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

• Nov 22nd 2011, 02:56 AM
alsn
The only non-abelian simple group of order 60 is A_5
Hi, assuming that a simple non-abelian group $G$ of order 60 can be embedded in $S_6$, I want to prove that $G$ cannot contain an odd permutation and is therefore a subgroup of $A_6$ (and then from that I can deduce that as it is order 60 it is $A_5$). How would I go about showing this?
• Nov 22nd 2011, 07:49 AM
Drexel28
Re: The only non-abelian simple group of order 60 is A_5
Quote:

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

Consider the sign map $\text{sgn}:G\to \mathbb{Z}_2$ where we're thinking of $G\hookrightarrow S_6$. Since $G$ is simple this is either zero or an embedding, and since $60\nmid 2$ I'm pretty sure it's clear that the map is zero. But, this tells us precisely that every element of $G$ is even.
• Nov 22nd 2011, 08:10 AM
alsn
Re: The only non-abelian simple group of order 60 is A_5
Quote:

Originally Posted by Drexel28
Consider the sign map $\text{sgn}:G\to \mathbb{Z}_2$ where we're thinking of $G\hookrightarrow S_6$. Since $G$ is simple this is either zero or an embedding, and since $60\nmid 2$ I'm pretty sure it's clear that the map is zero. But, this tells us precisely that every element of $G$ 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.
• Nov 22nd 2011, 09:05 AM
Tinyboss
Re: The only non-abelian simple group of order 60 is A_5
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.
• Nov 22nd 2011, 09:13 AM
Drexel28
Re: The only non-abelian simple group of order 60 is A_5
Quote:

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 $G$'s embedding in $S_6$ has even sign and so the embedding lives inside $A_6$.