
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
2)Let G be a simple group containing a subgroup of index .Prove that there is a monomorphism of the group, to the alternating group
Thanks, I hope some help

Re: Alternating Group
504 = 2^{3}*3^{2}*7
suppose, by way of contradiction, there was a monomorphism of G→A_{7}. we can thus consider G as a subgroup of A_{7}, and [A_{7}:G] = 5.
this in turn (letting A_{7} act on the coset space A_{7}/G) gives a homomorphism A_{7}→S_{5}.
since A_{7} is simple, the kernel of this homomorphism must be trivial or all of A_{7}. letting x be in A_{7}  G, we see x.G = xG ≠ G, so the action is not trivial (thus the kernel is not all of A_{7}), and A_{7} = 2,520 does not divide 120 = S_{5}, so the kernel is not trivial.
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we are given G, and H, with [G:H] = n > 2. letting G act on the coset space G/H gives a homomorphism G→S_{n}. since G is simple, and this action is nontrivial, the kernel must be trivial. so G can be regarded as a (simple) subgroup of S_{n}, 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 A_{3}, since S_{3} is not simple.
if n = 4, then G' is a simple subgroup of S_{4}. the only simple subgroups of S_{4} 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'∩A_{n}. since sgn: G'→{1,1} is a homomorphism with kernel G'∩A_{n}, we conclude G'∩A_{n} is a normal subgroup of G'. thus either G' = G'∩A_{n}, or G'∩A_{n} has index 2 in G'. but the latter condition implies G' is not simple, so we conclude that G' = G'∩A_{n}, that is: G' is a subgroup of A_{n}.
thus the monomorphism G→G' gives an injection of G into A_{n}.

Re: Alternating Group