Putnam 2007, Problem A-5

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 $n + 1$.

Solution:
In all solutions, let $G$ be a finite group of order $m$.

**First solution:** By Lagrange's theorem, if $m$ is not divisible by $p$, then $n = 0$. Otherwise, let $S$ be the set of $p$-tuples $(a_0, \ldots, a_{p-1}) \in G^p$ such that $a_0 \cdots a_{p-1} = e$; then $S$ has cardinality $m^{p-1}$, which is divisible by $p$. Note that this set is invariant under cyclic permutation, that is, if $(a_0, \ldots, a_{p-1}) \in S$, then $(a_1, \ldots, a_{p-1}, a_0) \in S$ also. The fixed points under this operation are the tuples $(a, \ldots, a)$ with $a^p = e$; all other tuples can be grouped into orbits under cyclic permutation, each of which has size $p$. Consequently, the number of $a \in G$ with $a^p = e$ is divisible by $p$; since that number is $n+1$ (only $e$ has order 1), this proves the claim.

**Second solution:** (by Anand Deopurkar) Assume that $n > 0$, and let $H$ be any subgroup of $G$ of order $p$. Let $S$ be the set of all elements of $G \setminus H$ of order dividing $p$, and let $H$ act on $G$ by conjugation. Each orbit has size $p$ except for those which consist of individual elements $g$ which commute with $H$. For each such $g$, $g$ and $H$ generate an elementary abelian subgroup of $G$ of order $p^2$. However, we can group these $g$ into sets of size $p^2 - p$ based on which subgroup they generate together with $H$. Hence the cardinality of $S$ is divisible by $p$; adding the $p-1$ nontrivial elements of $H$ gives $n \equiv -1 \pmod{p}$ as desired.

**Third solution:** Let $S$ be the set of elements in $G$ having order dividing $p$, and let $H$ be an elementary abelian $p$-group of maximal order in $G$. If $|H| = 1$, then we are done. So assume $|H| = p^k$ for some $k \geq 1$, and let $H$ act on $S$ by conjugation. Let $T \subset S$ denote the set of fixed points of this action. Then the size of every $H$-orbit on $S$ divides $p^k$, and so $|S| \equiv |T| \pmod{p}$. On the other hand, $H \subset T$, and if $T$ contained an element not in $H$, then that would contradict the maximality of $H$. It follows that $H = T$, and so $|S| = |T| = |H| = p^k \equiv 0 \pmod{p}$, i.e., $|S| = n+1$ is a multiple of $p$.

**Remark:** This result is a theorem of Cauchy; the first solution above is due to McKay. A more general (and more difficult) result was proved by Frobenius: for any positive integer $m$, if $G$ is a finite group of order divisible by $m$, then the number of elements of $G$ of order dividing $m$ is a multiple of $m$.