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

**alsn** · Nov 22nd 2011, 01:56 AM

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?

**Drexel28** · Nov 22nd 2011, 06:49 AM

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.

**alsn** · Nov 22nd 2011, 07:10 AM

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.

**Tinyboss** · Nov 22nd 2011, 08:05 AM

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.

**Drexel28** · Nov 22nd 2011, 08:13 AM

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 $A_6$ .

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