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Under the proposed isomorphism these become
\begin{equation*}
\{g(a), g(b)\}, \{g(a), g(c)\}, \{g(a), g(d)\}, \{g(c), g(d)\}
\end{equation*}
\begin{equation*}
\{c,d\}, \{c,b\}, \{c,a\}, \{b,a\}\text{,}
\end{equation*}
which are precisely the edges in \(G_2\text{.}\) Thus \(g\) is an isomorphism, so \(G_1 \cong G_2\)
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