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|G|=120, |A|=|B|=60, A, B normal in G. Then |A : A \cap B| <...

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|G|=120, |A|=|B|=60, A, B normal in G. Then |A : A \cap B| <...

Hi: Let G be a group, A, B subgroups, |G| = 120, |A| = |B| = 60, A and B normal in G. Is there any reason why the following should be true?: |A : A \cap B| < 3. Thanks.

f is not well-defined. If xB = yB, it could happen that xA and yA are different (unless you assume B \subseteq A).

Thank you for your post. I already saw that. Please see my other message.

Thanks a lot. It's clear like water. What I do not understand is this: there is a subgroup A of index 2 in S_5, and the author says: Let A_5 be the alternating group of degree 5 [nothing has been said about symmetric groups up to this point in the book]. Then also |S_5 : A_5| = 2, so A and A_5 are both normal subgroups of order 60 in S_5. IN PARTICULAR BY THE HOMOMORPHISM THEOREM |A : A \cap A_5| \leq 2. Why "in particular" and why by the homomorphism theorem?

Well I suppose he's referring to the isomorphism between A/(A \cap B) and AB/B, so same argument, AB = B or G. I just used |A \cap B||AB| = |A||B| which is more general, being true for any two subgroups, not necessarily normal.

f: G/B --> G/A, f(xB) = xA. Then f is a homomorphism and is onto, and its kernel is AB/B. By the homomorphism theorem (G/B)/(AB/B) isomorphic to G/A. Therefore G/A isomorphic to (G/B)/(A/A \cap B). Hence |(G/B)/(A/A \cap B)| = |G/A| and |B| = |A \cap B| from which |B : A \cap B| = 1. But I possitively know there is something wrong with this, for in some cases |B : A \cap B| = 2.

It is actually true for any finite group G and subgroups A and B both having index 2 in G (by the way, a subgroup of index 2 is always normal, so no need to assume that). We have |A \cap B||AB| = |A||B| thus |A|/|A \cap B| = |AB|/|B|. But AB is a subgroup of G containing B (and also A), so either AB = B or AB = G, thus |AB|/|B| = 1 or 2.

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