Exterior Powers of finite abelian group 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!
 A: 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.
