Prove/Disprove: If a monoid has a right inverse, then it is a group

Question

I have first started with an attempt to prove:

$$a^{-1}a=a^{-1}ae=a^{-1}aaa^{-1}$$ but it has lead me nowhere, so I tried to think about counterexample, but I do know much Monoids

See https://math.stackexchange.com/questions/433546/is-a-semigroup-g-with-left-identity-and-right-inverses-a-group — AnalysisStudent0414, Dec 11 '19 at 18:22
@AnalysisStudent0414 in my case it is a monoid, so $\forall a\in M: ae=ea=e$ — gbox, Dec 12 '19 at 11:01

score 1 · Accepted Answer · answered Dec 12 '19 at 11:07

1

I linked you the other question for the technique used there!

First, note that if $aa^{-1}=e$ then $$a^{-1}aa^{-1}a = a^{-1}a$$ so if $c=a^{-1}a$ then $cc=c$. Now you can multiply by the right inverse $c^{-1}$ to get $c=e$. So your monoid has a left inverse, that coincides with the right (i.e. it is a group).

answered Dec 12 '19 at 11:07

AnalysisStudent0414

8,262

Just answer another qeustion using this technique :) – gbox Dec 12 '19 at 11:11
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It's a standard and useful trick – AnalysisStudent0414 Dec 12 '19 at 11:12

Prove/Disprove: If a monoid has a right inverse, then it is a group

1 Answers1