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A group of order $3^a\cdot 5\cdot 11$ has a normal Sylow $3$-subgroup.

This is question 5C.7 in Isaacs's Finite Group Theory.

That section of the text is about transfer in finite groups, and proves results like Burnside's transfer lemma, that groups with all Sylows cyclic are metacyclic, etc. As for the problem, if $|G|=3^a\cdot5\cdot11$, I can, after checking a few cases and using induction, reduce to the case that $G$ is simple. But I don't see how to show, using what has come before, that $G$ can't be simple.

And actually, my bigger problem is that it seems like a lot of work for one problem. And even more problematic is the fact I haven't really used any results related to transfer! Is there another, easier route? Or perhaps is there a missing "abelian" in the problem?


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Also, in the future could you have your questions be completely self-contained? This is one of those cases where I opened up a few questions, switched tabs to read them, and completely missed rereading the title here. Embarrassingly, it wasn't until my third read-through that I realized the question was contained in the title. I know, I'm embarrassing. –  mixedmath Aug 9 '11 at 3:52
Well, I think I figured it out :) - if $P$ is a Sylow 5-group, then since $[N_G(P):C_G(P)]$ divides $4$, $P$ is central in its normalizer, so $G$ has a normal $5$-complement $N$, where $|N|=3^a\cdot11$; then counting shows $N$ has a normal Sylow 3-group, and we're done. –  user641 Aug 9 '11 at 4:28
@Steve: Instead of counting you could repeat the first part of your proof with $N$ and $11$. –  j.p. Aug 9 '11 at 19:33
@Steve,jug: Or apply Sylow's theorem with the prime 3 within the subgroup N, to deduce that N has a unique Sylow 3-subgroup –  Geoff Robinson Aug 13 '11 at 8:49

1 Answer 1

up vote 2 down vote accepted

As an alternative to your proof (given in a comment, already simplified by Geoff) you could also just use Sylow's theorem and no transfer (surely not intended by Isaacs):

By Sylow the number of $3$-Sylow subgroups has to be either $1$ or $55$. In the first case, you are done, so assume the latter case. Choose two $3$-Sylow subgroups $S\ne T$ such that their intersection $P = S\cap T$ is maximal among intersections of Sylow subgroups.

As $P < \mathrm{N}_S(P)$ and $P < \mathrm{N}_T(P)$, you get by the maximal choice of $P$ that the normalizer $H = \mathrm{N}_G(P)$ does not have a unique normal $3$-Sylow subgroup. As the order of $H$ is $55$ times a power of $3$, $H$ has as many $3$-Sylow subgroups as $G$, each of them properly containing $P$. As $P$ is a maximal intersection of $3$-Sylows of $G$, each $3$-Sylow subgroup of $H$ is contained in a unique $3$-Sylow subgroup of $G$, showing that $P$ is also contained in every $3$-Sylow subgroup of $G$. Hence $P$ is the intersection of all $3$-Sylow subgroups of $G$ and therefore normal in $G$.

Now look at $\bar{G} = G/P$: $\bar{G}$ has $55$ $3$-Sylow subgroups which intersect pairwise trivially (again by maximal choice of $P$) and are self-normalizing (i.e., $\mathrm{N}_{\bar{G}}(\bar{S}) = \bar{S}$). Therefore $\bar{G}$ contains exactly $54$ elements whose order are not a power of $3$.

By Sylow $\bar{G}$ has a unique $11$-Sylow subgroups (there are not enough $3'$-elements for more $11$-Sylow subgroups than that), which is normal and centralized by every non-trivial $3$-element as $\mathrm{Aut}(C_{11})$ is cyclic of order $10$, contradicting that the $3$-Sylow subgroups of $\bar{G}$ are self-normalizing.

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+1: Very nice, indeed! Your solution used only the kind of tools I understand, and yet left me with a feeling that I learned something. The use of a maximal intersection of two Sylow subgroups is something I should add to my repertoire. –  Jyrki Lahtonen Aug 24 '11 at 18:17
@Jyrki: Applying Sylow's theorem to the normalizer of a maximal intersection of Sylow subgroups extends slightly the range of small numbers $n$ for which the standard exercise "Show that there is no simple group of order $n$" can be solved by Sylow's theorem. –  Someone Aug 25 '11 at 9:16
Sorry, but could you explain why each 3-Sylow subgroup of $\bar{G}$ is self-normalizing? It is the only point that I understand not yet. Thanks for sharing this nice answer. –  awllower Aug 6 '12 at 6:54

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