae=eae=ae
further
cba=ce=c
and
cba=ea=a

have
ab=ba=e

Excuse me- what's the problem with my proof?

It's not his proof. When doing homework, one should present one's own work, not copy someone else's.

Palindrom said:
Excuse me- what's the problem with my proof?

i don't know. it looked kind of "sloppy". i guess it is ok, but i wanted something a bit more "structured"...

I agree with you both- but that's why I gave it in a sloppy way. It's not hard to formalize it, but it's your job...

## 1. What is a monoid?

A monoid is a mathematical structure consisting of a set and an associative binary operation that has an identity element. This means that for any two elements in the set, their operation will always result in another element in the set, and there exists an element in the set that when operated with any other element, will result in that same element.

## 2. What is a group?

A group is a mathematical structure consisting of a set and a binary operation that is associative, has an identity element, and has an inverse element for each element in the set. This means that for any two elements in the set, their operation will always result in another element in the set, there exists an element in the set that when operated with any other element, will result in the identity element, and each element has an inverse element that when operated together, will result in the identity element.

## 3. How do you prove that inv(G) is a group?

To prove that inv(G) is a group, we must show that it satisfies the four properties of a group: closure, associativity, identity, and inverse. This means that for any two elements in inv(G), their operation will result in another element in inv(G), the operation is associative, there exists an identity element in inv(G), and each element in inv(G) has an inverse element in inv(G).

## 4. How does the monoid G relate to the group inv(G)?

The monoid G is the set of elements that form the basis of inv(G). The group inv(G) is formed by taking the elements of G and adding the inverse element for each element in G. This means that inv(G) is an extension of G, where the inverse element is added to each element in G to form a group.

## 5. Why is it important to prove that inv(G) is a group?

Proving that inv(G) is a group is important because it allows us to use the properties of groups, such as the existence of an identity element and inverses, to solve problems and make calculations. It also helps us understand the relationship between monoids and groups, and how the addition of inverse elements can transform a monoid into a group.

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