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.