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Its a very basic question. I had been going through this pdf where :

There are no simple groups of orders 12 or 28: For order 12, count the number of Sylow 3-subgroups. By the conditions, there are 1 or 4. If 1, it is normal, so assume there are 4. Then since these groups are of order 3 (prime), they are disjoint except for the identity, so there are 4*2=8 elements of order 3. Then there are only 4 elements left, and they must comprise the Sylow 2-subgroup, which is of order 4, and therefore it must be unique and hence normal

I am not well verse with Sylows Theorem except for its results. I have a few questions to that end:

  1. In the last line, the author concluded, that "...comprise the Sylow 2-subgroup, which is of order 4, and therefore it must be unique and hence normal" Why cant there be 2 subgroups of 2 elements each. Is it because the identity element must be there in both the groups and hence has to be ONE group? Or am I missing something?

  2. Is it enough to show if there exists one p-Sylow subgroup which is normal, then because every Sylow subgroup is conjugate to each other, so every other subgroup is normal? But how does normality ensure its simple?

Apologies, for such questions which might be otherwise are trivial Soham

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1 Answer

up vote 4 down vote accepted
  1. By the definition of Sylow $p$-subgroups, a Sylow 2-subgroup of a group of order $12$ must have order $4$.

  2. If a group has a proper nontrivial normal subgroup, then it is not simple.

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can you please elaborate a bit more on 2 .Much thanks on 1. I missed it. Can you please elaborate on the relationship of normality with simplicity –  Soham Sep 27 '12 at 1:21
    
Much thank you. Thank you for your patience –  Soham Sep 27 '12 at 1:36
    
thanks will keep in mind –  Soham Sep 29 '12 at 18:30
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