I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem in this book:
On p.80
Problem 8:
Show that a simple abelian group $G \neq \{e\}$ is a cyclic group whose order is prime.
Did the author intend the following problem?
If this problem were the following, I could solve it:
Problem 8':
Show that a simple finite abelian group $G \neq \{e\}$ is a cyclic group whose order is prime.
Proof:
Because $G \neq \{e\}$, there is an element $g \in G$ which is not equal to $e$.
$H := \{g^i | i \in \mathbb{Z}\}$ is a subgroup of $G$ and $G$ is abelian.
So $H$ is a normal subgroup of $G$.
And $G$ is simple.
And $H \ni g \ne e$.
So $H = G$.
Let $n := \#H = \#G$.
Then, $n$ is prime.
If $n$ is not prime, then we can write $n = d d'$, $1 < d < n$, $1 < d' < n$.
Then $H' := \{(g^d)^i | i \in \mathbb{Z}\}$ is a subgroup of $G$ whose order is $d'$.
So, $H'$ is neither $\{e\}$ nor $G$.
Becasue $G$ is abelian, $H'$ is a normal subgroup of $G$.
And $G$ is simple.
This is a contradiction.
Best Answer
Suppose that $G$ is simple Abelian, if $x$ in $G$ has an infinite order the group generate by $2x$ is a proper normal subgroup since it does not contain $x$.
If $x$ has a finite order, there exists $y$ not in $H$ the group generated by $x$, and $H$ is normal and proper.