What is the defining property of a conjugate subgroup A^g?

[roam]

Homework Statement

Let G be a group and let $$A \leq G$$ be a subgroup. If $$g \in G$$, then $$A^g \subseteq G$$ is defined as

$$A^g = \{ a^g | a \in A \}$$ where $$a^g = g^{-1}ag \in G$$

Show that A^g is a subgroup of G.

The Attempt at a Solution

I will use the one step subgroup test. First I have to identify the property that distinguishes the elements of A^g (a defining condition). I don't see what's the binary operation so I can't tell what this property is...

If I knew this property, I would prove that the identity has this property, so that A^g is nonempty. Then I'd use the assumption that a and b have that property to show that ab^-1 has this property. Could anyone help me out to see what the property of this group is?

P.S. I think the identity for this group is "1", right?

 

[lanedance]

isn't it given by: $a^g = g^{-1}ag \in G$?

then from memory you need to show:
- the identity is part of the subgroup
- its closed under multiplication, ie if u & v are both in, then so is uv
- the inverse of any element is also part of subgroup

these whould all follow pretty straight forwarly from the group definition, remembering that A is a subgroup itself

 

[roam]

isn't it given by: $a^g = g^{-1}ag \in G$?

then from memory you need to show:
- the identity is part of the subgroup
- its closed under multiplication, ie if u & v are both in, then so is uv
- the inverse of any element is also part of subgroup

these whould all follow pretty straight forwarly from the group definition, remembering that A is a subgroup itself

I want to use the "one step subgroup test". For this test I don't have to check closure. I only need to check that ab^-1 is in the group whenever a and b are in the group.

To show that identity is in the group is it sufficient to show that g^-11g=1? This verifies that A^g in nonempty.

Suppose a=g^-1ag and b=g^-1bg are elements in A^g who satisfy the property.

b^-1 = gbg^-1

ab^-1 = g^-1ag . gbg^-1

What can I do next?

 

[HallsofIvy]

I want to use the "one step subgroup test". For this test I don't have to check closure. I only need to check that ab^-1 is in the group whenever a and b are in the group.

To show that identity is in the group is it sufficient to show that g^-11g=1? This verifies that A^g in nonempty.

Suppose a=g^-1ag and b=g^-1bg are elements in A^g who satisfy the property.

b^-1 = gbg^-1

No, this is wrong. You can show that $(gbg^{-1})^{-1}= gb^{-1}g^{-1}$ directly from the definition of "inverse".

ab^-1 = g^-1ag . gbg^-1

What can I do next?

 

[roam]

I made some mistakes, so I will start over.

We write two elements of A^g in the correct form (correct form here being $$a^g = g^{-1}ag$$):

a^g and b^g

Then I must show that $$a^g(b^g)^{-1}$$ also has the correct form; that is it can be written as $$g^{-1}(ab^{-1})g$$. I start here:

$$a^{g}(b^{g})^{-1} = (g^{-1}ag)(gb^{-1}g^{-1})$$

But how does one simplify $$(g^{-1}ag)(gb^{-1}g^{-1})$$ to get $$g^{-1}(ab^{-1})g$$? This is where I'm stuck...

 

[lanedance]

why do you assume $(b^{g})^{-1} = (gb^{-1}g^{-1})$ ?

 

[lanedance]

cos its not... try multiplying with $b^{g}$

 

[roam]

why do you assume $(b^{g})^{-1} = (gb^{-1}g^{-1})$ ?

Because $$b^g = g^{-1} b g$$

therefore $$(b^g)^{-1}= (g^{-1} b g)^{-1} = g b^{-1} g^{-1}$$

cos its not... try multiplying with $$b^{g}$$

could you show me? because I don't understand what you mean here...

 

[qbert]

just remember you need to switch orders when you invert (ab)^-1 = b^-1 a^-1.

 

[lanedance]

just remember you need to switch orders when you invert (ab)^-1 = b^-1 a^-1.

or just start with b^g and just see what you would have to multiply it by to get the identity