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What are the usual tricks is proving that a group is not simple? (Perhaps a link to a list?)

Also, I may well be being stupid, but why if the number of Sylow p groups $n_p=1$ then we have a normal subgroup?

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For your second question, you should notice that two subgroups are conjugate only if they are of the same order. So if I were to take a Sylow $p$-subgroup, $P$ say, and conjugate it by an element, $g\in G$, then $g^{-1}Pg$ is a subgroup of $G$ and it has the same order as $P$. As Sylow $p$-subgroups are defined by their order, if there is only one Sylow $p$-subgroup then $g^{-1}Pg=P$ for all $g\in G$. Thus, $P\lhd G$. –  user1729 Jan 25 '12 at 11:27
    
Thanks, @user1729, I have forgotten about that! –  John Jan 25 '12 at 11:34

2 Answers 2

Answering your first question exhaustively, informally, would mean, classifying all finite simple groups which was completed only in 2004, the claimed proof lying scattered among the several thousand pages of many journals.

However, second question is trivial: Sylow's theorem tell you that two Sylow $p$-subgroups are conjugate to each other. If there is only Sylow $p$-subgroup,say $H$, all its conjugates coincide. More specifically, $$gHg^{-1}=H~~~\forall~~g \in G$$

which is the definition for normality of a subgroups.

For the first question, I can however give you a nice reference: Dummit and Foote's Abstract Algebra. There are several techniques and theorems discussed and proved!

For a beginner, I'd personally suggest, he spends time learning, Poincare's Lemma that limits the cardinality of intersection of Conjugate subgroups; Sylow's Theorem. They can break ice in most situations. Look here for interesting stuff in this direction.

My personal experience: You can classify all simple groups upto 351 (I have done only till this number) with the help of these two theorems and looking at the conjugacy action of the group on the set of Sylow $p$-subgroups. I am not sure, if I needed other recipes here, but mostly not!

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Thank you. I am really just looking for the common tricks just as Using Sylow's to arrive at a contradiction in the number of elements, etc. Could you possibly help me out? Thanks again. –  John Jan 25 '12 at 11:32
    
I think Kannappan did help you out, with the pointer to Dummit and Foote, to Poincare, and that other link - if you have checked out all of those, and you're still not satisfied, maybe then it will be time to come back and ask Kannappan for more help. –  Gerry Myerson Jan 25 '12 at 11:59

To answer the second question, a subgroup is normal iff it coincides with its conjugates. The conjugate of a subgroup has the same cardinality as the subgroup. Hence every conjugate of a Sylow subgroup must be a Sylow subgroup. So, a $p$-Sylow subgroup is normal iff it is the only $p$-Sylow subgroup.

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Thanks! have forgotten about the cardinality rule. –  John Jan 25 '12 at 11:33

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