Isomorphism is Equivalence Relation

Theorem
Isomorphism is an equivalence on the set of algebraic structures.

This result applies to all algebraic structures: rings, groups, R-algebraic structures etc.

Proof
To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.

So, checking in turn each of the critera for equivalence:

Reflexive
From Identity Mapping is an Automorphism, the identity mapping $$I_S: S \to S$$ is an automorphism, which is an isomorphism from an algebraic structure onto itself.

So $$S \cong S$$, and isomorphism is seen to be Reflexive.

Symmetric
If $$S \cong T$$, then $$\exists \phi: S \to T$$ such that $$\phi$$ is an isomorphism.

Thus, from Inverse Isomorphism, $$\phi^{-1}: T \to S$$ is also an isomorphism, and $$T \cong S$$.

Thus we have shown that $$S \cong T \Longrightarrow T \cong S$$, so $$\cong$$ is symmetric.

Transitive
Let $$S_1 \cong S_2$$, and $$S_2 \cong S_3$$.

We can define these two isomorphisms:


 * $$\phi: S_1 \to S_2$$
 * $$\psi: S_2 \to S_3$$

We want to show that $$\psi \circ \phi$$ is an isomorphism between $$S_1$$ and $$S_3$$.

As $$\phi$$ and $$\psi$$ are both bijective, then the composition $$\psi \circ \phi$$ is likewise a bijection.

From Composition of Homomorphisms, we know that the composite of homomorphisms is itself a homomorphism.

So we have established that a composite of isomorphisms is a bijective homomorphism, and thus that $$\psi \circ \phi$$ is in fact an isomorphism, thus demonstrating that $$S_1 \cong S_3$$.

Thus we have shown that $$\cong$$ is transitive.


 * Thus isomorphism is reflexive, symmetric and transitive, and therefore an equivalence.