Using the Sylow Theorems, we can find out whether groups of certain order are simple or not. An example is the following:

$\textit{Prove that any group of order 351 is not simple}$

Given $|G| = 351 = 3^3*13$, we have that there exist Sylow 3-subgroups and Sylow 13-subgroups.

The number of these is given by the third Sylow theorem ($n(p)$ represents the number of Sylow P-subgroups):

$n(13) = 1+13*r$, and $n(3) = 1 + 3*r$ for $r=0$ or $r\in \mathbb{Z^+}$

and, also by the third Sylow theorem,

$n(13)|3^3$, and $n(3)|13$

And this gives us that $n(13) = 1$ or $n(13)=27$

If there is only one Sylow 13-subgroup, then it must be normal, and G is not simple. Therefore there would have to be 27 of them. These 27 groups then contain 12 elements which are distinct to each (because the identity is in all of them), of order 13 (because 13 is prime and any subgroup of prime order is cyclic). This leaves 27 elements to form the Sylow-3 subgroup, and so there can be only one of those, which then must be normal, and G is not simple.

The point is that in this example, there were exactly 27 elements left over after figuring out how many elements were in the Sylow 13-subgroups, allowing us to deduce that there is only one Sylow 3-subgroup. This was crucial.

If that were not the case, would I have still - using Sylow theory - been able to deduce that the group is simple? I suppose what I'm trying to ask is whether it is feasible to prove that a group is simple using similar techniques if we do not have a precise as recipe above - where everything worked very nicely.


This question is a bit hard to answer since it ostensibly depends on the order of the group in question.

Sylow's theorems can be used to prove things about general forms of group orders, for example the fact every group of order $pq,$ where $p>q$ are distinct primes, is a semi-direct product $C_{p}\rtimes C_{q},$ and furthermore if $q\nmid p-1,$ then this becomes a direct product and the group is cyclic. Via a pretty similar argument one can show that groups whose orders' are of the form $p^{2}q$ are not simple, and, though I haven't seen such proofs, I believe you can keep going like this quite some way with just Sylow's theorems (but the arguments get very long because there is a lot of case reduction).

If I recall correctly, $pq$ and $p^{2}q$ already covers most of the composite numbers up to order $60;$ the stragglers (except $48,$ see below) are then just exercises, and in this way one can prove that the only simple groups of order less than $60$ are those of prime order.

However, as with most things, the usefulness of Sylow's theorems depends not just on the situation but also on what other information and techniques you have at your disposal. If you know about the $\operatorname{Core}$ of a subgroup $H$ of a group $G$ (that is, the kernel of (the homomorphism induced by) the action of $G$ by left multiplication on the cosets of $H;$ equivalently, the largest subgroup of $H$ which is normal in $G$), then there are "middling" cases, like the following.

Exercise: Prove that every group of order $48$ has a normal subgroup of order $8$ or $16.$

Solution: Let $G$ be a group of order $48=2^{4}\cdot3.$ There exists a 2-Sylow $P$ of $G.$ Let $s_{2}$ be the number of 2-Sylows of $G.$ If $s_{2}=1$ then we are done: $P$ is of order 16 and is the unique 2-Sylow, so it is normal. From now on, assume $s_{2}\neq1,$ so that $P\not\trianglelefteq G.$ We will show that $\operatorname{Core}{(P)}$ is of order 8.

There is an embedding $G/\operatorname{Core}{(P)}\hookrightarrow S_{3}.$ Hence $48/\lvert{\operatorname{Core}{(P)}}\rvert\leq 6,$ so that $\lvert\operatorname{Core}{(P)}\rvert\geq 8.$ Since $P\not\trianglelefteq G$ we know $\operatorname{Core}{(P)}\neq P,$ so that $\lvert{\operatorname{Core}{(P)}}\rvert<16.$ But now we know that $\lvert{\operatorname{Core}{(P)}}\rvert$ divides $\lvert{P}\rvert=16$ and $8\leq\lvert{\operatorname{Core}{(P)}}\rvert<16.$ Hence $\lvert{\operatorname{Core}{(P)}}\rvert=8$ and we are done.

A very similar argument can be given to show, for example, that every group of order $108$ has a normal subgroup of order $9$ or $27.$

  • $\begingroup$ A nice answer. Thanks. $\endgroup$ – Matt Jan 19 '17 at 11:04

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