Thread: 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

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.

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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 non-trivial, 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 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'∩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.

thanks for your help!