Let [G:H]=2 and a, b not in H, then show that ab is in H.
Well, you have [G:H]=2.
This means there are exactly 2 cosets of H.
The first coset is simply H
The other coset contains both a and b (and through our assumption also ab).
Let's call the other coset aH.
So what can you conclude about b (and through our assumption about ab)?
starting from ur point, i was thinking lyk this...
b and ab both in aH; so b.(ab)^-1 is in H
hence a is in H and we reach a contradiction...
again if we suppose ab not in H, then it is in aH
so, ab=ah
now b= a^-1.a.b= a^-1.a.h = h belonging to H...contradiction...
Hmm, it seems you're on the right road.... but I don't completely get what you're doing.
If b and ab are (indeed) both in aH, why would b.(ab)^-1 be in H?
Are you applying a rule there?
Which rule?
And once you have your first contradiction, it follows that ab in not in aH.
Therefore it has to be in the only other coset, which is H.
What's the other contradiction for?
No no, the last part was just another alternate way I found to reach the cntradiction....
and about the first part,
i thought like this....
the cosets partition the group...ryt? the equivalence relation(R) on the parent group G is aRb iff ab^-1 is in H. Then we can get equivalnce classes of the right cosets; and two
members a and b of G are in same class iff ab^-1 is in H. ryt?
Now in our problm if we take H and Ha are the two cosets; then b and ab are in same class so b.(ab)^-1 should be in H...isn't it?
Ah okay.
Yep. That's right:Two members a and b of G are in same class iff ab^-1 is in H.That's an equivalent axiom for a subgroup!
It helps if you mention something like that to avoid leaving people (like me) in the dark.
And yes, if b and ab are in same class then b.(ab)^-1 should be in H.