# Relation Isomorphism Preserves Symmetry

## Theorem

Let $\left({S, \mathcal R_1}\right)$ and $\left({T, \mathcal R_2}\right)$ be relational structures.

Let $\left({S, \mathcal R_1}\right)$ and $\left({T, \mathcal R_2}\right)$ be (relationally) isomorphic.

Then $\mathcal R_1$ is a symmetric relation if and only if $\mathcal R_2$ 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.

Thus WLOG it suffices to prove only that if $\mathcal R_1$ is symmetric, then also $\mathcal R_2$ is symmetric.

So, suppose $\mathcal R_1$ is a symmetric relation.

Let $y_1, y_2 \in T$ such that $y_1 \mathrel {\mathcal R_2} y_2$.

Let $x_1 = \phi^{-1} \left({y_1}\right)$ and $x_2 = \phi^{-1} \left({y_2}\right)$.

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

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

- $x_1 = \phi^{-1} \left({y_1}\right) \mathrel {\mathcal R_1} \phi^{-1} \left({y_2}\right) = x_2$

As $\mathcal R_1$ is a symmetric relation it follows that:

- $x_2 \mathrel {\mathcal R_1} x_1$

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

- $y_2 = \phi \left({x_2}\right) \mathrel {\mathcal R_2} \phi \left({x_1}\right) = y_1$

So:

- $y_1 \mathrel {\mathcal R_2} y_2 \implies y_2 \mathrel {\mathcal R_2} y_1$

and so by definition $\mathcal R_2$ is a symmetric relation.

Hence the result.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $14.9 \ \text{(c)}$