Definition. A maximal subgroup of a group G is a proper subgroup $M$ of $G$ such that there are no subgroup $H$ of $G$ such that $M<H<G$.
Now, I want to solve the following problem:
Problem: $(\mathbb{Q},+)$ has no maximal subgroup.
There are solutions available here. But I find these hard to me…So I try to solve this own…
My Solution: Suppose that $M$ be a maximal subgroup of $(\mathbb{Q},+)$. Then by Fourth Isomorphism Theorem the subgroups of $\mathbb{Q}/M$ are only $\{0\}$ and $\mathbb{Q}/M$. And then $\mathbb{Q}/M$ must be a finite group, Since it can be proved that A group is finite if and only if it has finite number of subgroups (See here). Therefore $[\mathbb{Q}:M]=|\mathbb{Q}/M|<\infty$. Therefore $M$ becomes a proper subgroup of finite index of $(\mathbb{Q},+)$, and which, we know, is not possible. (*because, $(\mathbb{Q},+)$ has no proper subgroup of finite index). Hence we get a contradiction.
*Theorem. $(\mathbb{Q},+)$ has no proper subgroup of finite index
Proof: Suppose $H \le \mathbb{Q}$ such that $[\mathbb{Q}:H]=|\mathbb{Q}/H|=n$ (say) Now let $q \in \mathbb{Q}$. Then $(q+H)^n=H$ i.e. $nq+H=H$ i.e., $nq \in H$. Since $q$ is arbitrary in $\mathbb{Q}$ so this means $n\mathbb{Q} \subset H$. But look $n\mathbb{Q}=\mathbb{Q}$, since $q \mapsto nq$ is an automorphism of $\mathbb{Q}$. Hence $H=\mathbb{Q}$.
Is this solution correct? Thank you..
Best Answer
Your proof is good, but the first part can be simplified.
If $M$ is a maximal subgroup, the quotient $\mathbb{Q}/M$ is a simple abelian group, hence certainly finite (actually of prime order): any of its non identity element is a generator, so the group is cyclic and it cannot be infinite cyclic.
Now the task is to show the stronger statement that $\mathbb{Q}$ has no proper subgroup of finite index.
If $H$ is a subgroup of finite index $n$, then $n\mathbb{Q}\subseteq H$: indeed, for every $q\in\mathbb{Q}$, $n(q+H)=H$, which amounts to saying that $nq\in H$. However, $n\mathbb{Q}=\mathbb{Q}$, so $H=\mathbb{Q}$.
You can note that no divisible abelian group has proper subgroups of finite index, with exactly the same proof as for $\mathbb{Q}$.