Math Help - The only non-abelian simple group of order 60 is A_5

Hi, assuming that a simple non-abelian group of order 60 can be embedded in , I want to prove that cannot contain an odd permutation and is therefore a subgroup of (and then from that I can deduce that as it is order 60 it is ). How would I go about showing this?

Originally Posted by alsn

Hi, assuming that a simple non-abelian group of order 60 can be embedded in , I want to prove that cannot contain an odd permutation and is therefore a subgroup of (and then from that I can deduce that as it is order 60 it is ). How would I go about showing this?

Consider the sign map where we're thinking of . Since is simple this is either zero or an embedding, and since I'm pretty sure it's clear that the map is zero. But, this tells us precisely that every element of is even.

Originally Posted by Drexel28

Consider the sign map where we're thinking of . Since is simple this is either zero or an embedding, and since I'm pretty sure it's clear that the map is zero. But, this tells us precisely that every element of is even.

ahhh okay. so since all the elements of G map to zero and none map to 1, does this mean that all combinations of elements of G are even? think i get it.

Are you intending to make a distinction between "elements of G" and "combinations of elements of G"? The nature of a group is that combinations of elements are just elements themselves.

Originally Posted by alsn

ahhh okay. so since all the elements of G map to zero and none map to 1, does this mean that all combinations of elements of G are even? think i get it.

Like Tinyboss points out, this doesn't make sense per se. What I proved was that every element of 's embedding in has even sign and so the embedding lives inside .