If $a\cdot b = e$ and $x$ is inverse of $a$ , does this imply that $x = b$? $a,b,x$ are elements of a group .
$x$ is the inverse of $a$.
Here is my attempt to prove it :-
$a\cdot b = e$
$x\cdot (a\cdot b) = x\cdot e$
$(x\cdot a)\cdot b = x$
$e\cdot b = x$
$b = x$
Are my steps correct?
What I wanted to prove is that if $ab = e$, then $ba = e$
 A: Let $a,b\in G$ be such that $ab=e$, and let $x$ be the inverse of $a$. Then $ab=e=ax$, so $b=x$.
Edit: If $ab=e$, then
$$\begin{align}ba&=ba\\&=bea\\&=b(ab)a\\&=(ba)(ba).\end{align}$$
Let $c$ be the inverse of $ba$. Then $ba=(ba)(ba)$ implies that $c(ba)=c(ba)(ba)$, which implies that $e=ba$.
A: No, it does not imply that $x=a$. It implies that $x=b$. Maybe a typo?
We have $ab=e$ since $x=a^{-1}$ we get after multiplying both sides with $a^{-1}$:
$a^{-1}ab=a^{-1}e\Leftrightarrow eb=a^{-1}\Leftrightarrow b=a^{-1}=x$
A: It implies that $x = b$. 
Your reasoning is sound, and it is this exact reasoning that proves a group element's inverse is unique. You also proved this using only the identity, associativity, and inverse laws. Good job!
A: This is pretty standard, basic and elementary stuff; the kind of stuff one usually sees in the first few pages of a textbook on group theory; but essential stuff nevertheless.
Our OP neraj's proof that $x = b$ is, of course, unarguably flawless.  Lauds.
If we wish to see that
$ab = e \Longrightarrow ba = e \tag 1$
under the given hypotheses of the question, the simplest thing we can do is exploit the given that $x$ is the inverse of $a$, which by definition means that
$ax = xa = e; \tag 2$
of course we often write
$x = a^{-1} \tag 3$
under such circumstances; in any event, if
$ab = e, \tag 4$
then
$(ab)a = ea = a; \tag 5$
thus
$a(ba) = a, \tag 6$
whence, by (2),
$ba = e(ba) = (xa)(ba) = x((ab)a) = x(ea) = xa = e. \tag 7$
A: From $a\cdot b=e$ you draw $b=a^{-1}$, and the inverse is unique.
