Putnam 2007 A5: Finite group $n$ elements order $p$, prove either $n=0$ or $p$ divides $n+1$

Solution 1:

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.

Solution 2:

First of all by the Second Sylow Theorem we have that all Sylow p-groups are conjugates of each other. Furthermore we have that conjugation leaves the order of the element intact. For example let $P_1$ and $P_2$ be any Sylow p-groups. Then $\exists g \in G$ s.t. $gP_1g^{-1} = P_2$. Also if $x$ is an element of $P_1$ of order $p$, then $gxg^{-1}$ is an element of $P_2$ of order $p$. In other word this means that each Sylow p-group have the same number of elements of order $p$.

Thus as we the number of Sylow p-subgroups is $1 \pmod p$ the $n$ elements are partitioned into $kp+1$ Sylow p-subgroups, each containing the same number of elements.

For the second part you can use the notation used in the Sameer's answer and get that $n+1 \equiv |S| \equiv 0 \pmod p$

Solution 3:

While it is the case that the number of elements of order $p$ of a $p$-group is congruent to $-1$ modulo $p$, it is not true that the number of elements of order $p$ of a $p$-group must be of the form $p^k-1$ for some $k\in\mathbb{Z}_{>0}$. The solution you have is not entirely correct.

Here is a counterexample. Consider the dihedral group $$\begin{align}G&:=D_4=\big\langle a,b\,\big|\,a^4=1\,,\,\,b^2=1\,,\text{ and }bab^{-1}=a^{-1}\big\rangle \\&=\big\{a^r\,b^s\,\big|\,r\in\{0,1,2,3\}\text{ and }s\in\{0,1\}\big\}\,.\end{align}$$ Note that $|G|=8=2^3$. The elements of order $2$ of $G$ are listed below: $$a^2,b,ab,a^2b,a^3b\,.$$ That is, there are exactly $5$ elements of order $2$ in $G$. Clearly, $5\equiv-1\pmod{2}$, but it is not of the form $2^k-1$ for any $k\in\mathbb{Z}_{>0}$. The remaining $2$ non-identity elements of $G$ are $a$ and $a^3$, which are of order $4$.

Here is an alternative proof that the number $n$ of elements of order $p$ of a $p$-group $G$ satisfies $$n\equiv -1\pmod{p}\,.$$ Consider the set $A$ of elements of $G$ containing all $x\in G$ such that $x^p=1$. Clearly, $n=|A|-1$. We claim that $p$ divides $|A|$.

Let $Z$ denote the center of $G$. We note that the order of $A\cap Z$ must be a power of $p$, being a subgroup of an abelian $p$-group $Z\trianglelefteq G$. Thus, $p$ divides $|A\cap Z|$ (as every $p$-group has a nontrivial center, which contains an element of order $p$). It remains to verify that $p$ divides $|A\setminus Z|=|A|-|A\cap Z|$.

We can partition $B:=A\setminus Z$ into orbits of elements of $B$ under conjugation in $G$. Clearly, any conjugate of an element of $B$ is in $B$. Due to the Orbit-Stabilizer Theorem, the size of each orbit is a power of $p$, and since the orbit contains a noncentral element, its size is larger than $1$. Consequently, every orbit of elements in $B$ is of size $p^k$ for some integer $k\geq 1$. This proves that $|B|$ is divisible by $p$. (In fact, $|B|$ is also divisible by $p-1$.)

As an example, with $G$ being the dihedral group $D_4$ as above, we have $A=(A\cap Z)\cup B$, where $$A\cap Z=Z=\big\{1,a^4\big\} \text{ and }B=\big\{b,ab,a^2b,a^3b\big\}\,.$$ The partition of $B$ into conjugacy orbits (or conjugacy classes) is $$\{b,a^2b\}\cup\{ab,a^3b\}\,.$$