|G|=120, |A|=|B|=60, A, B normal in G. Then |A : A \cap B| < 3.

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.

EDIT: 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.

Re: |G|=120, |A|=|B|=60, A, B normal in G. Then |A : A \cap B| < 3.

Quote:

Originally Posted by

**ENRIQUESTEFANINI** f: G/B --> G/A, f(xB) = xA.

Shouldn't you verify that this function is well defined? It may be that there exist such that but in which case you won't have a well-defined function.

Re: |G|=120, |A|=|B|=60, A, B normal in G. Then |A : A \cap B| < 3.

Yes. The function seems to be not well defined. Instead, let M = A \cap B, f: G/M --> G/B, f(xM) = xB, Then f is well defined. And the homomorphism theorem leads me to G/B isomorphic to (G/M) / (BM/M) which gives |A \cap B| \leq |B| (\leq is "less than or equal"). That is, I get nothing.

The whole picture is this: There is a theorem that states

3.2.12 Let G be a group of order 60, which is not 5-closed. Then

(a) G is simple.

(b) The maximal subgroups of G are the normalizers of its Sylow subgroups.

When the proof ends, the author adds: Let G be as in 3.2.12 and U:= N_G(G_2) [here G_2 is some Sylow 2-subgroup]. Then |G : U| = 5, and 3.1.2 yields a monomorphism from G into the symmetric group S_5 [3.1.2 is a generalization of Cayley's theorem]. Hence G is isomorphic to a subgroup A of index 2 in S_5. Let A_5 be the alternating group of degree 5 [the reader knows nothing about symmetric groups at 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.

Fine up to here. But now he proceeds: In particular by the homomorphism theorem |A : A \cap A_5| \leq 2 [this is what I DO NOT UNDERSTAND], and the simplicity of A yields A = A_5. This shows that every group of order 60 that is not 5-closed is isomorphic to A_5.

Re: |G|=120, |A|=|B|=60, A, B normal in G. Then |A : A \cap B| < 3.

there is a theorem that goes: if G is a subgroup of S_{5} and H is a subgroup of G then either:

1. H∩A_{5} = H, or

2. [H:H∩A_{5}] = 2

(either half the elements of H are even, or all of them are even).

consider the homomorphism sgn:H→{1,-1}, where sgn(h) = 1 if h is even, sgn(h) = -1 if h is odd. what is its kernel? it is H∩A_{5} (the even elements of H).

since H/H∩A_{5} is isomorphic to a subgroup of {1,-1}, we have exactly two choices:

1. H/H∩A_{5} ≅ {1}, in which case H∩A_{5} = H (everything maps to the identity of {1,-1}).

2. H/H∩A_{5} ≅ {1,-1}, in which case H∩A_{5} is of index 2 (half the elements of H map to the identity).

Re: |G|=120, |A|=|B|=60, A, B normal in G. Then |A : A \cap B| < 3.

Someone told me how to prove it by using the identity . But I like your proof better as it makes use of the homomorphism theorem. Thanks a lot.