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.

]]>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.

]]>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.

]]>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.

]]>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**.

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].

We shall now continue to attempt to prove or otherwise discover how many more groups are not simple.