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

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SUMMARY

All simple groups of order 60 are isomorphic to the alternating group A5. The discussion establishes that a simple group G of order 60 possesses a subgroup of index 5, allowing G to act on the cosets via conjugation. A homomorphism from G to S5 is constructed, leading to the conclusion that the image subgroup H must be A5, as it cannot be S5 due to the order discrepancy. Thus, G is confirmed to be isomorphic to A5.

PREREQUISITES
  • Understanding of group theory concepts, particularly simple groups
  • Familiarity with the structure and properties of the alternating group A5
  • Knowledge of homomorphisms and their implications in group actions
  • Basic understanding of cosets and subgroup indices
NEXT STEPS
  • Study the properties of simple groups and their classifications
  • Learn about the action of groups on sets and the implications for subgroup structures
  • Explore the relationship between groups and their homomorphic images, particularly in symmetric groups
  • Investigate the significance of A5 in the context of group theory and its role in the classification of finite simple groups
USEFUL FOR

Mathematicians, particularly those specializing in group theory, educators teaching abstract algebra, and students preparing for advanced studies in algebraic structures.

glacier302
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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...
 
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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...
 
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.
 
i guess you should check that A5 is the only subgroup of order 60 in S5.
 
Thank you, I think I figured it out : )
 

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