Inner Automorphisms as a Normal Subgroup

In summary: Yes, I have tried many special cases. For example, suppose that \tau is the composition of two permutations, \tau^{-1}(g) and \tau^{-1}(h). Then \tau^{-1}(g) and \tau^{-1}(h) are both in the set of inner automorphisms, so \tau^{-1}(g) and \tau^{-1}(h) must also be in the set of inner automorphisms. However, this is not always the case. For example, if \tau is the composition of the permutations \tau^{-1} and \tau^{
  • #1
PingPong
62
0

Homework Statement



Let G be a group. We showed in class that the permutations of G which send products to products form a subgroup Aut(G) inside all the permutations. Furthermore, the mappings of the form [itex]\sigma_b(g)=bgb^{-1}[/itex] form a subgroup inside Aut(G) called the inner automorphisms and denoted Inn(g).

Prove that the inner automorphisms form a normal subgroup of Aut(G).

Homework Equations



None

The Attempt at a Solution



I attempted this problem one way, and my professor said I was going about it the wrong way - I did multiplications of the permutations and apparently I'm supposed to use function composition of the permutations. So this is what I have: I must show that, given a permutation [itex]\tau[/itex] that [itex]\tau \sigma_b \tau^{-1}[/itex] is in the set of inner automorphisms (at least, that's how we've been showing normality in class and my professor told me that this is at least correct).

So, I have
[tex]\tau\left(\sigma_b\left(\tau^{-1}\left(g\right)\right)\right)&=&\tau\left(b\tau^{-1}(g)b^{-1}\right)\\
&=&\tau(b)\tau\left(\tau^{-1}(g)\right)\tau(b^{-1})\\
&=&\tau(b)g\tau(b)^{-1}\\
&=&\sigma_{\tau(b)}(g)[/tex]

My question is, is [itex]\tau[/itex] equal to [itex]\tau^{-1}(g)[/itex] or is it [itex]\left(\tau(g)\right)^{-1}[/itex], or are these the same thing? I've done the problem assuming that [itex]\tau^{-1}[/itex] means [itex]\tau^{-1}(g)[/itex] and it seems to work, but I'm a bit uneasy about this. Can anybody confirm whether I've done this properly or not?
 
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  • #2
PingPong said:
So, I have
[tex]\tau\left(\sigma_b\left(\tau^{-1}\left(g\right)\right)\right)&=&\tau\left(b\tau^{-1}(g)b^{-1}\right)\\
&=&\tau(b)\tau\left(\tau^{-1}(g)\right)\tau(b^{-1})\\
&=&\tau(b)g\tau(b)^{-1}\\
&=&\sigma_{\tau(b)}(g)[/tex]
This is good
PingPong said:
My question is, is [itex]\tau[/itex] equal to [itex]\tau^{-1}(g)[/itex] or is it [itex]\left(\tau(g)\right)^{-1}[/itex], or are these the same thing?
[itex]\tau[/itex] is a mapping and [itex]\tau^{-1}(g)[/itex] is an element of the group on which the mapping is defined. Your question is not making sense. You have successfully shown that that the inner automorphisms are normal in the group of automorphisms. You can stop there.
 
Last edited:
  • #3
Mathdope said:
This is good

[itex]\tau[/itex] is a mapping and [itex]\tau^{-1}(g)[/itex] is an element of the group on which the mapping is defined. Your question is not making sense. You have successfully shown that that the inner automorphisms are normal in the group of automorphisms. You can stop there.

Okay, so I worded my question a bit strangely. Sorry :redface:

What I meant to ask was whether [itex]\tau^{-1}(g)=\left(\tau(g)\right)^{-1}[/itex] in general. I have a feeling that it's not and that's what I was uneasy about.
 
  • #4
PingPong said:
What I meant to ask was whether [itex]\tau^{-1}(g)=\left(\tau(g)\right)^{-1}[/itex] in general. I have a feeling that it's not and that's what I was uneasy about.
Have you tried any special cases? What happens when you apply tau to both sides?
 

Related to Inner Automorphisms as a Normal Subgroup

1. What is an inner automorphism?

An inner automorphism is a type of automorphism in abstract algebra that is defined by conjugation. This means that an inner automorphism takes an element in a group and maps it to another element by multiplying it on the left and right by a fixed element in the group.

2. What is a normal subgroup?

A normal subgroup is a subgroup of a group that is invariant under conjugation by all elements in the group. This means that if an element in the group is conjugated with an element in the normal subgroup, the result will still be in the normal subgroup.

3. How is an inner automorphism related to a normal subgroup?

An inner automorphism is a type of automorphism that is defined by conjugation, which is a property of normal subgroups. In fact, the set of all inner automorphisms forms a normal subgroup of the larger group of all automorphisms.

4. What is the significance of inner automorphisms as a normal subgroup?

The fact that inner automorphisms form a normal subgroup has important implications in group theory. It allows for the classification of groups based on the properties of their normal subgroups, and it also provides a way to study the structure of a group by looking at its normal subgroups.

5. Can you provide an example of inner automorphisms as a normal subgroup?

One example is the group of all invertible 2x2 matrices with real entries, known as GL(2,R). The set of all inner automorphisms of this group is isomorphic to the group of all real numbers under addition, and it forms a normal subgroup of GL(2,R). This example illustrates the relationship between inner automorphisms and normal subgroups.

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