8

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!

student
  • 3,887

1 Answers1

10

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.

Bach
  • 5,730
  • 2
  • 20
  • 41
DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • 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... – DonAntonio Aug 03 '12 at 02:13
  • @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). – student Aug 03 '12 at 16:54
  • Yes, you're right. Edited and thanks. – DonAntonio Aug 03 '12 at 22:08
  • @ DonAntoino I have to understood your point but how to conclude $G$ have normal subgroup of order 9 and 27. – user120386 Mar 25 '14 at 16:44
  • 9 or 27, @user120386 . $;|G|=108;,;;|S_4|=24;$ , and by the regular action of $;G;$ on the left cosets of $;P;$ we get a subgroup of $;P;$ which is normal in $;G;$ and at least of order $;\frac{|G|}{|S_4|}=4.5;$ and a divisor of $;27;$ ... – DonAntonio Mar 25 '14 at 17:00
  • @DonAntonio I have the same confusion as user120386 's... We can only deduce from the fact that the order of normal subgroup is $ \ge 5 $ and it is also a divisor of $ 108=2^2\times 3^3 $ to conclude that it is $\ge 6$ which is what the question asks us to prove. I didn't see any reason why we can deduce that the order is a divisor of $27$. – Bach Apr 22 '19 at 13:08
  • @user549397 Because the kernel of that homomorphism $;G\to S_4;$ is a subgroup of $;P;$ , of course. You may want to read about the regular action, the core of a group and etc. – DonAntonio Apr 22 '19 at 13:15
  • 1
    @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. – DonAntonio Apr 22 '19 at 16:08
  • @DonAntonio Aha! Finally, I get it... Thank you! – Bach Apr 22 '19 at 16:39
  • @DonAntonio Kernel of the homomorphism can only be of order $9$. Clearly $\mid kernel \mid \neq 1,3$ and also $kernel \neq P$, since P is not normal in $G$. Thus only option left is that $\mid ker \mid=9$. – user371231 Aug 13 '21 at 07:47