No simple groups of order 4851 or 5145 exist
Prove that there are no simple groups of order 4851 or 5145.
- Note that
. Suppose 
is a simple group of order 4851. By Sylow’s Theorem, we have 
and 
. Note that 
does not divide 
, so that no subgroup of has index less than or equal to 10. In particular, ![n_3(G) = [G:N_G(P_3)] \neq 7](https://web.archive.org/web/20150513005038im_/https://s0.wp.com/latex.php?latex=n_3%28G%29+%3D+%5BG%3AN_G%28P_3%29%5D+%5Cneq+7&bg=ffffff&fg=000000&s=0 "n_3(G) = [G:N_G(P_3)] \neq 7")
, so that 
. Let 
. We have 
. By Cauchy, there exists a Sylow 11-subgroup 
. Now 
is a subgroup, and since 3 does not divide 10 and 11 does not divide 8, by this previous exercise, 
is abelian. Thus we have 
. But we also know that 
, a contradiction.
- Note that
. Suppose is a simple group of order 5145. Note that does not divide 
since the largest power of 7 dividing is 
. Thus has no proper subgroups of index less than or equal to 20. In particular, from Sylow’s Theorem we have 
, a contradiction.
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