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Converse of Lagrange's Theorem is false

by math_nerd
Tags: converse, lagrange, theorem
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math_nerd
#1
Mar6-11, 04:40 AM
P: 22
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.
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Dick
#2
Mar6-11, 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.
math_nerd
#3
Mar6-11, 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 12|1,2,3,4,6,12. 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!

Dick
#4
Mar6-11, 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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