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 series.
Let $G$ be a finite group of order $n$.
If $G$ is simple, then $G \rhd \{e\}$, where $e$ is the identity element of $G$, is a composition series and we are done.

Thus, suppose that $G$ is not simple. Then it has a nontrivial proper normal subgroup.
Since $G$ is a finite group, there exists a maximal proper normal subgroup $H$.

Then the quotient $G/H$ is a simple group.
In fact, if $N$ is a proper normal subgroup of $G/H$, then the preimage of $N$ under the natural projection homomorphism $\pi:G \to G/H$ is a proper normal subgroup of $G$ containing $H$ by the fourth isomorphism theorem.
Since $H$ is maximal, the preimage $\pi^{-1}(N)$ must be $H$. This implies $N$ is trivial in $G/H$ and thus $G/H$ is simple.

Since $H$ is a proper subgroup of $G$, the order of $H$ is less than that of $G$.
Thus by the induction hypothesis, $H$ has a composition series
\[H=N_1 \rhd N_1 \rhd \cdots \rhd N_m=\{e\}.\]

The series
\[G=N_0 \rhd H=N_1 \rhd N_1 \rhd \cdots \rhd N_m=\{e\}\]
has a simple factors $N_i/N_{i+i}$, hence it is a composition series for $G$.

If the Quotient is an Infinite Cyclic Group, then Exists a Normal Subgroup of Index $n$
Let $N$ be a normal subgroup of a group $G$.
Suppose that $G/N$ is an infinite cyclic group.
Then prove that for each positive integer $n$, there exists a normal subgroup $H$ of $G$ of index $n$.
Hint.
Use the fourth (or Lattice) isomorphism theorem.
Proof. […]

Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8
Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.
Proof.
Let $G$ be a group of order $24$.
Note that $24=2^3\cdot 3$.
Let $P$ be a Sylow $2$-subgroup of $G$. Then $|P|=8$.
Consider the action of the group $G$ on […]

Infinite Cyclic Groups Do Not Have Composition Series
Let $G$ be an infinite cyclic group. Then show that $G$ does not have a composition series.
Proof.
Let $G=\langle a \rangle$ and suppose that $G$ has a composition series
\[G=G_0\rhd G_1 \rhd \cdots G_{m-1} \rhd G_m=\{e\},\]
where $e$ is the identity element of […]

Nontrivial Action of a Simple Group on a Finite Set
Let $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.
Proof.
Since $G$ acts on $X$, it […]

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 […]

Subgroup of Finite Index Contains a Normal Subgroup of Finite Index
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
Proof.
The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence […]

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 Subgroup of the Smallest Prime Divisor Index of a Group is Normal
Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.
Then prove that any subgroup of index $p$ is a normal subgroup of $G$.
Hint.
Consider the action of the group $G$ on the left cosets $G/H$ by left […]

## 1 Response

[…] You might also be interested in Any finite group has a composition series […]