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?


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.

  • 2
    $\begingroup$ 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

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....


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.