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Math Help - Alternating Group

  1. #1
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    Chile
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    Alternating Group

    Hello, I have the following exercises


    1)Let G a group of order 504.Prove that G is not injected into the group A_7




    2)Let G be a simple group containing a subgroup of index n> 2.Prove that there is a monomorphism of the group, to the alternating group A_n


    Thanks, I hope some help
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  2. #2
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    Tejas
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    Re: Alternating Group

    504 = 23*32*7

    suppose, by way of contradiction, there was a monomorphism of G→A7. we can thus consider G as a subgroup of A7, and [A7:G] = 5.

    this in turn (letting A7 act on the coset space A7/G) gives a homomorphism A7→S5.

    since A7 is simple, the kernel of this homomorphism must be trivial or all of A7. letting x be in A7 - G, we see x.G = xG ≠ G, so the action is not trivial (thus the kernel is not all of A7), and |A7| = 2,520 does not divide 120 = |S5|, so the kernel is not trivial.

    ********

    we are given G, and H, with [G:H] = n > 2. letting G act on the coset space G/H gives a homomorphism G→Sn. since G is simple, and this action is non-trivial, the kernel must be trivial. so G can be regarded as a (simple) subgroup of Sn, G'.

    now if n = 3, then |G'| ≤ 6, and since G' has a subgroup of index 3, |G'| ≥ 3. this means G is isomorphic to A3, since S3 is not simple.

    if n = 4, then G' is a simple subgroup of S4. the only simple subgroups of S4 are cyclic of order 2, and cyclic of order 3. neither of these kinds of subgroups contain subgroups of index 4, so in this case the theorem is vacuously true.

    so without loss of generality we may take n ≥ 5. now consider G'∩An. since sgn: G'→{-1,1} is a homomorphism with kernel G'∩An, we conclude G'∩An is a normal subgroup of G'. thus either G' = G'∩An, or G'∩An has index 2 in G'. but the latter condition implies G' is not simple, so we conclude that G' = G'∩An, that is: G' is a subgroup of An.

    thus the monomorphism G→G' gives an injection of G into An.
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  3. #3
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    Re: Alternating Group

    thanks for your help!
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