Any Finite Group Has a Composition Series

Group Theory Problems and Solutions

Problem 122

Let $G$ be a finite group. Then show that $G$ has a composition series.

 
LoadingAdd to solve later
Sponsored Links

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 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 $G$.

Related Question.

You might also be interested in
Infinite cyclic groups do not have composition series


LoadingAdd to solve later

Sponsored Links

More from my site

  • If the Quotient is an Infinite Cyclic Group, then Exists a Normal Subgroup of Index $n$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 8Every 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 SeriesInfinite 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 SetNontrivial 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 IsomorphicNormal 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 IndexSubgroup 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 GroupsIsomorphism 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 […]
  • Fundamental Theorem of Finitely Generated Abelian Groups and its applicationFundamental 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 […]

You may also like...

1 Response

  1. 09/27/2016

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

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Group Theory Problems and Solutions in Mathematics
Group of Order 18 is Solvable

Let $G$ be a finite group of order $18$. Show that the group $G$ is solvable.  

Close