# Subgroup math properties help

1. Aug 1, 2010

### roam

1. The problem statement, all variables and given/known data

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 Ag is a subgroup of G.

3. 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 Ag (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 Ag 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?

2. Aug 1, 2010

### lanedance

Re: Subgroups

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

3. Aug 2, 2010

### roam

Re: Subgroups

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 Ag in nonempty.

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

b-1 = gbg-1

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

What can I do next?

4. Aug 2, 2010

### HallsofIvy

Re: Subgroups

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

5. Aug 3, 2010

### roam

Re: Subgroups

I made some mistakes, so I will start over.

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

ag and bg

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...

6. Aug 3, 2010

### lanedance

Re: Subgroups

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

7. Aug 3, 2010

### lanedance

Re: Subgroups

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

8. Aug 3, 2010

### roam

Re: Subgroups

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

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

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

9. Aug 3, 2010

### qbert

Re: Subgroups

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

10. Aug 3, 2010

### lanedance

Re: Subgroups

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