Nondegeneracy of "symmetrization" of perfect group bihomomorphism

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Let $G_1, G_2 \leq G$ be finite abelian groups, s.t. $G=\langle G_1 \cup G_2 \rangle$. Let $f:G_1 \times G_2 \to \mathbb{C}^\times$ be a perfect bihomomorphism (i.e. it induces an isomorphism $G_1 \to G_2^\vee$) and let $Sym(f):(G_1 \times G_2)^{\times 2} \to \mathbb{C}^\times$ denote the symmetric bihomomorphism defined by $Sym(f)((x_1,x_2),(y_1,y_2)):=f(x_1,y_2)f(y_1,x_2)$. Moreover, assume that $Sym(f)|_{(G_1\cap G_2)^{\times 2}}$ is a perfect bihomomorphism, where $G_1\cap G_2$ is embedded in $G_1 \times G_2$ by $x \mapsto (x,-x)$, so that addition provides a short exact sequence $G_1 \cap G_2 \to G_1 \times G_2 \to G$.. under these assumptions, $Sym(f)$ factors through $G^{\times2}$, more precisely through a subextension $X \leq G_1 \times G_2$ of $G$ by the trivial group.

Question: Is $Sym(f)|_{G^{\times2}}$ perfect?