Infinite group which has every proper subgroup finite and the nontrivial semidirect product I have the following two doubts
Doubt 1 : We know that $\mathbb{Q}/ \mathbb{Z}$ and the infinite direct product $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \cdots$ are some examples of infinite groups in which every element has finite order. In fact, every element of the later group has the same finite order. For this each finite order element we can construct cyclic subgroup of finite order. But can we say from here that each proper subgroups of these groups are finite also? If answer is negative, then how to construct such groups?
Doubt 2 : I found that there are $5$ groups of order $18$ up to isomorphism. Among which there are two abelian groups namely, $\mathbb{Z}_{18}$ and $\mathbb{Z}_{3} \times \mathbb{Z}_{6}$ which is clear from the fundamental theorem of finite abelian groups and three non abelian groups. Now if I wish to find these non abelian groups using semidirect products, I could find that $\mathbb{Z}_{9} \rtimes \mathbb{Z}_{2}$ and $(\mathbb{Z}_{3} \times \mathbb{Z}_{3}) \rtimes \mathbb{Z}_{2}$ are two non abelian groups. But I am not being able to find another non abelian group using semidirect product. Regarding this matter, I wish to know that can every non abelian group be expressed as nontrivial semidirect product of groups?
 A: Both of $\mathbb Q / \mathbb Z$ and $\mathbb Z_2 \times \mathbb Z_2 \times \cdots$
have infinite, proper subgroups. For $\mathbb Q / \mathbb Z$, simply take the image under the quotient homomorphism $\mathbb Q \mapsto \mathbb Q / \mathbb Z$ of all fractions of the form $\frac{m}{2^k}$. For $\mathbb Z_2 \times \mathbb Z_2 \times \cdots$, simply take the kernel of the projection onto the first $\mathbb Z_2$ factor.
A: Regarding Doubt 1: Yes, both are torsion groups. But no, you cannot follow that. Consider the subgroup $$\lbrace 1 \rbrace \times \prod_{n \in \mathbb{N}_{>0}} C_2 \subset \prod_{n \in \mathbb{N}} C_2,$$ a proper infinite subgroup. For $\mathbb{Q}/\mathbb{Z}$ we can take the Prüfer-$p$-group $\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ for example. You just need a subgroup of $\mathbb{Q}$ that has infinitely many elements between $0$ and $1$ Containing the integers.
Regarding Doubt 2: Not every non-abelian group arises as a (non-trivial) semidirect product. An example for that is the quaternion group $Q_8$.
