Normalizer subgroup proof proving the inverse

In summary, the conversation discusses the existence of an inverse in relation to the normalizer in a group. The first argument presented is based on the definition of the normalizer and a proof found in a book. The second argument follows a similar route, but it is not clear why it is considered less valid.
  • #1
cmj1988
23
0
This specifically relates more towards the argument as to why an inverse exists.

First the problem

The normalizer is defined as follows, NG(H)={g-1Hg=H} for some g in NG(H). I get why identity exists and why the operation is closed. It is in arguing that an inverse exists that I have beef. Specifically this argument:

eHe= (g-1)-1g-1Hgg-1=(g-1)-1Hg-1

so g-1 is in NG(H)

This above proof I found in the book Groups, rings, and fields by D.A.R. Wallace.

I wondering why this is more valid than this:

H=gHg-1
g-1H=Hg-1
g-1Hg=H

So g-1 is in NG(H)

I'm sorry if this question is so dense someone has an aneurysm
 
Last edited:
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  • #2
I have another proof, but I'm not sure why this is any more valid. It follows the same route as the first one I presented up to this point:

gHg^{-1}
This is equivalent to
(g^{-1})^{-1}Hg^{-1}

Therefore g^{-1} is in N_G(H)
 

Related to Normalizer subgroup proof proving the inverse

1. What is a normalizer subgroup?

A normalizer subgroup is a subgroup within a larger group that contains all the elements that commute with every element in the larger group.

2. Why is it important to prove the inverse in a normalizer subgroup?

Proving the inverse in a normalizer subgroup is important because it helps to establish the structure and properties of the original group, and can provide insight into other subgroups within the group.

3. How do you prove the inverse in a normalizer subgroup?

To prove the inverse in a normalizer subgroup, you must show that for every element in the subgroup, there exists an inverse element that satisfies the definition of a normalizer subgroup.

4. What are the conditions for a subgroup to be considered a normalizer subgroup?

A subgroup must satisfy two conditions to be considered a normalizer subgroup: 1) it must be a subgroup of the larger group, and 2) it must contain all elements that commute with every element in the larger group.

5. Can any subgroup be a normalizer subgroup?

No, not every subgroup can be a normalizer subgroup. The subgroup must satisfy the two conditions mentioned above in order to be considered a normalizer subgroup.

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