Is \langle a,b \rangle Equal to \langle a,ab \rangle in a Group?

[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 GI 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?

 

[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.

 

[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.

 

[Punkyc7]

ok but what does$\langle$ a,b $\rangle$ actually mean. I really don't understand what I am suppose to do.

 

[Deveno]

ok but what does$\langle$ a,b $\rangle$ actually mean. I really don't understand what I am suppose to do.

$$\langle a,b \rangle$$

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?

 

[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?

 

[Dick]

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?

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.

 

[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

 

[Dick]

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

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.

 

[Punkyc7]

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

 

[Dick]

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.

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>.

 

[Punkyc7]

ok, how about this
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?

 

[Dick]

ok, how about this
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?

Yes, I think you are getting the idea.