Let $n$ be a positive integer.
Since $G/N$ is a cyclic group, let $g$ be a generator of $G/N$.
So we have $G/N=\langle g\rangle$.
Then $\langle g^n \rangle$ is a subgroup of $G/N$ of index $n$.

By the fourth isomorphism theorem, every subgroup of $G/N$ is of the form $H/N$ for some subgroup $H$ of $G$ containing $N$.
Thus we have $\langle g^n \rangle=H/N$ for some subgroup $H$ in $G$ containing $N$.

Since $G/N$ is cyclic, it is in particular abelian.
Thus $H/N$ is a normal subgroup of $G/N$.

The fourth isomorphism theorem also implies that $H$ is a normal subgroup of $G$, and we have
\begin{align*}
[G:H]=[G/N : H/N]=n.
\end{align*}
Hence $H$ is a normal subgroup of $G$ of index $n$.

Any Finite Group Has a Composition Series
Let $G$ be a finite group. Then show that $G$ has a composition series.
Proof.
We prove the statement by induction on the order $|G|=n$ of the finite group.
When $n=1$, this is trivial.
Suppose that any finite group of order less than $n$ has a composition […]

If Quotient $G/H$ is Abelian Group and $H < K \triangleleft G$, then $G/K$ is Abelian
Let $H$ and $K$ be normal subgroups of a group $G$.
Suppose that $H < K$ and the quotient group $G/H$ is abelian.
Then prove that $G/K$ is also an abelian group.
Solution.
We will give two proofs.
Hint (The third isomorphism theorem)
Recall the third […]

Normal Subgroups, Isomorphic Quotients, But Not Isomorphic
Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.
Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.
Proof.
We give a […]

Isomorphism Criterion of Semidirect Product of Groups
Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism.
The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation
\[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\]
where $a_i […]

A Simple Abelian Group if and only if the Order is a Prime Number
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.
Definition.
A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]

If the Quotient Ring is a Field, then the Ideal is Maximal
Let $R$ be a ring with unit $1\neq 0$.
Prove that if $M$ is an ideal of $R$ such that $R/M$ is a field, then $M$ is a maximal ideal of $R$.
(Do not assume that the ring $R$ is commutative.)
Proof.
Let $I$ be an ideal of $R$ such that
\[M \subset I \subset […]

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 […]

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 […]