# Isomorphisms of Tensor products of Sylow p-subgroups

Let $A$ be a finite abelian group of order $n$ and let $p^k$ be the largest power of the prime $p$ dividing $n$. Prove that $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A$ is isomorphic to the Sylow $p$-subgroup of $A$.

Proof:
We know that, if $M$ is a finite abelian group of order $a = p_1^{\alpha _1}p_2^{\alpha _2} \cdots p_k^{\alpha _k}$ then considered as a $\mathbb{Z}$-module, $M$ is annihilated by $(a)$, the $p_i$-primary component of $M$ is isomorphic to the direct product of its Sylow subgroups.
Here, $A$ is a finite abelian group of order $n$. As a $\mathbb{Z}$-module, $A$ is the direct sum of its Sylow $p$-subgroups, say $A = \oplus _{q\in Q}A_q$ where $Q$ is the finite set of primes dividing $n$. Tensor product commute with finite direct sums, so $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A \cong (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}(\oplus_{q\in Q} A)\cong \oplus _{q\in Q}(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q$ as $\mathbb{Z}$-homomorphism.
We show if $p\neq q$, then $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q = 0$. Since $p$ and $q$ are relatively prime (in this case) in the PID $\mathbb{Z}$, then there are $a,\, b \in \mathbb{Z}$ s.t. $ap^k + bq^l =1$. Let $\overline{x}\otimes y$ be a simple tensor in $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q$. Then, $\overline{x}\otimes y = \overline{x}\cdot 1 \otimes y = \overline{x}(ap^k+bq^l)\otimes y = \overline{x}ap^k\otimes y + \overline{x}bq^l\otimes y =$ $\overline{xp^k}a\otimes y + \overline{x}b\otimes q^ly = 0\otimes y + \overline{x}b\otimes 0 = 0$. Hence, $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_q = 0$.
Now we show $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \cong A_p$. Here, every simple tensor in $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p$ can be written as $\overline{1}\otimes x$. Then any tensor, $\overline{a}\otimes x = \overline{1}a\otimes x = \overline{1}\otimes ax$. Now define $\phi :(\mathbb{Z}/p^k\mathbb{Z})\times A_p \rightarrow A_p$ by $\phi (\overline{a},\,x)=ax$. Here, if $p^k$ divides $(a-b)$ then $(a-b)x =0 \Rightarrow ax = bx$. Hence $\phi$ is well-defined. It is easy to see that $\phi$ is a $\mathbb{Z}$-balanced map, so $\phi$ induces a group homomorphism $\Phi : (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \rightarrow A_p$ and $\iota$ is the necessary natural inclusion and $\Phi \circ \iota = \phi$.

Let $(\overline{a},x)\in (\mathbb{Z}/p^k\mathbb{Z})\times_{\mathbb{Z}}A_p$ then $(\Phi \circ \iota)(\overline{a},x)=\phi(\overline{a},x) \Rightarrow \Phi (\overline{a}\otimes x) = ax$.
It is easy to see that $\Phi$ is injective. Also, $\Phi$ is surjective, since for each $x\in A_p$, $\Phi(\overline{1}\otimes x)=x$. Thus $\Phi$ is an isomorphism, i.e. $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \cong A_p$.

Hence, $(\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A \cong (\mathbb{Z}/p^k\mathbb{Z})\otimes_{\mathbb{Z}}A_p \cong A_p$.