Show that for any prime $p$, $\mathbb{Q}/\mathbb{Z} = H \times K$, where $H=\left \{ \frac{a}{p^n}+\mathbb{Z}:a,n \in \mathbb{Z},n\geq0 \right \}$ is the Prüfer $p$-group and $K = \{c/d + Z : c, d \in \mathbb{Z}, (d, p) = 1\}$.
What I have done. First we need to prove that $H$ and $K$ are both subgroups of $\mathbb{Q}/\mathbb{Z}$. For $H$, when $n=0$, we have that the identity lies in $H$. Let $a_1/p^{n_1}+\mathbb{Z}, a_2/p^{n_2}+\mathbb{Z} \in H$. Then
$$ \left ( \frac{a_1}{p^{n_1}}+\mathbb{Z} \right ) +\left ( \frac {a_2}{p^{n_2}}+\mathbb{Z}\right )=\left ( \frac{a_1}{p^{n_1}}+\frac {a_2}{p^{n_2}} \right )+\mathbb{Z}=\frac{a_1p^{n_2}+a_2p^{n_1}}{p^{n_1+n_2}} \in H$$
Also, if $a/p^n+\mathbb{Z}\in H$, clearly $-a/p^n+\mathbb{Z}\in H$ is its inverse. Thus $H\leq\mathbb{Q}/\mathbb{Z}$. (Here I think that actually any $r+\mathbb{Z}\in H $ such that $a/p^n-r\in \mathbb{Z}$ is an inverse(?))
In the case of $K$, when $d=1$, we can see that the identity lies in $K$. Let $c_1/d_1+\mathbb{Z}, c_2/d_2+\mathbb{Z} \in K$. Then
$$ \left ( \frac{c_1}{d_1}+\mathbb{Z} \right ) +\left ( \frac {c_2}{d_2}+\mathbb{Z}\right )=\left ( \frac{c_1}{d_1}+\frac {c_2}{d_2} \right )+\mathbb{Z}=\frac{c_1d_2+c_2d_1}{d_1d_2} +\mathbb{Z}$$
since $(d_1,p)=1$ and $(d_2,p)=1$, we have that $(d_1d_2,p)=1$, so $\frac{c_1d_2+c_2d_1}{d_1d_2} +\mathbb{Z}\in K $. Also, $-c_1/d_1+\mathbb{Z}$ is the inverse for $c_1/d_1+\mathbb{Z}$. Thus, $K\leq \mathbb{Q}/\mathbb{Z}$. About normality, since $\mathbb{Q}/\mathbb{Z}$ is abelian, $H$ and $K$ are both normal subgroups.
My problem is to show that indeed $\mathbb{Q}/\mathbb{Z}$ is the external direct product of $H$ and $K$; that is, how can I prove that $\mathbb{Q}/\mathbb{Z}=HK$ and that $H\cap K=\{0+\mathbb{Z}\}$?
The correct spelling is $\mathbb{Q}/\mathbb{Z}=H+K$.
Hints: Let $a/b\in\mathbb{Q}$ and $b=p^kd$, where $(p,d)=1$. Then there exist integers $u$ and $v$ such that $p^ku+dv=1$. Hence it follows that $$ \frac{a}{b}=\frac{va}{p^k}+\frac{ua}{d}. $$ So $$ \frac{a}{b}+\mathbb{Z}=\left(\frac{va}{p^k}+\mathbb{Z}\right)+\left(\frac{ua}{d}+\mathbb{Z}\right). $$