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Prove that there is NO simple group of order $n$ for each the following integers: $n=88, n=96, n=132.$

I am supposed to solve this using Sylows theorems somehow.

Lets start with $n = 88$ and say we have a Group $G$ with $|G|=88=8 \cdot11=2^3 \cdot 11.$

How do I go on from here?

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    $\begingroup$ I would look at the number of Sylow 11-subgroups. Sylow's third theorem should tell you how many there are. Then Sylow's second theorem should tell you something about normality. $\endgroup$ – TheNumber23 Jan 6 '15 at 15:22
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For order $88$ the $n_{11}$ must be $1\mod 11$ and divide $8$ hence $n_{11}=1$ and the Sylow is unique,hence normal.

$96=2^5\times3$. It follows directly from Burnsides $p^aq^b$ theorem, i'll think on another solution. The solution provided by Dietrich is sweet, you should look at it.

$132=11\times3\times 2^2$, $n_{11}$ must be $1$ or $12$, if it's one you're done, if it is $12$ there are $12\times10=120$ elements of order $11$. $n_3$ must be $1\bmod 3$ and divide $44$. so it must be at least $4$ if it is not $1$. If it is $4$ there are $8$ elements of order $3$. this leaves $4$ elements not of orders $11$ or $3$, this is just enough for the $4$-Sylow subgroup which is forced to be unique.

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  • $\begingroup$ for 132 do you mean n11 must be 1 or 12 so if its one we are done and if its 12* there are 12×10 elements of order 11 $\endgroup$ – cf12418 Jan 6 '15 at 17:04
  • $\begingroup$ yeah, my bad. thanks. $\endgroup$ – Jorge Fernández Hidalgo Jan 6 '15 at 17:10
  • $\begingroup$ also what about n3 must be 1 mod 3 and divide 132* instead of 88? $\endgroup$ – cf12418 Jan 6 '15 at 18:39
  • $\begingroup$ it should say $44$ since it is $132/11$. But the argument does not change. $\endgroup$ – Jorge Fernández Hidalgo Jan 6 '15 at 18:57
  • $\begingroup$ Sure, no problem. Glad to help! $\endgroup$ – Jorge Fernández Hidalgo Jan 6 '15 at 19:15

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