This post has a pretty weird title, but the problem is easy to state and uses a few interesting mathematical concepts.  It’s worth going through.  Let’s start with the basics.


Problem 1.  Let S_{\alpha}(n) = 1^{\alpha} + 2^{\alpha} + \dots + n^{\alpha}.  Show that S_{\alpha}(n) is a polynomial for each \alpha\in {\mathbb N} and that the degree of the polynomial is \alpha + 1.


Indeed, for example, we have  that 1 + 2 + \dots + n = \frac{n(n+1)}{2}, as we learned in Calculus, and this is a polynomial of degree 2.  Similarly, 1^{2} + 2^{2} + \dots + n^{2} = \frac{n(n+1)(2n+1)}{6}, which is a polynomial of degree 3.  In the same respect, 1^{3} + 2^{3} + \dots + n^{3} = \frac{(n(n+1))^{2}}{4}, which is a polynomial of degree 4.

The associated polynomials in this case are given by Faulhaber’s formula:


Theorem (Faulhaber).  For \alpha\in {\mathbb N} we have \displaystyle \sum_{k=1}^{n}k^{\alpha} = \frac{1}{1 + \alpha}\sum_{j=0}^{\alpha}(-1)^{j}\binom{\alpha + 1}{j}B_{j}n^{\alpha + 1 - j}.


This formula looks terrifying, but it is not hard to apply in practice.  You may be wondering, though, what the B_{j}‘s in this formula stand for.  These are the strange and wonderful Bernoulli numbers, of course!  I always enjoy seeing these creatures, because they unexpectedly pop up in the strangest problems.  There are a number of ways to define these numbers, one of which is to just write them out sequentially, starting with B_{0}:


1, -\frac{1}{2}, \frac{1}{6}, 0, -\frac{1}{30}, 0, \frac{1}{42}, 0, -\frac{1}{30}, 0, \dots


But in this case it is not so easy to guess the next value.  The clever reader will notice that all of the odd numbered Bernoulli numbers (except the first) are zero, but other than that there does not seem to be a clear pattern.  Fortunately, we can construct a function which generates the values as coefficients; we’ll call this function (surprise!) a generating function.


Definition.  We define the sequence \{B_{j}\}_{j=0}^{\infty} by

\displaystyle \frac{t}{e^{t} - 1} = \sum_{j=0}^{\infty}B_{j}\frac{t^j}{j!}.


Notice that this will, in fact, generate the B_{j} as coefficients times \frac{1}{j!}.  Neat.  In practice, you can use a program like Mathematica to compute B_{j} for pretty large values of j; but, of course, there are lists available.  We can now use Faulhaber’s formula above, which gives us (assuming we have proven that the formula holds!) that the sums of powers of natural numbers form polynomials of degree \alpha + 1.


But something else happens that’s pretty interesting.  Let’s look at some of the functions.


\begin{array}{|c|c|}\hline \alpha & \mbox{Polynomial}\\\hline &\\ 1 & \frac{1}{2}n^{2} + \frac{1}{2}n\\ &\\ 2 & \frac{1}{3}n^{3} + \frac{1}{2}n^{2} + \frac{1}{6}n\\ & \\ 3 & \frac{1}{4}n^{4} + \frac{1}{2}n^{3} + \frac{1}{4}n^{2} \\ & \\ 4 & \frac{1}{5}n^{5} + \frac{1}{2}n^{4} + \frac{1}{3}n^{3} - \frac{1}{30}n\\ & \\ \hline \end{array}


Look at the coefficients in each of these polynomials.  Anything strange about them?  Consider them for a bit.


Problem.  Look at the coefficients.  What do you find interesting about them?  Note that, in particular, for a fixed \alpha, the coefficients of the associated polynomial sum to 1.  Convince yourself that this is probably true (do some examples!) and then prove that it is true.  Do this before reading the statements below.


Anecdote.  I spent quite a while trying to write down the "general form" of a polynomial with elementary symmetric polynomials and roots to try to see if I could prove this fact using some complex analysis and a lot of terms canceling out.  This morning, I went into the office of the professor to ask him about what it means that these coefficients sum up to 1.  He then gave me a one-line (maybe a half-line) proof of why this is the case. 


Hint.  What value would we plug in to a polynomial to find the sum of the coefficients?  What does plugging in this value mean in terms of the sum?





































[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.







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