Does AB=G Imply AB=BA in Group Theory?

[math8]

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

 

[tgt]

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

 

[HallsofIvy]

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

 

[sutupidmath]

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

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?

 

[morphism]

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?

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

 

[tgt]

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

it doesn't mean I can see it.

 

[sutupidmath]

ok, so what about this.


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!

 

[morphism]

ok, so what about this.


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 :

$$(a^{-1}b^{-1})^{-1}=ba \in AB$$

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:

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

$$ba=ab$$

This step is wrong. Can you see why?

 

[sutupidmath]

This step is wrong. Can you see why?

Hell YES! This happens when you assume that the same thing is going to happen in next step as well. We cannot do this, because we don't 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.