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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!

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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.

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