# Homework Help: Abstract Algebra Question

1. Feb 1, 2012

### Punkyc7

Show that $\langle$ a,b $\rangle$ = $\langle$ a,ab $\rangle$ = $\langle$ a^-1,b^-1 $\rangle$ for all a and b in a group G

I am not sure what this question is asking. Does this notation mean that a the cyclic group is generated by a,b and any combination of the two?

Last edited: Feb 1, 2012
2. Feb 1, 2012

### Dick

I don't think they said the group was cyclic and I don't think they said that any of those pairs generate the whole group. I think you are supposed to show they generate the same subgroup of G.

3. Feb 1, 2012

### Deveno

one way to show that <S> = <T> (where S and T are subsets of G), is to show every element of T is in <S>, and vice-versa.

4. Feb 1, 2012

### Punkyc7

ok but what does$\langle$ a,b $\rangle$ actually mean. I really dont understand what I am suppose to do.

5. Feb 1, 2012

### Deveno

$$\langle a,b \rangle$$

means the subgroup of G generated by a and b. this is, by definition, the smallest subgroup of G containing the set {a,b}, and will contain as subgroups the cyclic group generated by a, and the cyclic group generated by b.

you will have to use the fact that if a is in a (sub)group (as a generator, or product, or whatever), then so is a-1.

for example, it's pretty clear that ab is in the subgroup generated by a and b (ab is in <a,b> = <S>, where S = {a,b}), because ab is in this subgroup (by closure). the trick is, can you show that b is in the subgroup generated by a, and ab?

6. Feb 1, 2012

### Punkyc7

Would I say something like

a*$\langle$ a,b $\rangle$=$\langle$ a^2,ab $\rangle$

but since a generates a^2 we can leave it as a. Or would this be wrong?

7. Feb 1, 2012

### Dick

Wrong. Another way to look at <a,b> is that it is the set of all finite products of powers of a and b. Like a^3*b^(-1)*a^6*b^(-3). Etc etc etc. Change the powers and number of a's and b's anyway you like. Now pay more attention to what Deveno said in post 3.

Last edited: Feb 1, 2012
8. Feb 1, 2012

### Punkyc7

So since a and b generate some subgroup I can just say that a, ab must generate the group because any combination of the two elements will generate the subgroup

9. Feb 1, 2012

### Dick

That is so vague as to be almost meaningless. Read Deveno's post again. Deveno said something pretty specific that you seem to be missing.

10. Feb 1, 2012

### Punkyc7

he said there are two subgroups that are equal so they are the same subgroup...

11. Feb 1, 2012

### Dick

This one. E.g. show every element of {a,ab} is in <a,b> and show every element of {a,b} is in <a,ab>.

12. Feb 1, 2012

### Punkyc7

Consider the subgroup generated by <a,b>. Clearly a is in <a,b> and since a sub group is closed we know that ab is in <a,b>.

On the other hand consider the subgroup generated by <a,ab> since <a,ab> is a sub group its clear a is in <a,ab>. Also there exist and a^-1 so a^-1(ab) is in <a,ab> this implies that b is in<a,ab>.

Is that right?

13. Feb 2, 2012

### Dick

Yes, I think you are getting the idea.