On group cohomology In the paper Basic quasi-Hopf algebras of dimension $n^3$ by S. Gelaki, he writes the following at the bottom of page $167$:

Let $a$ be a generator of $\mathbb{Z}_n$ and $f\in \text{Hom}(\mathbb{Z}_n,\mathbb{Q}/\mathbb{Z})$ be given by $a\mapsto \frac{1}{n}$. Lift $f$ to $\mathbb{Q}$ and set $F:=df\in H^2(\mathbb{Z}_n,\mathbb{Z})$.

Shouldn't this be $f\in H^2(\mathbb{Z}_n,\mathbb{Q})$? He then continues by stating that the following lemma is well-known:

The map $H^1(\mathbb{Z}_n,k^*)=\text{Hom}(\mathbb{Z}_n,k^*)\rightarrow H^3(\mathbb{Z}_n,k^*)$ given by $\gamma\mapsto \gamma\cup F$ defines an isomorphisms of groups.

How should I interpret $\gamma\cup F$? I guess that we should interpret elements in $H^3(\mathbb{Z}_n,k^*)$ as functions from $\mathbb{Z}_n^3\rightarrow k^*$ and simply say that $\gamma\cup F(g_1,g_2,g_3):=\gamma(g_1)F(g_2,g_3)$. However I'm not familiar enough with group cohomology to be sure.  It's not immediately clear to me that the above should then be an isomorphism.
 A: Let $G$ be a group.  The connecting homomorphism associated to the short exact sequence of $G$-modules $$0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$$ is a map $d: H^*(G, \mathbb{Q}/\mathbb{Z}) \to H^{*+1}(G, \mathbb{Z})$.  So $F$ is indeed in $H^2(G, \mathbb{Z})$ as claimed.  The lift of $f$, after passing to cohomology, actually takes values in $\mathbb{Z} \subset \mathbb{Q}$.  
Now suppose $G = \mathbb{Z}_n$ is the cyclic group of order $n$, generated by $g \in G$.  The element $F$ constructed above turns out to be a generator of $H^2(G, \mathbb{Z})$ - it has order exactly $n$.  Here is an argument (from Cassels & Frohlich) to see that cupping with this element gives an isomorphism $H^1(G,M) \to H^3(G,M)$, and in fact $H^q(G,M) \to H^{q+2}(G,M)$ for all $q$ and all $G$-modules $M$.  Let $I_G = \ker \epsilon$ be the augmentation ideal of $\mathbb{Z}G$ and $T = g - 1$.  We have short exact sequences $$0 \to I_G \to \mathbb{Z}G \xrightarrow{\epsilon} \mathbb{Z} \to 0 \\ 0 \to \mathbb{Z} \to \mathbb{Z}G \xrightarrow{T} I_G \to 0.$$  
Examining the long exact sequence in cohomology, we deduce that $$H^2(G,\mathbb{Z}) \cong H^1(G,I_G) \cong \hat{H}^0(G,\mathbb{Z}).$$  So we just need to show that cupping with a generator of $\hat{H}^0(G,\mathbb{Z})$ is an automorphism of $\hat{H}^q(G,M)$.  By dimension-shifting again, it suffices to prove this for $q = 0$.  We have $\hat{H}^0(G,\mathbb{Z}) \cong \mathbb{Z}_n$, and cupping with the generator is represented by multiplication by a number $b$ coprime to $n$.  Since $\hat{H}^0(G,M)$ is $n$-torsion, multiplication by $b$ is an automorphism, and we are done.  
