# Classification of all groups up to Isomorphism having no proper non trivial subgroup of finite index.

Classification of all groups up to Isomorphism having no proper non trivial subgroup of finite index :

Case $$1$$ : $$|G|=n<\infty$$

Then for any prime $$p\mid n\implies \exists H\le G$$ such that $$|H|=p$$

Then $$G$$ has no proper non trivial subgroup of finite index iff $$n=p$$ iff $$G\cong \mathbb{Z_p}$$

Case $$2$$ : $$|G|=\infty$$

Case $$2$$(a) : $$G$$ is abelian.

Theorem: Let $$G$$ be a divisible abelian group then $$G$$ has no proper subgroup of finite index.

Proof: Let $$H\le G$$ be subgroup of finite index. Since $$G$$ is abelian, $$H\trianglelefteq G$$ and since $$G/H$$ is a finite divisible group, it must be trivial i.e $$H=G$$.

Converse: Let $$G$$ be an infinite abelian group with no proper subgroup of finite index then $$G$$ is divisible.

I don't know even converse is true or not, but i think...

Suppose $$G$$ is not divisible. Then $$\exists a(\neq e) \in G$$ and $$n_0\in\Bbb{N}$$ such that there doesn't exist any $$x\in G$$ for which $$x^{n_0}=a$$

Then may be $$H\cong G/{\langle a\rangle}$$ which is a proper subgroup of $$G$$ finite index $$n_{0}$$

Question: Is the attempt correct? Or there is a known counter example.

Is there any criteria for the Isomorphism between two divisible abelian groups?

Case $$2$$(b) : Beyond my knowledge. I will keep updating my progress.

It is hopeless to expect such a classification, because of the following fact:

Fact. Let $$G$$ an infinite group having a proper subgroup of finite index. Then $$G$$ is not simple.

In particular, any infinite simple group satisfies the required property, and answering your question would imply to know how to classify infinite simple groups, which is out of reach.

Proof of the fact. Let $$G$$ be an infinite group, and let $$H$$ be a proper subgroup of finite index . Then $$G$$ acts non trivially on $$G/H$$ by $$g\cdot (g'H)=gg'H$$.

This yields a non trivial group morphism $$G\to \mathfrak{S}(G/H)$$. Now, $$G$$ is infinite, and $$\mathfrak{S}(G/H)$$ is finite (and non trivial) by assumption, so the kernel is a proper normal subgroup of $$G$$.

If $$x^n=a$$ has no solutions in an abelian group $$G$$ ($$a\in G$$), then $$a\notin G^n$$. Then the group $$G/G^n$$ is nontrivial. It obviously has subgroups of finite index.

The following theorem is known. A divisible abelian group is the direct sum of groups each of which is isomorphic either to the additive group $$\mathbb{Q}$$ of rational numbers and or to a quasicyclic group $$\mathbb{Z}(p^\infty)$$. See for example Laslo Fuchs, Abelian Groups, 2015, Chapter 4.

• I guess you are talking about abelian groups in the first paragraph? Feb 26, 2023 at 8:12
• Yes of course following the author of the question. I clarified this in my answer. Feb 26, 2023 at 8:39