Sylow $p$-subgroups of $\mathbb{Q}/\mathbb{Z}$ I want to prove that $\mathbb{Q}/\mathbb{Z}=\bigoplus_{p} S(p)$ where $S_{p}$ is the Sylow $p$-subgroup of $\mathbb{Q}/\mathbb{Z}$ $$S(p)=\lbrace [q]\in\mathbb{Q}/\mathbb{Z}\mid \exists k\in\mathbb{N}~\text{such that } p^{k}[q]=[0]\rbrace.$$
At first instance I thought to look at a prime decomposition of rationals $\frac{a}{b}=p_{1}^{\alpha_{1}}\cdots p_{r}^{\alpha_{r}}$ for integers $\alpha_{j}$ but this doesn't seem to work since we have addition $+$ in $\mathbb{Q}/\mathbb{Z}$ and not a product operation.
 A: Let $T(p)$ be the subgroup of $\mathbb{Q}$ consisting of all rationals that can be written with a denominator which is a nonnegative power of $p$.
Note that this is indeed a subgroup: it contains $0=\frac{0}{p^0}$; and if $\frac{a}{p^n}$ and $\frac{b}{p^m}$ lie in $T(p)$, then $\frac{a}{p^n}-\frac{b}{p^m} = \frac{p^{m}a - p^nb}{p^{n+m}}\in T(p)$.
I claim that $\mathbb{Q}$ is the sum (not direct sum, just sum) of the $T(p)$ with $p$ ranging over all primes. Indeed, let $\frac{a}{b}\in\mathbb{Q}$, with $b\gt 0$. If $b=1$, then  write $b=p^0$ and the element lies in $T(p)$. If $b\gt 1$, take a prime factorization of $b$, $b=p_1^{r_1}\cdots p_k^{r_k}$, with $r_i$ positive integers. I want to find integers $q_i$ such that
$$\frac{a}{b} = \frac{q_1}{p_1^{r_1}}+\cdots + \frac{q_k}{p_k^{r_k}}.$$
Let $x_i = \frac{b}{p_i^{r_i}}$, $i=1,\ldots,k$. Then the right hand side is
$$\frac{q_1}{p_1^{r_1}}+\cdots + \frac{q_k}{p_k^{r_k}} = \frac{x_1q_1}{b}+\cdots \frac{x_kq_k}{b} = \frac{x_1q_1+\cdots+x_kq_k}{b}.$$
Thus, we want to find integers $q_1,\ldots,q_k$ such that $x_1q_1+\cdots+x_kq_k = a$.
Because $\gcd(x_1,x_2,\ldots,x_k)=1$, this is possible. Thus, we can find $q_1,\ldots,q_k$ satisfying the equality, and this proves that $\frac{a}{b}$ can be written in the desired form. That is, every rational is equal to a sum of rationals whose denominators are prime powers.
We now project down to $\mathbb{Q}/\mathbb{Z}$. Because $\mathbb{Q}$ is the sum of $T(p)$, then $\mathbb{Q}/\mathbb{Z}$ is the sum of the images of the $T(p)$; note that the $T(p)$ contain $\mathbb{Z}$.
In fact, the image $T(p)$ is precisely your $S(p)$. Indeed, if $q=\frac{a}{p^k}\in T(p)$, then $p^kq\in\mathbb{Z}$, so $p^k[q]=[0]$ in $\mathbb{Q}/\mathbb{Z}$, proving that $T(p)/\mathbb{Z}$ is contained in $S(p)$. And if $[q]\in S(p)$, then there exists $k$ such that $p^kq\in\mathbb{Z}$; thus, $q = \frac{a}{p^k}$ for some $a\in\mathbb{Z}$, proving that $q\in T(p)$, so $[q]\in T(p)/\mathbb{Z}$.
This proves that $\mathbb{Q}/\mathbb{Z}$ is the sum of the $S(p)$.
To verify the sum is direct, assume that we have pairwise distinct $p_1,\ldots,p_n$, and that $\frac{a_1}{p_1^{r_1}}\in S(p_1)\cap (S(p_2)+\cdots+S(p_n))$. Then there exist integers $z,a_2,\ldots,a_n$, and nonnegative integers $r_2,\ldots,r_n$ such that
$$\frac{a_1}{p_1^{r_1}}  = \frac{a_2}{p_2^{r_2}}+\cdots+\frac{a_n}{p_n^{r_n}} + z.$$
Letting $x=p_2^{r_2}\cdots p_n^{r_n}$, this means that we can write
$$\frac{a_1}{p_1^{r_1}} = \frac{y}{x}$$
for some integer $y$. This means that $a_1x = p_1^{r_1}y$, and thus that $p_1^{r_1}\mid a_1x$. Since $\gcd(p_1,x)=1$, it follows that $p_1^{r_1}\mid a_1$, and therefore that $\frac{a_1}{p_1^{r_1}}\in\mathbb{Z}$. That is, $[\frac{a_1}{p_1^{r_1}}] = [0]$, proving that the intersection is trivial and therefore that the sum is direct.
Thus, $\frac{\mathbb{Q}}{\mathbb{Z}} = \oplus_p S(p)$. What remains is to show that the $S(p)$ are Sylow $p$-subgroups of $\mathbb{Q}/\mathbb{Z}$. But indeed, these are the maximum $p$-subgroups of $\mathbb{Q}/\mathbb{Z}$: if $[\frac{a}{b}]$ has order a power of $p$, then $\frac{p^ka}{b}\in\mathbb{Z}$, so $\frac{a}{b}$ can be written as $\frac{n}{p^k}$ for some $n\in\mathbb{Z}$. Therefore, $[\frac{a}{b}] \in T(p)/\mathbb{Z}=S(p)$. Since $S(p)$ contains every element of order a power of $p$ of $\mathbb{Q}/\mathbb{Z}$, it is the Sylow $p$-subgroup, being a maximum (and hence maximal) $p$-subgroup.
