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 […]

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 […]

Prove that a Group of Order 217 is Cyclic and Find the Number of Generators
Let $G$ be a finite group of order $217$.
(a) Prove that $G$ is a cyclic group.
(b) Determine the number of generators of the group $G$.
Sylow's Theorem
We will use Sylow's theorem to prove part (a).
For a review of Sylow's theorem, check out the […]