Let $A$ be a finite $\mathbb{Z}$-module (i.e., a finite abelian group). My question is: for what $n\in \mathbb{Z}^{n\geq 2}$ the map 
\begin{align}
\alpha\_{n}:\bigwedge^nA&\to A^{\otimes n}\\
a\_1\wedge \cdots \wedge a\_n&\mapsto \sum\_{\pi\in \mathbb{S}\_n}(sig(\pi))a\_{\pi(1)}\otimes \cdots\otimes a\_{\pi(n)}
\end{align}
 is injective? For $n=2$, $\alpha\_2$ is injective. It follows, for example, just by induction on $r$, if $A=\mathbb{Z}/\mathbb{Z}m\_1\oplus \cdots \oplus\mathbb{Z}/\mathbb{Z}\_{m\_r}$.


Thanks!



Exterior Powers of finite abelian group
I think that this is always true. We can write $A$ as $A\_1\oplus\dotsb\oplus A\_r$, where each $A\_i$ is cyclic. Let $I(n)$ denote the set of sequences $(i\_1,\dotsc,i\_n)$ with $1\leq i\_1<i\_2<\dotsb <i\_n\leq r$. For $i\in I^n$ put $A(i)=A\_{i\_1}\otimes\dotsb\otimes A\_{i\_n}$, so $A^{\otimes n}=\bigoplus\_{i\in I^n}A(i)$. Put $A[n]=\bigoplus\_{i\in I(n)}A(i)$, so there is an evident inclusion $\iota\colon A[n]\to A^{\otimes n}$ and projection $\pi\colon A^{\otimes n}\to A[n]$. There is also a map $\mu\colon A^{\otimes n}\to\Lambda^nA$ sending $a\_1\otimes\dotsb\otimes a\_n$ to $a\_1\wedge\dotsb\wedge a\_n$. Because $A\_i$ is cyclic we have $\Lambda^2A\_i=0$, so $\Lambda^\*A\_i=\mathbb{Z}\oplus A\_i$. Moreover, for any $B$ and $C$, the multiplication map $\Lambda^\*(B)\otimes\Lambda^\*(C)\to\Lambda^\*(B\oplus C)$ is an isomorphism. Using this, we see that the map $\mu\iota\colon A[n]\to\bigwedge^nA$ is an isomorphism. On the other hand, it is also easy to see that the composite $\pi\alpha\mu\iota$ is the identity. It follows that $\alpha$ is injective.


