# Left and Right Inverses of Product

## Theorem

Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.

Let $x, y \in S$.

Let:

$(1): \quad x \circ y$ have a left inverse for $\circ$
$(2): \quad y \circ x$ have a right inverse for $\circ$.

Then both $x$ and $y$ are invertible for $\circ$.

## Proof

Let $z_L$ be the left inverse of $x \circ y$ and $z_R$ be the right inverse of $y \circ x$. Then:

 $\ds z_L \circ \paren {x \circ y}$ $=$ $\ds e_S$ $\ds \leadsto \ \$ $\ds \paren {z_L \circ x} \circ y$ $=$ $\ds e_S$

 $\ds \paren {y \circ x} \circ z_R$ $=$ $\ds e_S$ $\ds \leadsto \ \$ $\ds y \circ \paren {x \circ z_R}$ $=$ $\ds e_S$

Thus $y$ has both a left inverse $z_L \circ x$ and a right inverse $x \circ z_R$.

$z_L \circ x = x \circ z_R$

and $y$ has an inverse, that is, is invertible.

$\Box$

From the above, we have:

 $\ds \paren {z_L \circ x} \circ y$ $=$ $\ds e_S$ $\ds \leadsto \ \$ $\ds \paren {x \circ z_R} \circ y$ $=$ $\ds e_S$ $\ds \leadsto \ \$ $\ds x \circ \paren {z_R \circ y}$ $=$ $\ds e_S$

and:

 $\ds y \circ \paren {x \circ z_R}$ $=$ $\ds e_S$ $\ds \leadsto \ \$ $\ds y \circ \paren {z_L \circ x}$ $=$ $\ds e_S$ $\ds \leadsto \ \$ $\ds \paren {y \circ z_L} \circ x$ $=$ $\ds e_S$

Thus $x$ has both a left inverse $y \circ z_L$ and a right inverse $z_R \circ y$.

$y \circ z_L = z_R \circ y$

and $x$ has an inverse, that is, is invertible.

$\blacksquare$