Thread: The only non-abelian simple group of order 60 is A_5

Hi, assuming that a simple non-abelian group $\displaystyle $G$$ of order 60 can be embedded in $\displaystyle $S_6$$, I want to prove that $\displaystyle $G$$ cannot contain an odd permutation and is therefore a subgroup of $\displaystyle $A_6$$ (and then from that I can deduce that as it is order 60 it is $\displaystyle $A_5$$). How would I go about showing this?

Originally Posted by alsn

Hi, assuming that a simple non-abelian group $\displaystyle $G$$ of order 60 can be embedded in $\displaystyle $S_6$$, I want to prove that $\displaystyle $G$$ cannot contain an odd permutation and is therefore a subgroup of $\displaystyle $A_6$$ (and then from that I can deduce that as it is order 60 it is $\displaystyle $A_5$$). How would I go about showing this?

Consider the sign map $\displaystyle \text{sgn}:G\to \mathbb{Z}_2$ where we're thinking of $\displaystyle G\hookrightarrow S_6$. Since $\displaystyle G$ is simple this is either zero or an embedding, and since $\displaystyle 60\nmid 2$ I'm pretty sure it's clear that the map is zero. But, this tells us precisely that every element of $\displaystyle G$ is even.

Originally Posted by Drexel28

Consider the sign map $\displaystyle \text{sgn}:G\to \mathbb{Z}_2$ where we're thinking of $\displaystyle G\hookrightarrow S_6$. Since $\displaystyle G$ is simple this is either zero or an embedding, and since $\displaystyle 60\nmid 2$ I'm pretty sure it's clear that the map is zero. But, this tells us precisely that every element of $\displaystyle G$ 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 $\displaystyle G$'s embedding in $\displaystyle S_6$ has even sign and so the embedding lives inside $\displaystyle A_6$.