# $ab$ and $ba$ invertible elements imply $a$ and $b$ invertible in a Banach algebra

I encountered the following question as a question about linear operators. If $AB$ and $BA$ are invertible for some $A, B \in \frak{L}$$(E)$ where $E$ is a Banach space, then show that $A$ and $B$ are also invertible.

I worked this out by showing that $A$ is bijective and its inverse is linear and bounded. But that required making explicit the action of $A$ and $B$ as operators on an underlying space. So my question is whether there is an 'algebraic' way of showing the statement is true, i.e. whether if $ab$ and $ba$ are invertible elements in a Banach algebra imply $a$ and $b$ invertible.

Let $C$ be the inverse of $AB$ and $D$ be the inverse of $BA$.
Then: $I=C(AB)=(CA)B$ and $I=(BA)D=B(AD)$. With $X=CA$ and $Y=AD$ we therefore have
$XB=I=BY$.
This shows that $B$ is invertible. A similar proof shows that $A$ is invertible.