Exercise 3.4.1 (An abelian simple group is isomorphic to $Z_p$)

Proof. Let $G$ be an abelian simple group. Since $G$ is not trivial, there is some element $x\in G$ such that $x\ne 1$, so $|x|>1$.

• Suppose that $|x|$ is finite. As $G$ is abelian, every subgroup of $G$ is normal in $G$, thus $G$ has no other subgroup that $\{1\}$ or $G$. Then $⟨x⟩$ is a subgroup of $G$ with order greater than 1, therefore $G=⟨x⟩$. Let $n$ be the order of $x$. If $n$ were not prime, $n$ would be divisible by some integer $d$, where $1<d<n$. Then $⟨x^d⟩$ is a subgroup of $G=⟨x⟩$ of order $n∕d$, where $1<n∕d<n$, so $G$ would have a non trivial proper subgroup. This is a contradiction, so $n=p$ is prime, and $G=⟨x⟩\simeq Z_p$:

  $$G\simeq Z_p(p\text{ prime)}.$$

• Suppose now that $|x|=\infty$. Then $⟨x⟩$ is a non trivial subgroup of $G$. Since $G$ is simple $G=⟨x⟩$, where $⟨x⟩\simeq \mathbb{Z}$. Since $\mathbb{Z}\ne 2\mathbb{Z}$, then $H=⟨x^2⟩$ is a non trivial proper subgroup of $G$. This is impossible, so $G$ does not contain any element of infinite order: this case does not occur.

Every abelian simple group is isomorphic to $Z_p$ for some prime number $p$. □