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

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    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?
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    MHF Contributor Drexel28's Avatar
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    Re: The only non-abelian simple group of order 60 is A_5

    Quote Originally Posted by alsn View Post
    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.
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    Re: The only non-abelian simple group of order 60 is A_5

    Quote Originally Posted by Drexel28 View Post
    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.
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    Senior Member Tinyboss's Avatar
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    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.
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    MHF Contributor Drexel28's Avatar
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    Re: The only non-abelian simple group of order 60 is A_5

    Quote Originally Posted by alsn View Post
    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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