In A Course in Abstract Algebra Volume: Author(s): Vijay K. Khanna, S.K. Bhamri it is proven that:
$\mathbb{Q}$ with $+$ has no maximal normal subgroup.
Proof: Let $H$ maximal normal subgroup of $\mathbb{Q}$. Then $\mathbb{Q}/H$ is simple and so $\mathbb{Q}/H$ has no non trivial normal subgroup i.e., it will have no non trivial subgroup $(\mathbb{Q}$ being abelian, all subgroups are normal). Thus $\mathbb{Q}/H$ is a cyclic group or prime order $p$.
Let $x+H\in \mathbb{Q}/H$ by any element. Then $p(H+x)=H$ i.e., $H+px=H$ or that $px\in H$ for all $x\in \mathbb{Q}$.
Let $y\in\mathbb{Q}$ then $\frac{y}{p}\in \mathbb{Q}$. If $\frac{y}{p}=x$ then $y=px\in H$ or that $\mathbb{Q}\subseteq H\subseteq \mathbb{Q}$ a contradiction.
Question 1. Why a group $G$ simple and abelian is cyclic of orden $p$ prime? I proves the part of cyclic. Indeed, Let $G$ simple abelian, let $e\neq g\in G$ then $\langle g\rangle$ is normal in $G$ because $G$ is abelian and $\langle g\rangle=G$ because $G$ is simple. But, why $|\langle g\rangle |=p$ some $p$ prime?
Question 2. Why $px\in H$?
Thanks in advance.
Best Answer
If $|g|=pq$, $|g^p|=q$ if $|g|$ is infinite the group generated by $g^2$ is a subgroup.
For the second question, consider the quotient map $u:\mathbb{Q}\rightarrow\mathbb{Q}/H$, $u(px)=pu(x)=0$ since the order of $\mathbb{Q}/H$ is $p$. This implies that $px\in H$.