Any Subgroup of Index 2 in a Finite Group is Normal

Normal Subgroups Problems and Solutions in Group Theory

Problem 16

Show that any subgroup of index $2$ in a group is a normal subgroup.

LoadingAdd to solve later

Sponsored Links


Hint.

  1. Left (right) cosets partition the group into disjoint sets.
  2. Consider both left and right cosets.

Proof.

Let $H$ be a subgroup of index $2$ in a group $G$.
Let $e \in G$ be the identity element of $G$.

To prove that $H$ is a normal subgroup, we want to show that for any $g\in G$, $gH=Hg$.
If $g \in H$, then this is true. So we assume that $g \not \in H$.

Note that left cosets partition $G$ into two disjoint sets since the index is $2$.
Since $g \not \in H$, these are $eH$ and $gH$. (If $gH=H$, then $g \in H$.)

Similarly right cosets partition $G$ into two disjoint sets.
These disjoint right cosets are $He$ and $Hg$.

Because of these partitions, we have as sets
\[gH=G – eH=G-H=G-He=Hg.\] Therefore $H$ is a normal subgroup in $G$.


LoadingAdd 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 […]
  • Subgroup Containing All $p$-Sylow Subgroups of a GroupSubgroup Containing All $p$-Sylow Subgroups of a Group Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$. Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$. Then show that $N$ contains all $p$-Sylow subgroups of […]
  • Normal Subgroup Whose Order is Relatively Prime to Its IndexNormal Subgroup Whose Order is Relatively Prime to Its Index Let $G$ be a finite group and let $N$ be a normal subgroup of $G$. Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$. (a) Prove that $N=\{a\in G \mid a^n=e\}$. (b) Prove that $N=\{b^m \mid b\in G\}$.   Proof. Note that as $n$ and […]
  • Quotient Group of Abelian Group is AbelianQuotient Group of Abelian Group is Abelian Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$. Then prove that the quotient group $G/N$ is also an abelian group.   Proof. Each element of $G/N$ is a coset $aN$ for some $a\in G$. Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]
  • If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd OrderIf a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Let $G$ be a finite group of order $2n$. Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities […]
  • The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$ Let $p$ be a prime number. Let $G$ be a non-abelian $p$-group. Show that the index of the center of $G$ is divisible by $p^2$. Proof. Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$. Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]
  • The Preimage of a Normal Subgroup Under a Group Homomorphism is NormalThe Preimage of a Normal Subgroup Under a Group Homomorphism is Normal Let $G$ and $G'$ be groups and let $f:G \to G'$ be a group homomorphism. If $H'$ is a normal subgroup of the group $G'$, then show that $H=f^{-1}(H')$ is a normal subgroup of the group $G$.   Proof. We prove that $H$ is normal in $G$. (The fact that $H$ is a subgroup […]
  • The Product of a Subgroup and a Normal Subgroup is a SubgroupThe Product of a Subgroup and a Normal Subgroup is a Subgroup Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. The product of $H$ and $N$ is defined to be the subset \[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\] Prove that the product $H\cdot N$ is a subgroup of […]

You may also like...

Leave a Reply

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

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Group Theory Problems and Solutions in Mathematics
The Center of a p-Group is Not Trivial

Let $G$ be a group of order $|G|=p^n$ for some $n \in \N$. (Such a group is called a $p$-group.)...

Close