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This is a spin-off from a thread in this section, in which it was proved that all finite groups of order ≤ 120 are not simple, except the trivial group, the prime-ordered groups and a certain group of order 60.
We shall now continue to attempt to prove or otherwise discover how many more groups are not simple.
Last edited by JaneFairfax (2009-05-24 23:58:42)
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I would suggest to try to come up with arguments that handle a large number of orders, rather than go one by one. For example, any group of order pq will not be simple.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Yes, Im already familiar with a lot of results that apply to various classes of orders thanks mainly to you and Humphreys. Most orders I encounter when going through the orders one by one usually fit one of these familiar results, so I simply quote the relevant result and say, Thats that order done.
The reason for going through the orders one by one is that I would like to discover the least integer n such that proving that a group of order n is not simple really gets me stumped.
Of course, knowing beforehand that a group of a given order cannot be simple can help a lot, otherwise I might be wasting my time trying to prove the impossible. For instance, I might absent-mindedly start trying to prove that group of order 360 is not simple, which would get me absolutely nowhere since 360 is the order of A[sub]6[/sub].
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Let
. There are 1 or 12 Sylow 11-subgroups. If 12, then the union of these subgroups has elements and so there cant be 22 Sylow 3-subgroups in this case. So there are 1 or 4 Sylow 3-subgroups. If 4, then the union of all the Sylow 3- and 11-subgroups has elements. The remaining 3 elements must therefore be the nonidentity elements in the unique Sylow 2-subgroup.Hence a group of order 132 is not simple.
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Jane, let's just continually modify your opening post, keep them all in one place. If we have a general argument that works for multiple orders, or if an argument for a certain order is long, make a post below it and then just reference that post number. Also, please write the prime factorization for each order in the list!
I will be cleaning up this thread periodically, removing posts which don't contain material used in the proofs. Let me know if you object to this.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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The number of Sylow p groups must divide q and be congruent to 1 (mod p). The only such number is 1, and thus the Sylow p-subgroup is normal.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Jane, let's just continually modify your opening post, keep them all in one place. If we have a general argument that works for multiple orders, or if an argument for a certain order is long, make a post below it and then just reference that post number. Also, please write the prime factorization for each order in the list!
I will be cleaning up this thread periodically, removing posts which don't contain material used in the proofs. Let me know if you object to this.
Thats fine with me.
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Id just like to point out that there is a simple group of order 168, namely the projective special linear group of degree 2 over a field with 7 elements.
This is a note to myself (if not Ricky) so that when we reach 168, I wont try and bang my head against a brick wall by trying to prove any group of order 168 not simple.
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"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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