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?

  • 1
    $\begingroup$ Really? Why the downvote? $\endgroup$ – math54321 Jan 15 '20 at 5:33
  • $\begingroup$ Yes I know Sylow's theorem - how does that help here? $\endgroup$ – math54321 Jan 15 '20 at 5:52
  • 2
    $\begingroup$ This follows by the Burnside normal $p$-complement theorem. en.wikipedia.org/wiki/… $\endgroup$ – 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.

  • $\begingroup$ Thanks! I guess these normal p-complement theorems is about the easiest approach possible $\endgroup$ – math54321 Jan 15 '20 at 20:05
  • $\begingroup$ Yes, I have been thinking about another approach but without any luck. $\endgroup$ – Nicky Hekster Jan 15 '20 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.