Let [G:H]=2 and a, b not in H, then show that ab is in H.

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- June 3rd 2012, 01:07 PMsrv123index of subgroup
Let [G:H]=2 and a, b not in H, then show that ab is in H.

- June 3rd 2012, 03:13 PMILikeSerenaRe: index of subgroup
Hi again srv123! :)

Suppose that ab is not in H.

Then ab must be in the other coset.

Which other coset is there?

Can you draw any conclusions from there (that lead to a contradiction)? - June 4th 2012, 05:26 AMsrv123Re: index of subgroup
hello,

It has something to do with normality of H ryt?

i need the nxt clue:-(

ab is in abH if ab is not in H.. - June 4th 2012, 05:45 AMILikeSerenaRe: index of subgroup
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)? - June 4th 2012, 06:45 AMsrv123Re: index of subgroup
then b will be in H, ryt?

- June 4th 2012, 06:47 AMILikeSerenaRe: index of subgroup
Huh? :confused:

Can you expand a little? - June 4th 2012, 01:38 PMsrv123Re: index of subgroup
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... - June 4th 2012, 01:44 PMILikeSerenaRe: index of subgroup
Hmm, it seems you're on the right road.... but I don't completely get what you're doing. :confused:

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? - June 4th 2012, 01:59 PMsrv123Re: index of subgroup
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? - June 4th 2012, 02:12 PMILikeSerenaRe: index of subgroup
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.