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Converse of Lagrange's Theorem is false 
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#1
Mar611, 04:40 AM

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This is not a homework problem. In Gallian, there is an example given:
The group A_4 of order 12 has no subgroups of order 6. I can't seem to understand what this means in terms of how this is the "converse" of Lagrange's Theorem. 


#2
Mar611, 04:44 AM

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A converse of Lagrange's theorem would say for every divisor of the order of group, there's a subgroup of that order. It's false. Apparently.



#3
Mar611, 05:02 AM

P: 22

Hmm...okay. So here's a problem in the book that I think applies this concept. Prove that group order of 12 must have an element of order 2. This problem uses the converse, because 121,2,3,4,6,12. But by Lagrange we can say that order of 2 is definitely there. But to prove that order 3 is not a possibility, we can use the converse to make a contradiction: for every divisor of the order of the group, there's a subgroup of order 12. I hope this makes sense!



#4
Mar611, 05:53 AM

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Converse of Lagrange's Theorem is false
They didn't use a "converse to the Lagrange theorem". They used Cauchy's theorem, which is that if p is prime and divides the order of G then there is a subgroup of order p. There is no complete converse to the Lagrange theorem. Which is what they are trying to tell you. 6 isn't prime. There is a subgroup of order 3.



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