Problem: Let $G$ be a group of order $108 = 2^23^3$. Prove that $G$ has a proper normal subgroup of order $n \geq 6$.

My attempt: From the Sylow theorems, if $n_3$ and $n_2$ denote the number of subgroups of order $27$ and $4$, respectively, in $G$, then $n_3 = 1$ or $4$, since $n_3 \equiv 1$ (mod $3$) and $n_3~|~2^2$, and $n_2 = 1, 3, 9$ or $27$, because $n_2~|~3^3$.

Now, I don't know what else to do. I tried assuming $n_3 = 4$ and seeing if this leads to a contradiction, but I'm not even sure that this can't happen. I'm allowed to use only the basic results of group theory (the Sylow theorems being the most sophisticated tools).

Any ideas are welcome; thanks!

  • $\begingroup$ Are you allowed to use group actions? $\endgroup$ – Ink Aug 3 '12 at 0:04
  • $\begingroup$ @Brian if you mean things like the Class Equation, then yes. $\endgroup$ – student Aug 3 '12 at 0:30

Let $\,P\,$ be a Sylow $3$-subgroup. of $\,G\,$ and let the group act on the left cosets of $\,P\,$ by left (or right) shift. This action determines a homomorphism of $\,G\,$ on $\,S_4\,$ whose kernel has to be non-trivial (why? Compare orders!) and either of order $27$ or of order $9$ (a subgroup of $ \, P \, $, say) , so in any case the claim's proved.

  • $\begingroup$ Yes thanks, it was a typo and has already been corrected: it is a Sylow 3 subgroup. Whether the kernel of the induced homom. $\,\phi:G\to S_4\,$ by the action is a subgroup of $\,P\,$ of order 9 or $\,P\,$ itself we're cool since this is a proper subgroup of order greater than 6... $\endgroup$ – DonAntonio Aug 3 '12 at 2:13
  • $\begingroup$ @DonAntonio thanks, I understand it now. Still, the action is on the left cosets of $P$, rather than $P$ itself, right? It also follows that this homomorphism $\phi : G \rightarrow S_4$ cannot be surjective, since we would then have $G/\ker(\phi) \simeq S_4$, which can't happen (again because of the orders). $\endgroup$ – student Aug 3 '12 at 16:54
  • $\begingroup$ Yes, you're right. Edited and thanks. $\endgroup$ – DonAntonio Aug 3 '12 at 22:08
  • $\begingroup$ @ DonAntoino I have to understood your point but how to conclude $G$ have normal subgroup of order 9 and 27. $\endgroup$ – user120386 Mar 25 '14 at 16:44
  • 1
    $\begingroup$ @user549397 Sure. Check that the kernel of the homomorphism $\;G\to S_4\;$ associated with the action of $\;G\;$ on the left cosets of $\;P\;$ is precisely the core of $\;P\;$ , namely: $\;\bigcap_{g\in G} g^{-1}Pg\;$ . This group is characterized by being the maximal normal subgroup in $\;G\;$ which is contained in $\;P\;$ ... If you make it this far then you already got what you wanted. This is standard stuff, for example in Rotman's book. $\endgroup$ – DonAntonio Apr 22 at 16:08

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