Here's the lemma: Let $G$ be a finite group, $g\in G$ with image $\bar{g}\in G^{\text{ab}}$, and let $\chi:G\to\mathbb Q/\mathbb Z$ a homomorphism. Viewing $\bar{g}$ as an element of $H_T^{-2}(G,\mathbb Z)$ and $\chi\in H^1(G,\mathbb Q/\mathbb Z)$, we have $$\bar{g}\cup\chi = \chi(g)\in\mathbb Q/\mathbb Z,$$ noting that $H_T^{-1}(G,\mathbb Q/\mathbb Z)\simeq\frac{1}{\#G}\mathbb Z/\mathbb Z$. Here $H_T$ denotes Tate cohomology.
Here's the proof: consider the connecting homomorphisms $\delta$ and $\delta^\vee$ for the sequence $$0\to I_G\to\mathbb Z[G]\to\mathbb Z\to 0$$ and its $\mathbb Q/\mathbb Z$-dual $$0\to\mathbb Q/\mathbb Z\to\operatorname{Hom}_{\mathbb{Z}}(\mathbb Z[G],\mathbb Q/\mathbb Z)\to\operatorname{Hom}_{\mathbb Z}(I_G,\mathbb Q/\mathbb Z)\to 0.$$ The image of $\bar{g}$ in $H_T^{-1}(G,I_G)$ is the image of $g-1$ in $I_G/I_G^2$, and the inverse image of $\chi$ in $H_T^0(G,\operatorname{Hom}(I_G,\mathbb Q/\mathbb Z))$ is the class of the homomorphism $f$ that takes $h-1$ for $h\in G$ to $\chi(h)$. We then have $$\bar{g}\cup\chi = \delta(\bar{g})\cup(\delta^\vee)^{-1}(\chi)= f(g-1) = \chi(g).\square$$
I'm still getting used to cup products, so I'm having trouble understanding the computations. I understand that $\operatorname{Hom}_{\mathbb Z}(\mathbb Z[G],\mathbb Q/\mathbb Z) = \operatorname{CoInd}^G(\mathbb Q/\mathbb Z)$, and so is (Tate)-cohomologically trivial, so $$H_T^0(G,\operatorname{Hom}(I_G,\mathbb Q/\mathbb Z))\xrightarrow{\delta^\vee}H_T^1(G,\mathbb Q/\mathbb Z) = \operatorname{Hom}_{\operatorname{cts}}(G,\mathbb Q/\mathbb Z)$$ is an isomorphism, so $(\delta^\vee)^{-1}(\chi)$ is a well-defined class of $\operatorname{Hom}(I_G,\mathbb Q/\mathbb Z)^G/\operatorname{Norm}$.
(1) How do I know that it's the class of $f:g-1\mapsto \chi(g)$? (It's hard to imagine it's anything else, but is there a way to see this without tracing through the definition of the connecting map?)
Similarly, since $\mathbb Z[G]$ is projective, it's (Tate)-cohomologically trivial, so $H_T^{-2}(G,\mathbb Z)\xrightarrow{\delta}H_T^{-1}(G,I_G) = I_G/I_G^2$ is an isomorphism, so $\bar{g}$ is mapped to $g-1\bmod{I_G^2}$ (this one I think I get). So now the cup product $\bar{g}\cup\chi$ lives in $H_T^{-1}(G,\mathbb Q/\mathbb Z) = \mathbb Q/\mathbb Z$. I get the first part, that $\bar{g}\cup\chi = \delta(\bar{g})\cup(\delta^\vee)^{-1}(\chi) = (g-1)\bmod{I_G^2}\cup f$, but...
(2) Why is this in turn the composition $f(g-1)$?