examstudy
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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
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DFranklin
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Not familiar with monoids, but this looks relevant: https://math.stackexchange.com/quest...a-group/972164
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zetamcfc
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(Original post by 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
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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examstudy
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(Original post by 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.
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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examstudy
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(Original post by DFranklin)
Not familiar with monoids, but this looks relevant: https://math.stackexchange.com/quest...a-group/972164
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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zetamcfc
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(Original post by 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
Hmmm do you have a link to this?
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DFranklin
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(Original post by 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
Yeah - for some reason I thought you were talking left inverse (even though you said completely the opposite). My mistake, sorry.
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examstudy
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(Original post by zetamcfc)
Hmmm do you have a link to this?
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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Rowan281
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Yes. You just need to use associativity to prove that that a + (-a) = (-a) + a = 0 \forall a , where (-a) is the right inverse. Unless I'm missing something.
Last edited by Rowan281; 7 months ago
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DFranklin
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(Original post by 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)
As I understand it, the principle is not the same. Note the comment "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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zetamcfc
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(Original post by 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)
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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examstudy
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(Original post by 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.
Thank you, so assuming all elements have a right inverse, we can assume the monoid is a group?
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DFranklin
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(Original post by examstudy)
Thank you, so assuming all elements have a right inverse, we can assume the monoid is a group?
Yes.
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examstudy
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Thanks everyone, appreciate the help
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