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?

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.