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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.

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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?

Thanks!

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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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