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.

Non-Abelian Simple Group is Equal to its Commutator Subgroup
Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.
Definitions/Hint.
We first recall relevant definitions.
A group is called simple if its normal subgroups are either the trivial subgroup or the group […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

Commutator Subgroup and Abelian Quotient Group
Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.
Let $N$ be a subgroup of $G$.
Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.
Definitions.
Recall that for any $a, b \in G$, the […]

A Group of Order $20$ is Solvable
Prove that a group of order $20$ is solvable.
Hint.
Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.
See the post summary of Sylow’s Theorem to review Sylow's theorem.
Proof.
Let $G$ be a group of order $20$. The […]

The Number of Elements Satisfying $g^5=e$ in a Finite Group is Odd
Let $G$ be a finite group. Let $S$ be the set of elements $g$ such that $g^5=e$, where $e$ is the identity element in the group $G$.
Prove that the number of elements in $S$ is odd.
Proof.
Let $g\neq e$ be an element in the group $G$ such that $g^5=e$.
As […]

Every Cyclic Group is Abelian
Prove that every cyclic group is abelian.
Proof.
Let $G$ be a cyclic group with a generator $g\in G$.
Namely, we have $G=\langle g \rangle$ (every element in $G$ is some power of $g$.)
Let $a$ and $b$ be arbitrary elements in $G$.
Then there exists […]

Group of Order 18 is Solvable
Let $G$ be a finite group of order $18$.
Show that the group $G$ is solvable.
Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such […]