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Let $G$ have order $p^n m$ where p is a prime and $p \nmid m$. Suppose $m < 2p$. Show that $G$ has a normal subgroup of order $p^{n-1}$ or $p^n$.

I have tried to apply the Sylow Theorems but I can't see how to continue in the case when the Sylow $p$-subgroup is not normal. Let $P$ be a Sylow $p$-group. Then $\vert P \vert = p^n$. If $P$ is normal in $G$ we are done, otherwise $k := \vert \operatorname{Syl}_G(p) \vert \ge 2$. By Sylow's Theorem $k \mid m$. Using our assumption that $m < 2p$ we get that $m/k < p$.

Am I on the right track? Since $m=[G:P]$, $k=[G:N_G(P)]$ and $[G:P]=[G:N_G(P)][N_G(P):P]$ we get that $m/k = [N_G(P):P] < p$. Not sure if this is useful.

I know that $P$, being a $p$-group, has a normal subgroup of order $p^{n-1}$. I don't see why it would be normal in $G$ though.

I prefer hints to full answers.

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    $\begingroup$ If the Sylow subgroup is normal then we're done. Otherwise the number of Sylow subgroups divides $m<2p$ and is congruent to $1$ mod $p$. It follows that $m=p+1$. That's seems like it'll be useful, so I guess you could call it a hint, but I don't have time to finish to find out how useful it is! $\endgroup$ – Matt Samuel Jun 18 '16 at 20:10
  • $\begingroup$ Can I ask where did you get this problem? $\endgroup$ – MonsieurGalois Jun 18 '16 at 20:15
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    $\begingroup$ Assume that no Sylow subgroup is normal. Consider $O_p(G)$, intersection of all Sylow $p$-subgroups. $O_p(G)$ is the kernel of the action of $G$ on the cosets of a Sylow $p$-subgroup which has $p+1$ elements by the hint above by @MattSamuel . Then $[G:O_p(G)]$ divides $|S_{p+1}| = (p+1)!$. Say $O_p(G) = p^a$ and using number theoretic arguments to get $a = n-1$. $\endgroup$ – Levent Jun 18 '16 at 20:36
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If $P$ is not normal and there are $p+1$ Sylow $p$ subgroups. Suppose there is a normal subgroup $Q$ of order $p^{n-1}$, then $$Q=\bigcap_{P\in\operatorname{Syl}_p(G)}P.$$ Hence $$\left|\bigcup_{P\in\operatorname{Syl}_p(G)}P\right|=p^{n+1}.$$

You should be able to show that if there is no such $Q$, then the union of the Sylow $p$ subgroups is too big and that the same is true if $G$ has more than $p+1$ Sylow $p$ subgroups.

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