Show that all simple groups of order 60 are isomorphic to A5.

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Prove that if G is a simple group of order 60, then G is isomorphic to A5.

So far, I have shown that there is a subgroup of G with index 5. I know that with this information I should be able to show that G is isomorphic to A5, but I'm not sure how...
 
35
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Prove that if G is a simple group of order 60, then G is isomorphic to A5.

So far, I have shown that G has a subgroup of index 5. If I call these 5 cosets K1, K2, K3, K4, and K5, then G acts by conjugation on these cosets, and there is a homomorphism from G to S5. I'm not sure how to proceed from there...
 
11
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I think this is right.
As you said, there's a homomorphism from G to S5. The image of the G is a subgroup of S5, call it H. This subgroup can't be S5 itself, since S5 has 120 elements, and A5 has only 60. If |H| < 60 then the kernel of the homomorphism is more than just the identity, and is normal in G. But G is simple, so |H|=60, i.e. H=A5, so then G=A5.
 

mathwonk

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i guess you should check that A5 is the only subgroup of order 60 in S5.
 
35
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Thank you, I think I figured it out : )
 

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