Show that a finite abelian group is cyclic if and only if every Sylow-p-subgroups are cyclic. For the first part i used the result of "subgroup of cyclic group is cyclic", then it is clear that every sylow-p-subgroup of G is also cyclic.
But in the converse i'm a little bit stuck.
 A: First, a useful lemma: If $g,h\in G$ satisfy $gh=hg$ and $\gcd(\lvert g\rvert,\lvert h\rvert)=1$ then $\lvert gh\rvert=\lvert g\rvert\cdot\lvert h\rvert$. To see why this is the case, note that if $(gh)^m=1$ then $g^mh^m=1$ so $g^m=h^{-m}$. Let $x=g^m=h^{-m}$. Then $x=g^m\in\langle g\rangle$ and $h^{-m}\in\langle h\rangle$ so $x\in\langle g\rangle\cap\langle h\rangle$. However, $\langle g\rangle$ and $\langle h\rangle$ have coprime orders so Lagrange's theorem shows that $\langle g\rangle\cap\langle h\rangle=1$. Then $x=1$ so $g^m=h^m=1$. In particular, $m$ must be both a multiple of $\lvert g\rvert$ and a multiple of $\lvert h\rvert$. Since $\lvert g\rvert$ and $\lvert h\rvert$ are coprime, $m$ must be a multiple of $\lvert g\rvert\cdot\lvert h\rvert$. Conversely, $(gh)^{\lvert g\rvert\cdot\lvert h\rvert}=(g^{\lvert g\rvert})^{\lvert h\rvert}(h^{\lvert h\rvert})^{\lvert g\rvert}=1^{\lvert h\rvert}1^{\lvert g\rvert}=1$.
More generally, induction on $k$ shows that if $g_1,\ldots,g_k\in G$ commute and have pairwise coprime orders (meaning that $\gcd(\lvert g_i\rvert,\lvert g_j\rvert)=1$ for all $i\neq j$) then $\lvert g_1\cdots g_k\rvert=\lvert g_1\rvert\cdots\lvert g_k\rvert$.

Now onto the solution: Consider the prime factorization $\lvert G\rvert=p_1^{a_1}\cdots p_k^{a_k}$. For each $1\leq i\leq k$, let $g_i$ be a generator of a Sylow $p_i$-subgroup of $G$. By the above result,
$$\lvert g_1\cdots g_k\rvert=\lvert g_1\rvert\cdots\lvert g_k\rvert=p_1^{a_1}\cdots p_k^{a_k}=\lvert G\rvert$$
which shows that $G$ is cyclic. 
