Coset Question

1. Feb 12, 2012

Punkyc7

Let H and K be Subgroups
show Ha$\cap$Ka = (H$\cap$K)a for all a $\in$G

pf

Let x$\in$Ha$\cap$Ka

Then x$\in$Ha and x$\in$Ka

Can I just say that x $\in$(H$\cap$K)a ? Or am I missing something.

2. Feb 12, 2012

tiny-tim

hi Punkyc7!
nooo

your next word should be "∃"

3. Feb 12, 2012

Punkyc7

huh?

4. Feb 12, 2012

tiny-tim

sorry, that character doesn't show up on some computers

your next words should be "there exists a y such that …"

5. Feb 12, 2012

Punkyc7

Ok so there exist a y such that y is in Ha and Ka. Then is it right?

6. Feb 12, 2012

tiny-tim

no!!

x is in Ha and Ka, so there exists a y in … such that … ?

7. Feb 12, 2012

Punkyc7

bear with me..

There exist a y in (H$\cap$K)a such that x=ya?

8. Feb 12, 2012

tiny-tim

not quite

9. Feb 12, 2012

Punkyc7

is it a y in G? such that x=ya? Then im not sure if I need the a anymore.

10. Feb 12, 2012

tiny-tim

no!

once again

11. Feb 12, 2012

Punkyc7

ok Im sure where the y is floating around but if its not in G or (H$\cap$k)a. The only place left would have to be just H$\cap$K right

12. Feb 12, 2012

tiny-tim

yes!!!!!

now, can you see why?

13. Feb 12, 2012

Deveno

suppose y is in both Ha and Ka.

what does this mean?

it means y = ha, for some h in H, and y = ka, for some k in K.

for this h, and this k:

ha = ka.

can you think of something to do with this? (perhaps multiply both sides by something?)

*****

that is only HALF the problem, though. the other half means you suppose:

y is in (H∩K)a.

what can you do with this?