Is an inverse element of binary operation unique? If yes then how?

Question

I am trying to prove it but not getting any clue how to start it! $$a*b=b*a=e,$$ $$a*c=c*a=e$$ How to show $b=c$?

Answer 1

If the binary operation is associative you have that for all $a\in G$ if $b,c\in G$ are elements such that

$ac=ca=e$ and $ab=ba=e$

then

$c=ce=c(ab)=(ca)b=eb =b$

so

$c=b$

Answer 2

It is not unique. Take $S = \{a,b,c,e\}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.

Answer 3

Well, I think you are refering to the shortening rule: $$a*b=a*c \Rightarrow b=c, \quad b*a=c*a\Rightarrow b=c.$$ If you work in a ring $R$ and $0\ne a\in R$ is not a zero divisor, then $a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $a\in R$ is invertible!