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
- 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-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
- 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=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.
FAQs
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}.
Answer: A3 is the normalizer of the alternating group A3.