I seek to prove that a group G of order 189 is not simple.
So, for contradiction, I assume G is simple. $|G|=189=3^3 7$. Now, by the Sylow theorems,
$n(7)=1+7k$ divides $3^3=27$. But this is only true when $n(7)=1$, thus G has a normal Sylow 7-subgroup, and so G is not simple.
Is this correct? This is the first time I've encountered this situation where $n(p)=1$ because $n(p)$ couldn't be $m$, which is $3^3=27$ in this case. All other examples I've done, it can usually be $m$ and then I prove it isn't by counting the number of elements.