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

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.

• 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. Oct 12, 2018 at 12:31
• Nice @Nicky! Why did I miss that :-) Oct 12, 2018 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....