Normalizer of a Group: Definition, Example, Subgroup Proof

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.

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

1. N(H) is a subgroup of G.
2. 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^-1a^-1 as we know that (ab)^-1=b^-1a^-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^-1Ha.

⇒ a^-1H(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

1. If H is a subgroup of G, then the normalizer N(H) is the largest subgroup of G in which H is normal.
2. If H is a normal subgroup of G, then by definition the normalizer of H is the group G itself. That is, N(H)=G.
3. 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=S₃ be the permutation group of order 6. We know that S₃ has only one subgroup of order 3 which is the Alternating group A₃. Then by the above Property3, we can say that A₃ is a normal subgroup of S₃.

Now, by the above Property2, we obtain that the normalizer of the Alternating group A₃ is A₃ itself. In other words, N(A₃)=A₃.

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