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I followed the template from my last post to show that 726 isn't simple, could you tell me if it's correct?

Abelian case: $a \in G / \{1\}$. If $\langle a\rangle \neq G$, then we are done. If $\langle a\rangle = G$, then $\langle a^{66}\rangle $ is a proper normal subgroup of $G$. General case: WLOG we can assume $G \neq Z(G)$. $\langle 1\rangle \neq Z(G)$ which is a proper normal subgroup of $G$. Done. Otherwise $|Z(G)|= 1$. $$ 726 = 1 + \sum_{**}\frac{|G|}{|C_G(x)|} $$ There must be some $a\in G$ such that 11 does not divide $$ \frac{|G|}{|C_G(a)|} $$ It follows that $\frac{|G|}{|C_G(a)|} = 2 $ or $3$ or $6$ $\Rightarrow [G:C_G(a)] = 2$ or $3$ or $6$ $\Rightarrow 726 \mid2!$ or $726\mid3!$ or $726 \mid6!$.

Therefore group of order 726 is not simple.

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Your argument is correct, though a few more words of explanation might make it clearer. –  Geoff Robinson Feb 14 '12 at 8:53
    
Apparently you use a theorem saying that if $G$ has a subgroup $H$ of index $n$, then $G$ has a normal subgroup of index that is a factor of $n!$. So unless $|G|$ is a factor of $n!$, it must have a non-trivial normal subgroup. For full credit in an exam, as a teacher, I would insist that you repeat this argument each and every time. While it is a standard argument, it is not necessarily mentioned on each and every first course on group theory. In an informal setting, like a problem session, it would be ok. But as a TA I would ask a clarifying question, when you reach this point :-) –  Jyrki Lahtonen Feb 14 '12 at 9:16
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You could greatly simplify the argument for the abelian case: A finite abelian group is simple if and only if it's cyclic of prime order. –  kahen Feb 14 '12 at 9:54
    
As others have pointed out, this is definitely unclearly worded. So I'd suggest you do take a look at my answer to your previous query, where I'd written clearly as to what theorems have gone in where! :) –  user21436 Feb 14 '12 at 10:23
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up vote 4 down vote accepted

As a matter of interest, it is a well-known general fact that if $G$ is a finite group of order greater than $2$ whose order is divisible by $2$ but not $4,$ then $G$ is not simple. A proof can be found in many algebra texts. If $|G| =n$, let $\sigma: G \to S_n$ be the group homomorphism defined by Cayley, so $g\sigma$ acts by (say) right multiplication on $G$. Since $|G|$ is even, $G$ contains an element $t$ of order $2,$ by Cauchy's theorem. Then $t\sigma$ acts as a product of $\frac{n}{2}$ $2$-cycles, so acts as an odd permutation, as $\frac{n}{2}$ is odd. Hence the elements $g \in G$ such that $g\sigma$ is an even permutation form a normal subgroup of index $2$ in $G,$ and this subgroup is normal, proper and non-trivial. The result also is a consequence of Burnside's transfer theorem, but that is a relatively sophisticated tool.

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