# Another question from Elements of Abstract Algebra by Clark - transformation groups

1. Jul 6, 2006

### learningphysics

This is question 53$$\gamma$$. Given a group G of transformations that acts on X... and a subgroup of G, Go (g * x = x for all x for each g in Go), show that the quotient group G/Go acts effectively on X.

A group G "acts effectively" on X, if g * x = x for all x implies that g = e, where g is a member of G.

I don't see how the quotient group G/Go can act on X... Each member of the quotient group, is itself a set of transformations. For example, take Go which is a member of G/Go. It seems to me that Go * x (where x belongs to X) is undefined, since Go is not a one to one correspondence from X to X (each member of Go is, but Go itself isn't).

I'd appreciate any help. Thanks.

2. Jul 6, 2006

### George Jones

Staff Emeritus
Hint.

Let h be in G, so [h] = hG0 is in G/G0. Is [h]*x := h*x well defined?

3. Jul 6, 2006

### learningphysics

Hi George... Thanks for the reply.

It seems to me like it is not well defined. There may be two different functions... say h1 and h2, such that [h1] = [h2]... So is [h1]*x = h1*x or h2*x ?

4. Jul 6, 2006

### George Jones

Staff Emeritus
In general, yes. However, in this case, we know something about the G-action of elements of G0. Can this be exploited to show that G/G0 action that I gave is well defined?

5. Jul 6, 2006

### matt grime

You should try to verify this yourself, it is quite straightfoward. What does [h1]=[h2] mean? That there is a k in Go such that h1k=h2. What was the definition of Go?

6. Jul 6, 2006

### learningphysics

Ah.... I see now... h1*x = h2*x. Thanks George and Matt.

Last edited: Jul 6, 2006
7. Jul 6, 2006

### George Jones

Staff Emeritus
Careful - this isn't isn't necessarily true. But what is true?

Maybe you just made a typo.

8. Jul 6, 2006

### learningphysics

Hmm.... If [h1]=[h2] then there's a k in Go such that h1= h2k so

h1 * x = h2k * x
h1 * x = h2 * (k * x), then since k is in Go, k * x = x

h1 * x = h2 * x

I'm probably making a really stupid mistake somewhere. Sorry guys... I appreciate the patience.

9. Jul 6, 2006

### George Jones

Staff Emeritus
Sorry - my mistake.

Edit: I was thinking of the end result, i.e, the G/Go action.

10. Jul 6, 2006

### mathwonk

if you call all the elements of G that act trivially, "the identity", then the only elements that act trivially after that are called the identity.