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The answer is $\ \displaystyle\bigoplus_{i<j}\ \mathbb{Z}/\!\gcd(n_i,n_j)$. The reason is that for $G$ finite, $H^2(G,U(1))$ is the dual abelian group of $H_2(G,\mathbb{Z})$. Now use the Künneth formula.
The answer is $\ \displaystyle\bigoplus_{i<j}\ \mathbb{Z}/\!\gcd(n_i,n_j)$.
The answer is $\ \displaystyle\bigoplus_{i<j}\ \mathbb{Z}/\!\gcd(n_i,n_j)$. The reason is that for $G$ finite, $H^2(G,U(1))$ is the dual abelian group of $H_2(G,\mathbb{Z})$. Now use the Künneth formula.
