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I have to show that there is no simple group of order $96$ using the sylow theorems.

I know that $96 = 2^5\cdot 3$, and from the third sylow theorem $n_2 = 1$ or $3$ and $n_3 = 1$ or $4$ or $16$.

I have seen some proofs of this statement that use a so called index factorial theorem but I haven't learned about this. Is there a way to prove this using just the sylow theorems?

  • 3
    $n_3$ may also be equal to $4$.2012-05-08
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    One approach is to use my argument [here](http://math.stackexchange.com/questions/134437/subgroups-of-finite-nonabelian-simple-groups/134441#134441): if $G$ were simple and had $3$ $2$-Sylows, it would embed in $S_3$.2012-05-08
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    Or, by the same argument, if $G$ were simple and had a subgroup of index $3$ (such as a $2$-Sylow), it would embed in $S_3$. This may be what you call the index factorial theorem.2012-05-08
  • 0
    Thanks Olivier, I missed that. @ChrisEagle: That was one of the proofs I had seen but I don't really understand it. Isn't there a simpler way?2012-05-08
  • 4
    @jb88, the simpler way is for you to spent a little time and try to understand Chris's argument, which is actually quite simple!2012-05-08
  • 0
    The index factorial theorem comes up in the proof of this problem: [a PDF from google search](http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&ved=0CF0QFjAD&url=http%3A%2F%2Fwebpages.csus.edu%2F~dp555%2Fm210a%2Fhw%2Fhw09.pdf&ei=fqWpT6vBGarY2QX8xbmmAg&usg=AFQjCNGcktPhR4noA4laD2y5y-1L9LVtZw&sig2=VpB5XRV7hO1QhFsa6nEyVA)2012-05-08
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    Perhaps relevant: http://math.stackexchange.com/questions/64249/is-there-a-general-result-that-groups-of-order-2n-cdot-3-are-solvable/64254#642542012-05-08

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