Group of order $3k$ has subgroup of index $3$ - simple proof

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?

• Really? Why the downvote? – math54321 Jan 15 '20 at 5:33
• Yes I know Sylow's theorem - how does that help here? – math54321 Jan 15 '20 at 5:52
• This follows by the Burnside normal $p$-complement theorem. en.wikipedia.org/wiki/… – verret Jan 15 '20 at 8:50

If $$|G|=3k$$, with gcd$$(6,k)=1$$, this implies not only that $$3$$ is the smallest prime dividing the order of $$G$$, but also that $$3$$ is the highest power of $$3$$, dividing the order. Hence, a Sylow $$3$$-subgroup $$P$$ is cyclic. A well-know theorem (non-trivial, based on transfer theory, see for example M.I. Isaacs, Finite Group Theory, (5.14) Corollary) implies that $$G$$ has a normal $$3$$-complement, that is, there exists an $$N$$ normal in $$G$$, such that $$G=PN$$ and $$P \cap N=1$$. Hence $$|G:N|=3$$ and we are done.