# Relation Isomorphism Preserves Symmetry

## Theorem

Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.

Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.

Then $\RR_1$ is a symmetric relation if and only if $\RR_2$ is also a symmetric relation.

## Proof

Let $\phi: S \to T$ be a relation isomorphism.

By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.

Without loss of generality, it suffices to prove only that if $\RR_1$ is symmetric, then also $\RR_2$ is symmetric.

So, suppose $\RR_1$ is a symmetric relation.

Let $y_1, y_2 \in T$ such that $y_1 \mathrel {\RR_2} y_2$.

Let $x_1 = \map {\phi^{-1} } {y_1}$ and $x_2 = \map {\phi^{-1} } {y_2}$.

As $\phi$ is a bijection it follows from Inverse Element of Bijection that $y_1 = \map \phi {x_1}$ and $y_2 = \map \phi {x_2}$.

As $\phi^{-1}$ is a relation isomorphism it follows that:

- $x_1 = \map {\phi^{-1} } {y_1} \mathrel {\RR_1} \map {\phi^{-1} } {y_2} = x_2$

As $\RR_1$ is a symmetric relation it follows that:

- $x_2 \mathrel {\RR_1} x_1$

As $\phi$ is a relation isomorphism it follows that:

- $y_2 = \map \phi {x_2} \mathrel {\RR_2} \map \phi {x_1} = y_1$

So:

- $y_1 \mathrel {\RR_2} y_2 \implies y_2 \mathrel {\RR_2} y_1$

and so by definition $\RR_2$ is a symmetric relation.

Hence the result.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$