Does $ab=1$ imply $ba=1$ in a ring? Let $R$ be a ring with unity $1$ and let $a,b\in R$ be not zero divisors. Is there any counterexample for:
$$ab=1\quad\Rightarrow \quad ba=1$$
 A: If you drop the condition "no zero-divisors", there are counterexamples:
There are rings with elements which are left-invertible, but not right-invertible. Standard example: Consider a vector space $V$ with the countable infinite basis $\{v_1,v_2,\ldots\}$ and let $R$ be the endomorphism ring of $V$. Now the endomorphism $b$ induced by $v_1\mapsto v_2$, $v_2\mapsto v_3$, $v_3\mapsto v_4$ etc. is left-invertible, but not right-invertible. However, it is a right zero divisor in $R$.
(Explanation: I missed the condition "no zero-divisors" in the first place.)
A: Hint $\ $ Exploit conjugation $\rm\ \ 1\!-\!ba\, =\, b\, (1\!-\!ab)\, b^{-1}.\ $ [Multiply it by $\rm\:b\:$ to eliminate $\rm\:b^{-1}]$
Remark $\ $ To learn more about such look up "Dedekind finite" rings. You might also enjoy this tantalizing Halmos problem: explain why the following formal power series "proof" of the fact that if $\rm\:(1\!-\!ab)^{-1}\:$ exists then $\rm\:(1\!-\!ba)^{-1}\:$ exists, and has the value
$$\rm\begin{eqnarray} (1\!−\!ba)^{−1} &=&\rm 1+ba+b\color{#C00}{ab}a+b\color{#0A0}{abab}a+\ \cdots \\
&=&\rm 1+b(1\!+\,\color{#C00}{ab}\ \ \,+\ \ \color{#0A0}{abab}\ \ +\ \cdots)\,a\\ &=&\rm 1+b(1\!−\!ab)^{-1} a\end{eqnarray}$$
A: Assume $ab=1$ and $b$ is not a zero-divisor.
Then $bab=b\cdot ab=b\cdot 1=b$, hence $(ba-1)b=0$, which implies $ba-1=0$ as $b$ is not a divisor of zero. Thus $ba=1$.
A: Let $(M, \cdot, e)$ be a monoid. We have a morphism from $M$ to the monoid of $\mathcal{F}(M)$ of functions from $M$ to $M$, by $a \mapsto L_a$, where $L_a(b) = a \cdot b$. Clearly it maps invertible elements to invertible elements in $\mathcal{F}(M)$, that is, to bijective functions. Conversely, if $a\in M$ is such that $L_a$ is bijective, then $a$ is invertible in $M$. Indeed, since $L_a$ is surjective, there exists  $b\in M$ such that $L_a(b) = e$, that is,
$a b = e$. To show that $a$ is invertible, we need to check that $b a= e$, or, since $L_a$ injective, $a( b a) = a e$. By associativity the LHS equals $(a b) a = e a = a = a e$, and we are done.
If $M$ is the multiplicative monoid of an associative ring $R$, then:

*

*$L_a$ is surjective if and only if $a b = e$ for some $b$ ( $a$ has a right inverse). Indeed, $a b = e$ implies $a b c = c$.


*$L_a$ is injective if and only if $a$ is not a left zero divisor
