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Mathematics 8 Online
OpenStudy (zzr0ck3r):

if H is a subgroup of G let a~b iff a^(-1)*b is in H if H is a subgroup of G let a~b iff a*b^(-1) is in H are these the same two equivalence relations? if so prove, if not give counter example.

OpenStudy (zzr0ck3r):

I don't think they are but I cant think of a good counter example.

OpenStudy (kinggeorge):

I couldn't resist giving this a try. Are \(a,b\in H\) or are they in \(G\)?

OpenStudy (zzr0ck3r):

G

OpenStudy (kinggeorge):

Then I'm pretty sure these are not the same equivalence relations. To find the counterexample, I think using \(S_4\) would work, although we might need to go to \(S_5\).

OpenStudy (kinggeorge):

If \(a=(123)\) and \(b=(234)\), then \(a^{-1}b=(134)\). But \(ab^{-1}=(124)\). So if we let \(H=\langle(134)\rangle\), then \(a^{-1}b\in H\), but \(ab^{-1}\notin H\).

OpenStudy (kinggeorge):

This all in \(G=S_4\).

OpenStudy (kinggeorge):

I'll be back in a few minutes to answer any remaining questions.

OpenStudy (zzr0ck3r):

nah good idea, I did not even think about permutations good idea:) ty again

OpenStudy (kinggeorge):

You're welcome.

OpenStudy (kinggeorge):

It's usually a good idea to try things with the symmetric group. Especially since every group is isomorphic to a subgroup of the symmetric group.

OpenStudy (zzr0ck3r):

hmm, that must be the great theorem that we are getting too. very n eet

OpenStudy (zzr0ck3r):

How do we know ab^-1 its not in H without producing all the elements?

OpenStudy (kinggeorge):

Well, H doesn't have many elements, so you can list the out. \(H=\{e,(134),(143)\}\). You could also notice that (134) doesn't act on 2, while (124) does. So the group generated by (134) can't act on 2 at all. Thus, (124) is not in H.

OpenStudy (zzr0ck3r):

Great I was just looking for some insight ty

OpenStudy (kinggeorge):

You're welcome.

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