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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?

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    $\begingroup$ 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
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    $\begingroup$ This follows by the Burnside normal $p$-complement theorem. en.wikipedia.org/wiki/… $\endgroup$ – verret Jan 15 '20 at 8:50
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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.

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  • $\begingroup$ 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

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