Let $R = \mathbb{Z}[G]$ be the group ring of the group $G = \{1, \sigma\}$ of order $2$. Let $$M = \mathbb{Z}e_1 + \mathbb{Z}e_2$$ be a free $\mathbb{Z}$-module of rank $2$ with basis $e_1$ and $e_2$. Define
$$\sigma(e_1) = e_1 + 2e_2$$ and $$\sigma(e_2) = -e_2.$$ We want to prove that this makes $M$ into an $R$-module and that the $R$-module $$ \bigwedge^2(M)$$ is a group of order $2$ with $e_1 \wedge e_2$ as generator.
My reasoning: The group ring $\mathbb{Z}[G]$ will by definition be all (I believe) finite sums $$z_1g_1+ \ldots + z_ng_n \quad(z_i \in \mathbb{Z},g_i \in G)$$ which in our case will reduce to something on the form $$z_1'\cdot 1 + z'_2\sigma \quad(z_1',z_2' \in \mathbb{Z}).$$
I believe we only need to check $\sigma$:s action on $M$, since $\sigma$ generates the group $G$, and since $M$ is naturally a $\mathbb{Z}$-module.
First we check $$\sigma(e_1)+\sigma(e_2) = e_1+2e_2-e_2 = e_1+e_2$$ so that for this to be an $R$-module action, we need $$\sigma(e_1)+\sigma(e_2) = \sigma(e_1+e_2) = e_1+e_2.$$
We find that $$(\sigma)(m_1+m_2) = (\sigma)((ae_1+be_2)+(ce_1+de_2)) = (\sigma)((a+c)e_1+(b+d)e_2) = (a+c)e_1+(b+d)e_2$$
and $$\sigma(m_1)+\sigma(m_2) = \sigma(ae_1+be_2)+ \sigma(ce_1+de_2) = ae_1+be_2+ce_1+de_2 = (a+c)e_1+(b+d)e_2.$$
We furthermore have $$(z_1+z_2\sigma+z'_1+z'_2\sigma)(m) = (z_1+z_2\sigma+z'_1+z'_2\sigma)(ae_1+be_2)$$$$ = (z_1+z'_1)(ae_1+be_1)+(z_2+z'_2\sigma)(ae_1+be_2) = (z_1+z'_1)(ae_1+be_1)+(z_2+z'_2)(ae_1+be_2) $$$$= (z_1+z_2)(ae_1+be_2)+(z'_1+z'_2)(ae_1+be_2) = (z_1+z_2\sigma)(ae_1+b_e1)+(z'_1+z'_2\sigma)(ae_1+be_1)$$
as well as $$1 \cdot m = 1 \cdot (ae_1+be_2) = ae_1+be_2$$ where $1 \in \mathbb{Z}$ is the identity in $$\mathbb{Z}[G].$$
To not overwhelm the reader, I´ll refrain from checking the last axiom, but I am quite confident it holds. Assuming it holds, we can conclude that $M = \mathbb{Z}e_1+\mathbb{Z}e_2$ is indeed an $\mathbb{Z}[G]$-module.
For the last statement, recall that we want to show that $$\bigwedge^2(M)$$ is a group of order $2$ with $e_1 \wedge e_2$ as generator.
Here, we note (I think) that $$M \otimes M$$ has basis $$\{e_i \otimes e_j\}_{i,j = 1}^{2}.$$ We note that there is a surjective $R$-module homomorphism $$\phi:M^{\otimes 2} \twoheadrightarrow \bigwedge^2(M)$$ that sends $$m_1 \otimes m_2 \longmapsto m_1 \wedge m_2.$$
Now, we can rewrite this as $$(ae_1+be_2) \otimes (ce_1+de_2) \longmapsto ae_1+be_2 \wedge ce_1+de_2 = ae_1 \wedge de_2+be_2 \wedge ce_1$$ $$ = ae_1 \wedge de_2 - ce_1 \wedge be_2 = (ad-bc)(e_1 \wedge e_2).$$
Since the map is surjective, it is clear that $$e_1 \wedge e_2$$ generates $$\bigwedge^2(M).$$
Now, since we showed that $M$ is a $\mathbb{Z}[G]$-module, it follows by properties of the tensor-product that $M^{\otimes 2}$ is an $\mathbb{Z}[G]$ module, and furthermore that the quotient-module $\bigwedge^2(M)$ is an $\mathbb{Z}[G]$-module.
Now, if I am thinking right, we then find that $$e_1 \wedge e_2 \underbrace{=}_{\text{since} \ 2e_2 \wedge e_2 = 0} (e_1+2e_2) \wedge e_2 = \sigma(e_1) \wedge e_2$$ so that $$e_1 \wedge e_2+e_1 \wedge e_2 = e_1 \wedge e_2+\sigma(e_1) \wedge e_2 = e_1 \wedge e_2 + e_1 \wedge \sigma(e_2)$$ $$= e_1 \wedge e_2 + e_1 \wedge -e_2 = e_1 \wedge e_2-e_1 \wedge e_2 = 0.$$
Now, I believe I am allowed to say that $$\sigma(e_1) \wedge e_2 = e_1 \wedge \sigma(e_2)$$ since $$\sigma(e_1) \otimes e_2 = e_1 \otimes \sigma(e_2)$$ so that for that for $\phi$ to be well-defined, this has to hold. Why is it the case that $$\sigma(e_1) \otimes e_2 = e_1 \otimes \sigma(e_2)$$ one might ask. I believe this follows from the fact that all cyclic groups $G$ are commutative, hence $\mathbb{Z}[G]$ is commutative, hence we can give $M$ the bimodule-structure $(\mathbb{Z}[G],\mathbb{Z}[G])$ so that for $r \in \mathbb{Z}[G]$ and $m \in M$ we have $r \cdot m = m \cdot r$. Then it follows that $$r (m_1 \otimes m_2) = rm_1 \otimes m_2 = m_1 \otimes rm_2.$$
Since we have already shown that $e_1 \wedge e_2$ generates $\bigwedge^2(M)$ as a $\mathbb{Z}[G]$-module, it follows that it is a group of order $2$, with the the elements $\{0,e_1 \wedge e_2\}$.
Is my thinking correct?