Can I show that y is in Ha\capKa?

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Let H and K be Subgroups
show Ha[itex]\cap[/itex]Ka = (H[itex]\cap[/itex]K)a for all a [itex]\in[/itex]G

pf

Let x[itex]\in[/itex]Ha[itex]\cap[/itex]Ka

Then x[itex]\in[/itex]Ha and x[itex]\in[/itex]Ka

Can I just say that x [itex]\in[/itex](H[itex]\cap[/itex]K)a ? Or am I missing something.
 
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hi Punkyc7! :smile:
Punkyc7 said:
Let x[itex]\in[/itex]Ha[itex]\cap[/itex]Ka

Then …

nooo :redface: …

your next word should be "∃" :wink:​
 
sorry, that character doesn't show up on some computers :redface:

your next words should be "there exists a y such that …" :smile:​
 
Ok so there exist a y such that y is in Ha and Ka. Then is it right?
 
Punkyc7 said:
Ok so there exist a y such that y is in Ha and Ka. Then is it right?

no!

x is in Ha and Ka, so there exists a y in … such that … ? :smile:
 
bear with me..

There exist a y in (H[itex]\cap[/itex]K)a such that x=ya?
 
is it a y in G? such that x=ya? Then I am not sure if I need the a anymore.
 
Punkyc7 said:
is it a y in G? such that x=ya?

no!

once again :smile:​
 
ok I am sure where the y is floating around but if its not in G or (H[itex]\cap[/itex]k)a. The only place left would have to be just H[itex]\cap[/itex]K right
 
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?