show that $S_7$ has no subgroup of order 11 how to show that $S_7$ has no sub group of order 11
Is it a proof by contradiction, whereby, let H be the subgroup of order 11,
assume H is a subset of subset of $S_7$, then by lagrange's theorem we state it is not a subset since [...]

Contradiction would work here.  Suppose $H \leq S_7$ has order $11$.  By Lagrange's theorem, $11 = |H| \mid |S_7| = 7!$, which is impossible.  Hence, no such $H$ exists.
