A Simple Abelian Group if and only if the Order is a Prime Number

Problem 290

Let $G$ be a group. (Do not assume that $G$ is a finite group.)
Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.

A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is either the trivial group or $G$ itself.

Proof.

$(\implies)$ If $G$ is a simple abelian group, then the order of $G$ is prime.

Suppose that $G$ is a simple abelian group. Then $G$ is a nontrivial group by definition.

We first show that $G$ is a finite group.
Let $g\in G$ be a nonidentity element of $G$. Then the group $\langle g \rangle$ generated by $g$ is a subgroup of $G$. Since $G$ is an abelian group, every subgroup is a normal subgroup.

Since $G$ is simple, we must have $\langle g \rangle=G$. If the order of $g$ is not finite, then $\langle g^2 \rangle$ is a proper normal subgroup of $\langle g \rangle=G$, which is impossible since $G$ is simple.
Thus the order of $g$ is finite, and hence $G=\langle g \rangle$ is a finite group.

Let $p$ be the order of $g$ (hence the order of $G$).
Seeking a contradiction, assume that $p=mn$ is a composite number with integers $m>1, n>1$. Then $\langle g^m \rangle$ is a proper normal subgroup of $G$. This is a contradiction since $G$ is simple.

Thus $p$ must be a prime number.
Therefore, the order of $G$ is a prime number.

$(\impliedby)$ If the order of $G$ is prime, then $G$ is a simple abelian group.

Let us now suppose that the order of $G$ is a prime.

Let $g\in G$ be a nonidentity element. Then the order of the subgroup $\langle g \rangle$ must be a divisor of the order of $G$, hence it must be $p$.

Therefore we have $G=\langle g \rangle$, and $G$ is a cyclic group and in particular an abelian group.
Since any normal subgroup $H$ of $G$ has order $1$ or $p$, $H$ must be either trivial $\{e\}$ or $G$ itself. Hence $G$ is simple. Thus, $G$ is a simple abelian group.

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