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Given G is a set with associative binary opeartion.

Given for any,$a,b \in G$ , there exists $a'$ in G such that $ be=b, $. And $a'a=e$ First i try to show that $eb=b$.

Given,

$be=b$

$beb=bb$

$b'beb=b'bb$

$b'beb=eb=b$

$b'beb=b$

$eb=b$

Now for any element c of set G i want to show that $ce=c$

$ce=cc'e=ec=c$

To show right inverse :

Given $a'a=e$, $a'aa'=ea'$ $a'eaa'=ea'$

$a'aa'=a'$

$a'aa'=a'e$

$aa'=e$ (i believe cancellation laws can be proved by using left inverse, so using them)

Is this correct ? Thank you

Anne Bauval
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Sophie Clad
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  • You assumed $eb=b$ to prove $eb=b.$ (When writing "$b'beb=eb=b$") – Anne Bauval Apr 09 '23 at 10:35
  • @AnneBauvalno. started with $be=b$ – Sophie Clad Apr 09 '23 at 10:38
  • Your equalities $ce=cc'e=ec$ are wrong. And the sentence just before is strange (We have $ce=c$ by hypothesis, for every element $c$ in $G$, so you do not have to show it). Anyway, please look at the proposed duplicate. The answer gives a counterexample to what you wanted to prove (up to exchanging right and left, which is easy). – Anne Bauval Apr 09 '23 at 10:38
  • @AnneBauval I have edited question. its different question now – Sophie Clad Apr 09 '23 at 11:41
  • its unique right identity element instead – Sophie Clad Apr 09 '23 at 11:50
  • You should not change a question when you already received comments and link to a duplicate. Better post your new question in a new post. Anyway your new question would also be a duplicate: https://math.stackexchange.com/questions/3163347/is-a-semigroup-with-unique-right-identity-and-left-inverse-a-group I rolled back to keep the body consistent with the title and the closure box. @KurtG. – Anne Bauval Apr 09 '23 at 12:46
  • As for your proof attempt: even with the stronger hypothesis of uniqueness of the right identity (which you don't use), you still have the flaw indicated in my 1st comment, and the oddities of the 2nd one. – Anne Bauval Apr 09 '23 at 12:58
  • @AnneBauval i am not sure what exactly is meaning of uniqueness of right identity..can you help ? – Sophie Clad Apr 09 '23 at 16:32
  • Didn't the answer there suffice? (uniqueness of right identity is used explicitely there) – Anne Bauval Apr 09 '23 at 17:14

0 Answers0