1

I know the question has been asked here before, but i came up with a proof and was hoping that someone could tell me if it is correct.

$105=3*5*7$ by standard arguments we can show that either $n_5$ or $n_7$ are 1. Regardles, it implies that $HK$ is a subgroup where $H,K$ are some 5,7-sylow subgroups. It is normal as its index is 3. Also it must be cyclic by again, standard arguments. Now if 3-sylow is normal. Then $HKP\equiv HK\times P\equiv Z_{35}\times Z_3$ Thus $G$ is abelian, not only that, it must be $Z_{35}\times Z_3$

Is this correct?

2132123
  • 1,565

1 Answers1

0

The "if 3-Sylow is normal" is not obvious, because $n_3=7$ must be ruled out. Doing a standard counting argument doesn't seem to give it to you because (if n_p is the number of Sylow P-subgroups) $n_3=7$, $n_5=21$ and $n_7=1$ does not get more than 105 elements. Instead, it turns out the subgroup of order 35 which you found is normal. See Normal subgroup of prime index.

SamLopez
  • 33