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I have already shown that, by Sylow's third theorem, $p = 2$ or $p = 3$. I also believe that the number of $7$-Sylow subgroups should be $2^{3k}$ for $k = 0,1,2,\dots$ if $p = 2$ or $3^{6k}$ for $k = 0,1,2,\dots$ if $p = 3$. However, I don't know where to go from here to hopefully derive a contradiction, and force the existence of $1$ Sylow $7$-subgroup or $1$ Sylow $p$-subgroup. Any hints would be appreciated.

Zubin Mukerjee
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John
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Either the $p$-Sylow subgroup is normal (then the group is not simple), or there are 7 conjugates. Then there is a nontrivial homomorphism $G\to S_7$ (conjugation action, image acts transitively). If we assume the group is simple it must be isomorphic to the image of the homomorphism. But the only divisors of 7! that have the form $7\cdot p^n$ with $p=2,3$ are $14, 21, 28, 56, 63, 112$. But the number of 7-Sylow subgroups would have to be in $1,8,15,...$, leaving 56 as only possibility. In this case there would be 8 Sylow subgroups with $8\cdot 6=48$ elements of order $7$. But then there is only element space for one 2-Sylow subgroup.

ahulpke
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  • Thanks! Just one concern though, are 14, 21, 28, 63, and 112 not immediately ruled out because the smallest power of 2 congruent to 1 mod 7 is 2^3 and the smallest power of 3 congruent to 1 mod 7 is 3^6? Is that what you meant by standard Sylow counting? – John Dec 23 '14 at 01:50
  • I've modified the end of the answer -- I hope this explains better. – ahulpke Dec 23 '14 at 15:47