I need to know which of the following group has a proper subgroup which is NOT cyclic
$1. \mathbb{Z}_{15}\times \mathbb{Z}_{77}$
$2. S_3$
$3. (\mathbb{Z},+)$
$4.(\mathbb{Q},+)$
any finitely generated subgroup and also $(\mathbb{Z},+)$ of $4$ is cyclic so $4$ is out, $2\mathbb{Z}$ is proper cyclic subgroup of $(\mathbb{Z},+)$ so $3$ is also out. $S_3$ has proper subgroup $A_3$ of order $3$ so cyclic as it is prime order. so $2$ is also out. so our answer is $1$, and $\mathbb{Z}_5\times \mathbb{Z}_7$ or $\mathbb{Z}_5\times \mathbb{Z}_{11}$is not cyclic but a proper subgroup of $1$, am I right?
Best Answer
Hints:
$$\begin{align*}\bullet&\;\;\;\Bbb Z_n\times\Bbb Z_m\cong\Bbb Z_{mn}\;,\;\;\text{if}\;\;(m,n)=1\\{}\\ \bullet&\;\;\;\forall\,x\in G\;,\;\;\langle x\rangle\;\;\text{is a cyclic subgroup of}\;\;G\\{}\\ \bullet&\;\;\;(\Bbb Z,+)\;\;\text{is a cyclic group}\\{}\\ \bullet&\;\;\;(\Bbb Q,+)\;\;\text{ is $\bf{not}$ a cyclic group}\end{align*}$$