To make a subgroup normal in a group, we need to consider its normalizer. In this article, we will learn about normalizer of a subgroup with applications.

Table of Contents

## Normalizer of a Subgroup

Let H be a subgroup of a group G. Then the normalizer of H in G is defined as follows:

N(H) = {g∈G: gHg^{-1}=H}.

We will prove that

- N(H) is a subgroup of G.
- H is a normal subgroup of N(H).

## Normalizer is a Subgroup Proof

Let G be a group and H be a subgroup of G. First, we prove that the normalizer N(H) is a subgroup of G. Let a, b ∈ N(H). We will show that ab, a^{-1} ∈ N(H).

As a, b ∈ N(H), by the definition of the normalizer, we have that

aHa^{-1}=H and bHb^{-1}=H **…(*)**

Now, abH(ab)^{-1}

= abHb^{-1}a^{-1} as we know that (ab)^{-1}=b^{-1}a^{-1}

= a(bHb^{-1})a^{-1}

= aHa^{-1} = H **by (*)**

Thus, abH(ab)^{-1} = H implies that ab ∈ N(H).

Now, as aHa^{-1}=H, we have that H=a^{-1}Ha.

⇒ a^{-1}H(a^{-1})^{-1} = H. This is because (a^{-1})^{-1}=a.

Hence, a^{-1} ∈ N(H).

Thus, we have shown that ab, a^{-1} ∈ N(H) for all a, b ∈ N(H). This proves that N(H) is a subgroup of G.

## Prove that H is Normal in N(H)

Let h ∈ H be an element. As H is a subgroup, we must have that

hHh^{-1} = H.

⇒ h ∈ N(H)

Hence, H ⊆ N(H).

We have shown above that N(H) is a subgroup of G. As H ⊆ N(H), we conclude that H is a subgroup of the normalizer N(H).

Let a ∈ N(H) be an element of the normalizer of H. Then by the definition of N(H), we have that

aHa^{-1}=H.

This relation holds for any element a in N(H). Hence, we deduce that H is a normal subgroup of the normalizer N(H) of H.

## Properties of Normalizer of Groups

- If H is a subgroup of G, then the normalizer N(H) is the largest subgroup of G in which H is normal.
- If H is a normal subgroup of G, then the normalizer of H is H itself. That is, N(H)=H.
- If a finite group G contains exactly one subgroup H of a particular order, then that subgroup H is normal in G. This is because both gHg
^{-1}and H have the same order and as G contains exactly one subgroup of order |H|, we conclude that gHg^{-1}=H for all g in G. Thus, H is a normal subgroup of G.

## Normalizer of a Group Example

Let G=S_{3} be the permutation group of order 6. We know that S_{3} has only one subgroup of order 3 which is the Alternating group A_{3}. Then by the above Property3, we can say that A_{3} is a normal subgroup of S_{3}.

Now, by the above Property2, we obtain that the normalizer of the Alternating group A_{3} is A_{3} itself. In other words, N(A_{3})=A_{3}.

## FAQs

**Q1: What is the normalizer in group theory?**

Answer: If H is a subgroup of G, then the normalizer of H, denoted N(H), is defined by the set N(H) = {g∈G: gHg^{-1}=H}.

**Q2: What is the normalizer of the alternating group A**

_{3}?Answer: A_{3} is the normalizer of the alternating group A_{3}.