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I am trying to prove a simple group of order 168 has no subgroup of order 14, the hint ask me show how many Sylow 7 subgroup it has and what is the order of the normalizer of such Sylow 7 subgroup. I have showed there is 8 Sylow 7 subgroup and the order of its normalizer is 21. but I am not sure how is that related to show G has no subgroup of 14.

can anyone help? Thanks

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    $\begingroup$ Any group of order $14$ has a normal subgroup of order $7$, which is inconsistent with a Sylow $7$-normalizer having order $21$. $\endgroup$ – Derek Holt Mar 5 '14 at 17:33
  • $\begingroup$ Derek, excellent answer! +1 $\endgroup$ – Nicky Hekster Mar 5 '14 at 17:36
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Highlights (for you to check/prove):

Suppose $\;H\le G\;,\;\;|G|=168\;,\;\;G$ simple, $\;|H|=14\;$ , and let $\;P\le H\;,\;\;|P|=7\;$ , say $\;P=\langle p\rangle\;$.

Since $\;G\;$ cannot have an element of order $\;14\;^{(*)}$ , it must be that $\;H\cong D_{14}\;$ , from where we deduce that $\;\exists\,s\in H\;,\;\;ord(s)=2\;$ , s.t.

$$sps=p^{-1}\implies s\in N_G(P)$$

But this is impossible since $\;|N_G(P)|=21\;$ ...

$\;^{(*)}\;$ This is proved basically in the same way we reached the above contradiction...

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