1. Sep 30, 2008

### math8

If A and B are subgroups of G and AB=G does it follow that AB=BA? If yes, why?

2. Sep 30, 2008

### tgt

I can't see why you can't have BA={1}.

3. Sep 30, 2008

### HallsofIvy

Staff Emeritus
tgt, can you give an example in which AB= G and BA= {1}?

4. Sep 30, 2008

### sutupidmath

What does AB mean, i mean what operation are you performing between A and B here? is this supposed to be the same operation of the group G, or?

5. Oct 1, 2008

### morphism

AB is the set of elements of the form ab (with a in A & b in B).

6. Oct 1, 2008

### tgt

it doesn't mean I can see it.

7. Oct 1, 2008

### sutupidmath

Like said AB is the set of all elements such a is in A and b is in B. Then from this follows that also $$a^{-1}b^{-1}\in AB, since, a^{-1}\in A, b^{-1}\in B$$ this comes from the fact that both A,B are subgroups , so they do have inverses.

NOw since AB=G, it means also that the inverse of $$a^{-1}b^{-1}$$ is in AB. that is :

$$(a^{-1}b^{-1})^{-1}=ba \in AB$$ So, from here we have:

$$(a^{-1}b^{-1}) (a^{-1}b^{-1})^{-1}=e$$ or

$$(a^{-1}b^{-1}) ba=e$$ Now multiplying with a from the left side we get

$$a(a^{-1}b^{-1}) ba=ae=>b^{-1}(ba)=a$$ now multiplying with b we get

$$ba=ab$$ since ab, was chosen randomly from AB it means that AB=BA, doesn't it?

I am not sure that this is absolutely right, someone else like halls or morphism, might comment on this!

8. Oct 1, 2008

### morphism

Your proof should have stopped here - this is all you need to show. AB=BA is an equality of sets.

As for the rest of your post:
This step is wrong. Can you see why?

9. Oct 1, 2008

### sutupidmath

Hell YES! This happens when you assume that the same thing is gonna happen in next step as well. We cannot do this, because we dont know whether A, B are commutative. so by doing what i did in the last step, i should have multiplied by b from the left side, which would get us nowhere, since we would end up with ba=ba... and this is not what we wanted.

Well, i first also thought that i should stop here
$$(a^{-1}b^{-1})^{-1}=ba \in AB$$

but, like i said, wasn't sure enought, so i kept goin, but if i would have noticed earlier that my last step was nothing productive, then i probbably would have realized that this
$$(a^{-1}b^{-1})^{-1}=ba \in AB$$ is all we need.. which also makes sens, sice this shows that for every element ab in AB, where a is in a,and b is in be, also ba is in AB, which means that b is in A, and b is in B. now sice, like said, ab were randomly chosen, that is for any ab, from AB it means that AB=BA.

Last edited: Oct 1, 2008