# Group theory question.

1. Oct 17, 2006

### mehrts

Let n be in |N. Let G denote S_n , the symmetric group on n
symbols. Let W be a subset of {1, 2, ..., n}.

Write down VERY simple
necessary and sufficient conditions on |W|,

for G_W to equal G_(W).

We know G_W < G_(W) < G , but now what ?

Last edited: Oct 17, 2006
2. Oct 17, 2006

### AKG

Care to define GW and G(W)?

3. Oct 17, 2006

### mehrts

Oops sorry..............

4. Oct 18, 2006

### AKG

Okay, this is a very simple problem. Have you tried anything?

5. Oct 18, 2006

### mehrts

We just started the topic and all I need is a hint on how to start the problem. :(

6. Oct 18, 2006

### matt grime

That jpeg is far too small to read for me. What are G_W and G_(W)? The hint is, as ever, start by writing out the definitions and what you want to prove.

7. Oct 18, 2006

### mehrts

Click on the jpeg to see a bigger picture.

8. Oct 18, 2006

### matt grime

G_W ={ g in G : g(w)=w for all w in W}

G_(W) = { g in G : g(W)=W}

why not just type it? The first is the subgroup of G that fixes W elementwise (i.e. fixes every element of W), and the second the subgroup that fixes W setwise (i.e. permutes the elements of W amongst themselves), so they're different precisely when there is something in G(W) that is not in G_W. So, writing out what the definition means that the answer is.....

9. Oct 18, 2006

### AKG

Say n = 4, W = {1,4}. What are GW and G(W)?

10. Oct 18, 2006

### mehrts

For example,

Let S = {1, 2, 3, 4}.

If W = {1}

G_W = G_(W) = {(1),(2 3 4),(2 4 3),(3 4),(4 2),(2 3)}.

IF W = {1, 2}

G_W = {(1), (3 4)}

G_(W) = {(1), (1 2), (3 4), (1 2)(3 4)}

If W = {1, 2, 3}

G_W = {(1)}.

G_(W) = {(1), (2 3), (3 1), (1 2), (1 2 3),(1 3 2)}.

So can I conclude that |W|= 1 if G_W = G_(W) ?

Last edited: Oct 18, 2006
11. Oct 19, 2006

### AKG

From this alone, no you can't conclude that. Ultimately, you can, but you have to provide a real proof. Also note that you are asked to find a necessary and sufficient condition for GW = G(W). This means you need to conclude not only that |W| = 1 if GW = G(W), but also that GW = G(W) if |W| = 1.

12. Oct 19, 2006

### mehrts

Thanks.
So their is only one neccessary and sufficient condition then ?
Yup, the second part of the question was asking to prove the conditions are neccessary and sufficient. :)

13. Oct 21, 2006

### mehrts

I think the correct answer would be that |W| = 0 or 1. Since the empty set contains the identity mapping. Is this correct ?

14. Oct 21, 2006

### AKG

You're right, |W| = 0 or 1. The empty set does not contain the identity mapping, the empty set contains nothing, that's why it's called the empty set. So prove that if |W| = 0 or 1, then GW = G(W), and also prove the converse, i.e. if GW = G(W), then |W| = 0 or 1.

15. Oct 22, 2006

### mehrts

Thanks alot. :)