Putnam 2007 Question A5:
"Suppose that a finite group has exactly $n$ elements of order $p$, where $p$ is a prime. Prove that either $n=0$ or $p$ divides $n+1$."
I split this problem into two cases: where $p$ divides $|G|=m$, and where $p$ does not divide $m$. The latter case is trivial – by Lagrange's Theorem, $n=0$ as the order of an element must divide the order of the group. The first case appears more complicated and my idea is to use Sylow Theory, and I came across an interesting solution using this on https://blogs.haverford.edu/mathproblemsolving/files/2010/05/Putnam-2007-Solutions.pdf:
"There are 1 + kp Sylow p-subgroups and, because they are all conjugate and every element of order p is contained in some Sylow p-subroup, they partition the n elements of order p into 1 + kp equal-size collections. The number of elements of order p in any p-group is always one less than a power of p, implying n + 1 ≡ (1 + kp)(-1) + 1 ≡ 0 modulo p."
The places I am stuck are:
1) Why the fact that all Sylow p-subgroups being conjugate and every element of order $p$ contained in some Sylow p-subgroup (I understand why both these facts are true), implies that the Sylow p-subgroups partition the $n$ elements of order $p$ into $1+kp$ equal-size collections – heck I don't even understand what is meant by this…
2) Why the number of elements of order $p$ in any p-group is always one less than a power of p.
If anyone has other ways to show that $p$ dividing $m$ implies that $p$ divides $n+1$, that would also be greatly appreciated 🙂
Thanks
Best Answer
Here is a proof of the result, via an (famous/very clever/pretty) idea James McKay used to prove Cauchy's theorem.
Let $S$ denote the set of $p$-tuples $(a_1, a_2, \cdots, a_p)$ where $a_i \in G$ and $a_1 a_2 a_3 \cdots a_p = 1$. Note that $\mathbb{Z}/p\mathbb{Z}$ acts on $S$ by cyclic rotations. Thus, we have $$|S| = \#\{\text{orbits of size 1}\} + \#\{\text{orbits of size p}\} \cdot p $$ Orbits of size 1 correspond to tuples of the form $(x, x, x, \cdots, x)$, i.e. elements $x\in G$ of order $p$, excepting the orbit corresponding to the trivial tuple $(1, 1, 1, \cdots 1)$. Thus, $$\#\{\text{orbits of size 1}\} = \#\{\text{elements of order }p\} + 1 = n+1$$ On the other hand, note that $|S| = |G|^{p-1}$ because the first $p-1$ elements of any $p$-tuple can be chosen totally arbitrarily, and then the last element of the tuple is fixed. Taking the equation for $|S|$ modulo $p$, we see that if $p$ divides $|G|$, then $p$ divides $n+1$, as desired.