Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8

Normal Subgroups Problems and Solutions in Group Theory

Problem 568

Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.

 
FavoriteLoadingAdd to solve later

Sponsored Links

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 the left cosets $G/P$ by left multiplication.
This induces a permutation representation homomorphism
\[\phi: G\to S_{G/P},\] where $S_{G/P}$ is a group of bijective maps (permutations) on $G/P$.

This homomorphism is defined by
\[\phi(g)(aP)=gaP\] for $g\in G$ and $aP\in G/P$.


Then by the first isomorphism theorem, we see that
\[G/\ker(\phi) \cong \im(\phi) < S_{G/P}.\] This implies that the order of $G/\ker(\phi)$ divides the order of $S_{G/P}$. Note that as $|G/P|=3$, we have $|S_{G/P}|=|S_3|=6$. Thus, we must have $4\mid |\ker{\phi}|$.

Also note that $\ker(\phi) < P$. To see this let $x\in \ker(\phi)$. Then we have \[xP=\phi(x)(P)=\id(P)=P.\] Here $\id$ is the identity map from $G/P$ to itself. Hence $x\in P$. It follows that $|\ker(\phi)|$ divides $|P|=8$.


Combining these restrictions, we see that $|\ker(\phi)|=4, 8$.
Being the kernel of a homomorphism, $\ker(\phi)$ is a normal subgroup of $G$.
Hence the group $G$ of order $24$ has a normal subgroup of order $4$ or $8$.


FavoriteLoadingAdd to solve later

Sponsored Links

More from my site

  • 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 […]
  • A Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is NormalA Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is Normal Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer. Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$. (Michigan State University, Abstract Algebra Qualifying […]
  • 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 […]
  • Group Homomorphisms From Group of Order 21 to Group of Order 49Group Homomorphisms From Group of Order 21 to Group of Order 49 Let $G$ be a finite group of order $21$ and let $K$ be a finite group of order $49$. Suppose that $G$ does not have a normal subgroup of order $3$. Then determine all group homomorphisms from $G$ to $K$.   Proof. Let $e$ be the identity element of the group […]
  • Any Finite Group Has a Composition SeriesAny 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 […]
  • Abelian Normal subgroup, Quotient Group, and Automorphism GroupAbelian Normal subgroup, Quotient Group, and Automorphism Group Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$. Let $\Aut(N)$ be the group of automorphisms of $G$. Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime. Then prove that $N$ is contained in the center of […]
  • A Subgroup of the Smallest Prime Divisor Index of a Group is NormalA 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 […]
  • 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...

Leave a Reply

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

More in Group Theory
Normal Subgroups Problems and Solutions in Group Theory
Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4

Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$....

Close