Isomorphism of groups is an equivalence relation

Dependencies:

1. Isomorphism on Groups
2. Inverse of a group isomorphism is a group isomorphism
3. /sets-and-relations/equivalence-relation
4. /sets-and-relations/composition-of-bijections-is-a-bijection

Isomorphism of groups is an equivalence relation.

Proof

A group $G$ is isomorphic to itself via $\phi: G \mapsto G ; \phi(x) = x$. So isomorphism is reflexive.

If $G \cong H$ via $\phi$, then $H \cong G$ via $\phi^{-1}$. So isomorphism is symmetric.

Let $G \cong H$ via $\phi$ and $H \cong K$ via $\varphi$.

\[ (\varphi\phi)(g_1g_2) = \varphi(\phi(g_1g_2)) = \varphi(\phi(g_1)\phi(g_2)) = \varphi(\phi(g_1))\varphi(\phi(g_2)) = (\varphi\phi)(g_1)(\varphi\phi)(g_2) \]

The composition of 2 bijections is a bijection, so $\varphi\phi$ is a bijection.

Therefore, $\varphi\phi$ is an isomorphism from $G$ to $K$. This means isomorphism is transitive.

Dependency for: None

Info:

• Depth: 3
• Number of transitive dependencies: 5

Transitive dependencies:

1. /sets-and-relations/composition-of-bijections-is-a-bijection
2. /sets-and-relations/equivalence-relation
3. Group
4. Isomorphism on Groups
5. Inverse of a group isomorphism is a group isomorphism