If $a\cdot b = e$ and $x$ is inverse of $a$ , does this imply that $x = b$?

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$a,b,x$ 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 $ba = e$

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0
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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),

$ba = e(ba) = (xa)(ba) = x((ab)a) = x(ea) = xa = e. \tag 7$

3
On

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 $e=ba$.

3
On

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 $a^{-1}$:

$a^{-1}ab=a^{-1}e\Leftrightarrow eb=a^{-1}\Leftrightarrow b=a^{-1}=x$

1
On

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!

0
On

From $a\cdot b=e$ you draw $b=a^{-1}$, and the inverse is unique.