Abstract Algebra Question

  • Thread starter Punkyc7
  • Start date
  • #1
Punkyc7
420
0
Show that [itex]\langle[/itex] a,b [itex]\rangle[/itex] = [itex]\langle[/itex] a,ab [itex]\rangle[/itex] = [itex]\langle[/itex] a^-1,b^-1 [itex]\rangle[/itex] 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:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
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
Deveno
Science Advisor
Gold Member
MHB
2,725
6
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
Punkyc7
420
0
ok but what does[itex]\langle[/itex] a,b [itex]\rangle[/itex] actually mean. I really dont understand what I am suppose to do.
 
  • #5
Deveno
Science Advisor
Gold Member
MHB
2,725
6
ok but what does[itex]\langle[/itex] a,b [itex]\rangle[/itex] actually mean. I really dont understand what I am suppose to do.

[tex]\langle a,b \rangle[/tex]

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
Punkyc7
420
0
Would I say something like

a*[itex]\langle[/itex] a,b [itex]\rangle[/itex]=[itex]\langle[/itex] a^2,ab [itex]\rangle[/itex]

but since a generates a^2 we can leave it as a. Or would this be wrong?
 
  • #7
Dick
Science Advisor
Homework Helper
26,263
619
Would I say something like

a*[itex]\langle[/itex] a,b [itex]\rangle[/itex]=[itex]\langle[/itex] a^2,ab [itex]\rangle[/itex]

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.
 
Last edited:
  • #8
Punkyc7
420
0
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
Dick
Science Advisor
Homework Helper
26,263
619
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.
 
  • #10
Punkyc7
420
0
he said there are two subgroups that are equal so they are the same subgroup...
 
  • #11
Dick
Science Advisor
Homework Helper
26,263
619
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>.
 
  • #12
Punkyc7
420
0
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?
 
  • #13
Dick
Science Advisor
Homework Helper
26,263
619
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.
 

Suggested for: Abstract Algebra Question

  • Last Post
Replies
1
Views
527
  • Last Post
Replies
6
Views
428
  • Last Post
Replies
5
Views
911
Replies
7
Views
1K
Replies
8
Views
627
Replies
8
Views
782
Replies
25
Views
996
Replies
15
Views
730
Replies
4
Views
371
Top