# Isomorphism Preserves Inverses

## Theorem

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an isomorphism.

Let $\left({S, \circ}\right)$ have an identity $e_S$.

Then $x^{-1}$ is an inverse of $x$ for $\circ$ iff $\phi \left({x^{-1}}\right)$ is an inverse of $\phi \left({x}\right)$ for $*$.

That is, iff:

- $\phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$

## Proof 1

Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ has an identity $e_S$.

From Epimorphism Preserves Identity, it follows that $\left({T, *}\right)$ also has an identity, which is $\phi \left({e_S}\right)$.

Let $y$ be an inverse of $x$ in $\left({S, \circ}\right)$.

Then:

\(\displaystyle \phi \left({x}\right) * \phi \left({y}\right)\) | \(=\) | \(\displaystyle \phi \left({x \circ y}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({e_S}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({y \circ x}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({y}\right) * \phi \left({x}\right)\) | $\quad$ | $\quad$ |

So $\phi \left({y}\right)$ is an inverse of $\phi \left({x}\right)$ in $\left({T, *}\right)$.

As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.

Thus the result for $\phi \left({x}\right)$ can be applied to $\phi^{-1} \left({\phi \left({x}\right)}\right)$.

$\blacksquare$

## Proof 2

We have that an isomorphism is a homomorphism which is also a bijection.

By definition, an epimorphism is a homomorphism which is also a surjection.

That is, an isomorphism is an epimorphism which is also an injection.

Thus Epimorphism Preserves Inverses can be applied.

$\blacksquare$