Suppose $G$ is a group of order $3k$ with $gcd(k,6) = 1$ (so $2 \nmid k, 3 \nmid k$). Why does $G$ always have a subgroup of index $3$?
If there is a subgroup of index $3$ then it will be normal - this question is about the existence of such a subgroup though.
The result actually follows from Feit-Thompson and existence of Hall subgroups but this is too overpowered. I'm looking for an simple proof, something like:
If $G$ is a counterexample with $|G|$ minimal and $H$ is a nontrivial normal subgroup with $3 \nmid |H|$, then $G/H$ has a subgroup of index $3$, and its preimage in $G$ would then be index $3$ in $G$. So WLOG every nontrivial normal subgroup of $G$ has order divisible by $3$. How can one finish the argument?