TITLE: $\Bbb{Z}/p^k \Bbb{Z} \otimes_{\Bbb{Z}} A $ is isomorphic to the Sylow $p$-subgroup of $A$
QUESTION [6 upvotes]: Let $A$ be a finite abelian group of order $n$ and let $p^k$ be the largest power of the prime $p$ dividing $n$. Then $\Bbb{Z}/p^k \Bbb{Z} \otimes_{\Bbb{Z}} A $ is isomorphic to the Sylow $p$-subgroup of $A$.
Hints on proving this?

REPLY [5 votes]: Since $A$ is finite, it is torsion, so by a theorem we have that
$$
\bigoplus_{p \ prime} A(p) \approx A
$$
where $A(p)$ is the subgroup of all elements $x$ with $ord(x) = p^e$ for some $e$, i.e. all $p$-power order elements.
Then 
$$
\Bbb{Z}/p^k \Bbb{Z} \otimes A \approx \bigoplus_{q \ prime} \Bbb{Z}/p^k \Bbb{Z} \otimes A(q)
$$
Each $A(q)$ has $q$-power order say $q^e$.  Then for all $z \otimes a \in M = \Bbb{Z}/p^k \Bbb{Z} \otimes A(q)$ for $\gcd(p,q) = 1$, we have $z \otimes a = (1\cdot z) \otimes a = 1 \otimes (z \cdot a)$.  But if $z = cq^e$, we get $1 \otimes 0 = 0\otimes 0$, showing that the whole tensor product module is $0$.  Well since $q^e$ is a unit modulo $p^k$ we have that $z   = cq^e$ always has a solution $c$.  So we're done here.
Since now we have that the direct sum is equal to $A(p)$, we're done since $A(p)$ can easily be seen to be a Sylow $p$-subgroup.