# Abstract algebra

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It is true that if you have a semi group with a right identity and right inverse you have a group. So if you have a monoid, is it true that with a right inverse the monoid is a group? Since by definition the monoid must have an identity element (and therefore a right identity) and a right inverse, so why can’t you just apply the rule above?

Thank you

Thank you

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#2

Not familiar with monoids, but this looks relevant: https://math.stackexchange.com/quest...a-group/972164

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#3

(Original post by

It is true that if you have a semi group with a right identity and right inverse you have a group. So if you have a monoid, is it true that with a right inverse the monoid is a group? Since by definition the monoid must have an identity element (and therefore a right identity) and a right inverse, so why can’t you just apply the rule above?

Thank you

**examstudy**)It is true that if you have a semi group with a right identity and right inverse you have a group. So if you have a monoid, is it true that with a right inverse the monoid is a group? Since by definition the monoid must have an identity element (and therefore a right identity) and a right inverse, so why can’t you just apply the rule above?

Thank you

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(Original post by

Yes. Are you trying to prove this, or do you just need the statement of the result? For your second question, of course, as you can just run through pretty much the same argument, but you won't have to prove a left inverse exists as you are already given that.

**zetamcfc**)Yes. Are you trying to prove this, or do you just need the statement of the result? For your second question, of course, as you can just run through pretty much the same argument, but you won't have to prove a left inverse exists as you are already given that.

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(Original post by

Not familiar with monoids, but this looks relevant: https://math.stackexchange.com/quest...a-group/972164

**DFranklin**)Not familiar with monoids, but this looks relevant: https://math.stackexchange.com/quest...a-group/972164

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#6

(Original post by

No I am not trying to prove it, it is just out of curiosity since when I googled it many discussions said it was

**examstudy**)No I am not trying to prove it, it is just out of curiosity since when I googled it many discussions said it was

**only possible for finite monoids, I’m not sure why**
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#7

(Original post by

My question is if both the right inverse and right identity exist, rather than a left and a right(since you have a monoid- you know it has an identity within the set). It stems from the idea that the existence of a right inverse and right identity in a semi group implies a group, just wondered if this applies to monoids given we know the identity element exists

**examstudy**)My question is if both the right inverse and right identity exist, rather than a left and a right(since you have a monoid- you know it has an identity within the set). It stems from the idea that the existence of a right inverse and right identity in a semi group implies a group, just wondered if this applies to monoids given we know the identity element exists

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Hmmm do you have a link to this?

**zetamcfc**)Hmmm do you have a link to this?

I know they are talking about the existence of a left inverse instead of right but the principal is the same. I am suggesting effectively that the right inverse existing means the monoid is a group and therefore the left inverse must exist... and be the right inverse itself (uniqueness)

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#9

Yes. You just need to use associativity to prove that that , where (-a) is the right inverse. Unless I'm missing something.

Last edited by Rowan281; 7 months ago

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#10

(Original post by

https://math.stackexchange.com/quest...o-the-right-in

I know they are talking about the existence of a left inverse instead of right but the principal is the same. I am suggesting effectively that the right inverse existing means the monoid is a group and therefore the left inverse must exist... and be the right inverse itself (uniqueness)

**examstudy**)https://math.stackexchange.com/quest...o-the-right-in

I know they are talking about the existence of a left inverse instead of right but the principal is the same. I am suggesting effectively that the right inverse existing means the monoid is a group and therefore the left inverse must exist... and be the right inverse itself (uniqueness)

*Note, however that the group axioms as usually stated are redundant (this is generally stated as a remark or given as an exercise in basic algebra textbooks), in the sense that it is equivalent to state that there is a left identity, and that every element has a left inverse.*" (where obviously "left" can be replaced everywhere with "right" to get the case we're wanting to show is a group).

The finiteness specifically comes in from the argument needed to go from the existence of the left inverse to the existence of the right inverse.

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#11

**examstudy**)

https://math.stackexchange.com/quest...o-the-right-in

I know they are talking about the existence of a left inverse instead of right but the principal is the same. I am suggesting effectively that the right inverse existing means the monoid is a group and therefore the left inverse must exist... and be the right inverse itself (uniqueness)

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(Original post by

I am not sure if I'm reading this right, but in the link we are not assuming any other element has a right inverse. Whereas we need this for the standard proof using idempotency https://math.stackexchange.com/quest...edirect=1&lq=1 see the second and third answers.

**zetamcfc**)I am not sure if I'm reading this right, but in the link we are not assuming any other element has a right inverse. Whereas we need this for the standard proof using idempotency https://math.stackexchange.com/quest...edirect=1&lq=1 see the second and third answers.

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#13

(Original post by

Thank you, so assuming all elements have a right inverse, we can assume the monoid is a group?

**examstudy**)Thank you, so assuming all elements have a right inverse, we can assume the monoid is a group?

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