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**JaneFairfax****Member**- Registered: 2007-02-23
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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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**Ricky****Moderator**- Registered: 2005-12-04
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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.

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**JaneFairfax****Member**- Registered: 2007-02-23
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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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**JaneFairfax****Member**- Registered: 2007-02-23
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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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**Ricky****Moderator**- Registered: 2005-12-04
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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.

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**Ricky****Moderator**- Registered: 2005-12-04
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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.

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**JaneFairfax****Member**- Registered: 2007-02-23
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Ricky wrote:

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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**JaneFairfax****Member**- Registered: 2007-02-23
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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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**Ricky****Moderator**- Registered: 2005-12-04
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