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Show a group of order $4563=3^3\cdot13^2$ is not simple.

I am confused when dealing with these Sylow's game. Especially when the order becomes large. It seems there is no common rules to solve this kind problems.
My try: $n_{13}=1+13t$ and $n_{13}$ divides $3^3=27$, hence $t=0,2$. If $t=0$ then we are done. If $t=2$, There are $27$ 13-subgroups of order 169. Then consider the conjugation action of G on the set of 13-subgroups. Prove it induced a homomorphism between $G$ and $S_{27}$. And want to claim a contradiction by $|G|$ does not divide $27!$. But sadly it does divide. Any other method can get this figured out?

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The next things to try are often:

  • to consider the intersections of Sylow groups, and
  • to use the fact that groups of order $p^2$, $p$ a prime, are always abelian.

Let's see. Assume that we have two Sylow $13$-subgroups that intersect non-trivially, say $P_1\cap P_2$ and $x\neq 1_G, x\in P_1\cap P_2$ with $|P_1|=|P_2|=13^2$. Then $x$ is centralized by both $P_1$ and $P_2$. This means that $C_G(x)$ has order that is a proper multiple of $13^2$. The size of the conjugacy class of $x$, $[G:C_G(x)]$, is thus a proper factor of $[G:P_1]=27$, and therefore at most nine. If $x$ is in the center, $G$ cannot be simple. Otherwise we get a non-trivial homomorphism from $G$ to $S_9$ by the conjugation action on the conjugacy class of $x$. Again, we are done.

So let's assume that all those twenty-seven Sylow $13$-subgroups intersect trivially. In that case their union contains $(27\cdot 168)+1=|G|-26$ elements. Now we can look at Sylow $3$-subgroups. Their non-identity elements must all be among those $26$ elements not belonging to any Sylow $13$. This means that there is room for only a single Sylow $3$-subgroup, which is then normal concluding the proof.

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    $\begingroup$ If $P_1$ and $P_2$ are different Sylow $13$-subgroups, then $P_1 \cap P_2 \subsetneq P_1$. Now observe that $|G| \gt |P_1P_2|=\frac{|P_1||P_2|}{|P_1 \cap P_2|}$. Hence, $|P_1 \cap P_2| \gt \frac{169}{27} \gt 6.25$. Thus $ |P_1 \cap P_2|=13$. This can replace/skip your last argument, Jyrki. $\endgroup$ Oct 12 '18 at 12:31
  • $\begingroup$ Nice @Nicky! Why did I miss that :-) $\endgroup$ Oct 12 '18 at 19:46
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The Burnside Transfer Theorem does this even more briefly. A 13-Sylow cannot be in the center of its normalizer, so, as every group of order $13^2$ is abelian, it must be properly contained in its normalizer, which thus has index at most $9$. And we again conclude that either the 13-Sylow is normal or there is a non-trivial homomorphism to $\mathfrak{S}_9$.

Of course, the Burnside $p^aq^b$ theorem is faster still....

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