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Francesco Polizzi
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Both calculations are correct.

In general, there is an isomorphism $$H^2(G, \, \mathbb{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$

Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283