A Very Basic Introduction to the Sylow Theorems.

May 18, 2012

[Note: It’s been a while!  I’ve now completed most of my pre-research stuff for my degree, so now I can relax a bit and write up some topics.  This post will be relatively short just to “get back in the swing of things.”]


In Group Theory, the big question used to be, “Given such-and-such is a group, how can we tell which group it is?”




The Sylow Theorems (proved by Ludwig Sylow, above) provide a really nice way to do this for finite groups using prime decomposition.  In most cases, the process is quite easy.  We’ll state the theorems here in a slightly shortened form, but you can read about them here.  Note that subgroup which is of order p^{\beta} for some \beta is unsurprisingly called a p-subgroup. A p-subgroup of maximal order in G is called a Sylow p-subgroup.


Theorem (Sylow).  Let G be a group such that |G| = p^{\alpha}m for p\not| m.  Then,

  1. There exists at least one subgroup of order p^{\alpha}.
  2. The Sylow p-subgroups are conjugate to one-another; that is, if P,Q are Sylow p-subgroups, then there is some g\in G such that gPg^{-1} = Q.  Moreover, for all g\in G, we have that gPg^{-1} is a Sylow p-subgroup.
  3. The number of Sylow p-subgroups of G, denoted by n_{p}, is of the form n_{p} \equiv 1\mbox{ mod } p.  In other words, n_{p} divides m.


This first part says that the group of Sylow p-subgroups of G is not empty if p divides the order of G.  Note that this is slightly abbreviated (the second part is actually more general, and the third part has a few extra parts) but this will give us enough to work with.


Problem: Given a group  |G| = pq for p,q prime and p < q, is G ever simple (does it have any nontrivial normal subgroups)?  Can we say explicitly what G is?


We use the third part of the Sylow theorems above.  We note that n_{q} | p and n_{q} \equiv 1\mbox{ mod } q, but p < q so this immediately implies that n_{q} = 1 (why?).  So we have one Sylow q-subgroup; let’s call it Q.  Once we have this, we can use the second part of the Sylow theorem: since for each g\in G we have gQg^{-1} is a Sylow q-subgroup, but we’ve shown that Q is the only one there is!  That means that gQg^{-1} = Q; this says Q is normal in G.  We have, then, that G isn’t simple.  Bummer.

On the other hand, we can actually say what this group is.  So let’s try that.  We know the Sylow Q-subgroup, but we don’t know anything about the Sylow P-subgroups.  We know that n_{p} \equiv 1\mbox{ mod }p and n_{p} | q, but that’s about it.  There are two possibilities: either n_{p} = 1 or n_{p} = q.

For the first case, by using the modular relation, if p does not divide q-1 then this forces n_{p} = 1; this gives us a unique normal Sylow p-subgroup P.  Note that since the orders of our normal subgroups multiply up to the order of the group, we have PQ \cong P\times Q \cong G; in other words, G \cong {\mathbb Z}_{p}\times {\mathbb Z}_{q}.

For the second case, n_{p} = q.  We will have a total of q subgroups of order p and none of these are normal.   This part is a bit more involved (for example, see this post on it), but the punch line is that it will be the cyclic group {\mathbb Z}_{pq}.


I’ll admit that the last part is a bit hand-wavy, but this should at least show you the relative power of the Sylow theorems.  They also come in handy when trying to show something either does or does not have a normal subgroup.  Recall that a simple group has no nontrivial normal subgroups.


Question.  Is there any simple group with |G| = 165?


I just picked this number randomly, but it works pretty well for this example.  We note that |G| = 165 = 3\cdot 5\cdot 11.  Let’s consider, for kicks, n_{11}.  We know n_{11} must divide 3\cdot 5 = 15 and it must be the case that n_{11} \equiv 1\mbox{ mod } 11; putting these two facts together, we get n_{11} = 1.  This immediately gives us a normal subgroup of order 11, which implies there are no simple groups of order 165.


Question.  Is there any simple group with |G| = 777?


Alas, alack, you may say that 777 is too big of a number to do, but you’d be dead wrong.  Of course, 777 = 3\cdot 7\cdot 37.  Use the same argument as above to show there are no simple groups of this order.


Question.  Is there any simple group with |G| = 28?


Note that 28 = 2^{2}\cdot 7 so we need to do a little work, but not much.  Just for fun, let’s look at n_{7}.  We must have that it is 1 modulo 7 and it must divide 2^{2} = 4.  Hm.  A bit of thinking will give you that n_{7} = 1, which gives us the same conclusion as above.


Of course, there are examples where this doesn’t work nicely.  Think about the group of order 56, for example.  In order to work with these kinds of groups, one must do a bit more digging.  We will look into more of this later.


5 Responses to “A Very Basic Introduction to the Sylow Theorems.”

  1. the big question used to be, “Given such-and-such a group, how can we tell what it is?”

    What does that mean?

    “Given X, how can we tell what X is?” You just told me: it’s X.

    • James said

      I’ve edited to reflect what I meant by this a bit more clearly; the idea was something like: “Given that G is a group, which group is it?”

      This might still seem a bit silly, but the idea is something like this: pretend you know something is a finite group (i.e., it has the structure of a group). For example, there are many situations [homology, cohomology, etc.] where some things are groups and the goal is to find out *which* groups they are. If you’re given every single element, you still may not be able to say what the group is [there are, for example, 47 non-isomorphic groups of order 120]. Sylow allows us to “limit” the number of possibilities for our group. For example, Sylow tells us that every group of order 49 is abelian. Even if it does not tell us this much, it allows us to look at Sylow Subgroups, which behave rather well with respect to our usual methods of probing groups [representations, etc].

  2. $p \not| m$ ← This means not divisible by $m$?

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