[proofplan]
We verify the three defining properties of an [equivalence relation](/page/Equivalence%20Relation). Reflexivity follows from the identity map on each group. Symmetry follows by proving that the inverse of a [group isomorphism](/page/Group%20Isomorphism) is again a group isomorphism. Transitivity follows by composing two group isomorphisms and checking that the composite is a group isomorphism.
[/proofplan]
[step:Use identity maps to prove reflexivity]
Let $G \in \mathcal{C}$. Write the group operation on $G$ multiplicatively, and let $1_G$ denote its identity element. Define the identity map
\begin{align*}
\operatorname{id}_G: G \to G,\qquad g \mapsto g.
\end{align*}
For all $g_1,g_2 \in G$,
\begin{align*}
\operatorname{id}_G(g_1g_2)=g_1g_2=\operatorname{id}_G(g_1)\operatorname{id}_G(g_2).
\end{align*}
Thus $\operatorname{id}_G$ is a [group homomorphism](/page/Group%20Homomorphism). It is bijective, with inverse equal to itself, so it is a group isomorphism. Therefore $G \sim G$. Since $G \in \mathcal{C}$ was arbitrary, $\sim$ is reflexive on $\mathcal{C}$.
[/step]
[step:Invert an isomorphism to prove symmetry]
Let $G,H \in \mathcal{C}$ and assume $G \sim H$. By definition of $\sim$, there exists a group isomorphism $\varphi: G \to H$. Since $\varphi$ is bijective, it has an inverse map
\begin{align*}
\varphi^{-1}: H \to G.
\end{align*}
We check that $\varphi^{-1}$ is a group homomorphism. Let $h_1,h_2 \in H$. Define $g_1 := \varphi^{-1}(h_1)$ and $g_2 := \varphi^{-1}(h_2)$, so $\varphi(g_1)=h_1$ and $\varphi(g_2)=h_2$. Since $\varphi$ is a homomorphism,
\begin{align*}
\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)=h_1h_2.
\end{align*}
Applying $\varphi^{-1}$ to both sides gives
\begin{align*}
\varphi^{-1}(h_1h_2)=g_1g_2=\varphi^{-1}(h_1)\varphi^{-1}(h_2).
\end{align*}
Thus $\varphi^{-1}$ is a group homomorphism. Since $\varphi^{-1}$ is bijective, it is a group isomorphism $H \to G$. Hence $H \sim G$.
[guided]
We must show that the relation is symmetric: starting from $G \sim H$, we must construct an isomorphism in the reverse direction. The assumption $G \sim H$ means, by definition of the relation, that there is a group isomorphism
\begin{align*}
\varphi: G \to H.
\end{align*}
Because an isomorphism is bijective, the inverse function exists:
\begin{align*}
\varphi^{-1}: H \to G.
\end{align*}
It remains to verify the group-homomorphism property for this inverse map; bijectivity alone is not enough.
Let $h_1,h_2 \in H$. Since $\varphi$ is bijective, there are unique elements $g_1,g_2 \in G$ such that
\begin{align*}
\varphi(g_1)=h_1
\end{align*}
and
\begin{align*}
\varphi(g_2)=h_2.
\end{align*}
Equivalently, $g_1=\varphi^{-1}(h_1)$ and $g_2=\varphi^{-1}(h_2)$. Now use the fact that $\varphi$ is a homomorphism:
\begin{align*}
\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)=h_1h_2.
\end{align*}
The element $g_1g_2$ is therefore the unique preimage of $h_1h_2$ under $\varphi$. Hence
\begin{align*}
\varphi^{-1}(h_1h_2)=g_1g_2=\varphi^{-1}(h_1)\varphi^{-1}(h_2).
\end{align*}
This proves that $\varphi^{-1}:H\to G$ is a group homomorphism. Since it is also bijective, it is a group isomorphism. Therefore $H \sim G$, which proves symmetry.
[/guided]
[/step]
[step:Compose isomorphisms to prove transitivity]
Let $G,H,K \in \mathcal{C}$ and assume $G \sim H$ and $H \sim K$. By definition of $\sim$, there exist group isomorphisms
\begin{align*}
\varphi: G \to H
\end{align*}
and
\begin{align*}
\psi: H \to K.
\end{align*}
Define the composite map
\begin{align*}
\psi \circ \varphi: G \to K,\qquad g \mapsto \psi(\varphi(g)).
\end{align*}
For all $g_1,g_2 \in G$, using first that $\varphi$ is a homomorphism and then that $\psi$ is a homomorphism,
\begin{align*}
(\psi \circ \varphi)(g_1g_2)=\psi(\varphi(g_1g_2))=\psi(\varphi(g_1)\varphi(g_2)).
\end{align*}
Therefore
\begin{align*}
(\psi \circ \varphi)(g_1g_2)=\psi(\varphi(g_1))\psi(\varphi(g_2))=(\psi \circ \varphi)(g_1)(\psi \circ \varphi)(g_2).
\end{align*}
Thus $\psi \circ \varphi$ is a group homomorphism. Since $\varphi$ and $\psi$ are bijective, their composite $\psi \circ \varphi$ is bijective. Hence $\psi \circ \varphi$ is a group isomorphism $G \to K$, so $G \sim K$. Therefore $\sim$ is transitive on $\mathcal{C}$.
[/step]
[step:Conclude the relation is an equivalence relation]
We have shown that $\sim$ is reflexive, symmetric, and transitive on the specified collection $\mathcal{C}$ of groups. These are precisely the defining properties of an equivalence relation. Hence group [isomorphism is an equivalence relation](/theorems/2406) on $\mathcal{C}$.
[/step]
