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\title[Congruences associated with families of nilpotent subgroups]{Congruences associated with families of nilpotent subgroups and a theorem of Hirsch}
\author{\firstname{Stefanos} \lastname{Aivazidis}\IsCorresp}
\address{Department of Mathematics \& Applied Mathematics, University of Crete, Greece}
\email{s.aivazidis@uoc.gr}
\author{\firstname{Thomas} \lastname{M\"uller}}
%\address{Queen Mary \& Westfield College, London, UK}
\address{Department of Mathematics, University of Vienna, Austria}
\email{muellet4@univie.ac.at}
\thanks{The first author is partially supported by the Hellenic Foundation for Research and Innovation, Project HFRI-FM17-1733. This research was supported through the program ``Research in Pairs'' by the Mathematisches Forschungsinstitut Oberwolfach in 2019.}
\CDRGrant[Hellenic Foundation for Research and Innovation]{HFRI-FM17-1733}
\subjclass{20D20, 20D60}
\keywords{\kwd{Nilpotent systems of subgroups}
\kwd{congruences}}
\begin{abstract}
Our main result associates a family of congruences with each suitable system of nilpotent subgroups of a finite group. Using this result, we complete and correct the proof of a theorem of Hirsch concerning the class number of a finite group of odd order.
\end{abstract}
\begin{document}
\maketitle
\section{Introduction} A celebrated result of Burnside~\cite[p.~295]{Burnside}, established using the then still recent (ordinary) character theory of finite groups, states that if $G$ is a group of odd order then the class number $k(G)$ of $G$ satisfies the congruence $k(G) \equiv \vert G\vert \pmod{16}$. The starting point of the work reported here was our desire to correct and complete the proof of an apparently little known theorem of Hirsch~\cite{Hirsch} providing a beautiful non-trivial refinement of Burnside's result.
\begin{theo}\label{Thm:HirschClass}
Let $N = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ be the order of a group $G,$ where $p_1, p_2,\ldots, p_k$ are odd primes, and $\alpha_1, \alpha_2,\ldots, \alpha_k$ are positive integers. Let $d$ be the greatest common divisor of the numbers $p_i^2 - 1$ with $1\leq i\leq k$. Then $N \equiv k(G)\pmod{2d}$.
\end{theo}
Burnside's result follows from Theorem~\ref{Thm:HirschClass}, since $p^2\equiv 1\pmod{8}$ for $p$ odd, so that $8\mid d$. To quickly finish our description of this line of development, we mention that, using some of the easier ideas of Hirsch~\cite{Hirsch}, Poland~\cite{Poland} subsequently obtains an alternative congruence for $k(G)$ which, when combined with Theorem~\ref{Thm:HirschClass}, yields the following further strengthening of Burnside's theorem.
\begin{theo}[{\cite[Cor.~3.10]{Poland}}]\label{Thm:Poland}
In the notation of Theorem~\ref{Thm:HirschClass}, we have, for $N = \left\lvert G\right\rvert$ odd,
\begin{equation}
k(G) \equiv N \pmod{\lcm\{2d, \tau\}},
\end{equation}
where $\tau = \gcd_{i\leq i\leq k} (p_i-1)^2$.
\end{theo}
Hirsch's proof of Theorem~\ref{Thm:HirschClass}, though ingenious and mostly correct, exhibits several gaps, and one non-trivial error. The most serious gap we found is the lack of any justification offered for the following claim, which Hirsch~\cite{Hirsch} makes in passing, and then bases the more subtle final part of his argument on:
\begin{enonce}
{Claim}\label{claim}
The number of Sylow $p$-subgroups of a group $G$ of odd order, containing a given $p$-element, is odd.
\end{enonce}
It is not too hard (though not completely trivial) to supply a proof of Claim~\ref{claim}, if one is willing to invoke the Odd Order Theorem, and to exploit the solubility of $G$; however, the paper~\cite{FTOdd} establishing Burnside's Odd Order Conjecture only appeared in 1963, hence was not available to Hirsch in the late 1940's. This raises the question whether an elementary proof of Claim~\ref{claim} can be given, which does not presuppose solubility of the group involved. This is indeed the case but, as often happens, starting out with a rather specialised problem has led us to the discovery of a more general result, relating congruences to certain systems of nilpotent subgroups in an arbitrary finite group, which appears to be of interest in itself; cf. Section~\ref{Sec:Setup}, in particular Theorem~\ref{Thm:Main}.
\section{Set-up and main result}\label{Sec:Setup}
Let $G$ be a finite group, and let $\set_G$ be a collection of nilpotent subgroups of $G$ satisfying:
\begin{enumerate}\alphenumi
\item\label{(a)}
$1\in \set_G$ and $G\not\in \set_G$;
\item\label{(b)}
$S\in\set_G$ and $g\in G$ implies $S^g\in\set_G$;
\item\label{(c)}
$S\in\set_G$ and $R\leq S$ implies $R\in\set_G$;
\item\label{(d)}
for each subgroup $K\leq G$ (in particular for $G$ itself), the maximal elements of the subposet
\[
\set_K\coloneqq \left\{S\in \set_G: S \leq K\right\}\, \subseteq \set_G
\]
(with inclusion as partial order) form a single $K$-conjugacy class $\mathcal{M}_K$;
\item\label{(e)}
for any two subgroups $K, L$ of $G$ with $K \leq L$, we have $(K:S) \big\vert (L : T)$, where $S\in\mathcal{M}_K$ and $T\in\mathcal{M}_{L}$.
\end{enumerate}
For clarification, we note that the concept of a maximal element used here refers to the context of \emph{poset theory}: if $(P, \leq)$ is a poset, an element $a\in P$ is termed \emph{maximal}, if $a\leq p$ for some $p\in P$ implies $a=p$ (i.e., $a$ is not strictly covered). In particular, if $P$ has a greatest element $a_0$, then $a_0$ is the only maximal element of $P$.
The following is now easily checked.
\begin{lemm}\label{Lem:Inherit}
Let $G$ be a finite group, and let $\set_G$ be a system of nilpotent subgroups of $G$ satisfying Conditions \eqref{(a)}{\rm--}\eqref{(e)}. Let $K\leq G,$ and suppose that $K\not\in\set_G$. Then the pair $(K, \set_K)$ again satisfies Conditions \eqref{(a)}{\rm--}\eqref{(e)}, where $\set_K$ is defined as above.
\end{lemm}
\begin{exams}\leavevmode
\begin{enumerate}
\item\label{(I)}
Suppose that $\left\lvert\pi(G)\right\rvert>1$. For a prime $p\in\pi(G)$, let $\set_G$ be the collection of all $p$-subgroups of $G$. For $K \leq G$, the collection $\mathcal{M}_K$ of maximal elements of the poset $(\set_K, \subseteq)$ consists precisely of the Sylow $p$-subgroups of $K$. Conditions~\eqref{(a)}{\rm--}\eqref{(c)} are clear, while Conditions~\eqref{(d)}{\rm--}\eqref{(e)} hold by Sylow's theorem.
\item\label{(II)}
Let $\pi$ be a set of primes, and let $G$ be a soluble group whose order is not a $\pi$-number. Suppose that $G$ contains a nilpotent Hall $\pi$-subgroup $H$. Then each $\pi$-subgroup of $G$ is nilpotent, and is contained in some conjugate of $H$; in particular, the Hall $\pi$-subgroups of $G$ form a single conjugacy class (by Wielandt's corresponding theorem in~\cite{Wielandt}). Let $\set_G$ be the collection of all $\pi$-subgroups of $G$. For $K \leq G$, the maximal elements of the poset $(\set_K, \subseteq)$ are the Hall $\pi$-subgroups of $K$. Again, Conditions~\eqref{(a)}{\rm--}\eqref{(c)} are clear, while Conditions~\eqref{(d)} and~\eqref{(e)} hold thanks to P. Hall's characterisation of soluble groups; cf. in particular~\cite{PHall}.
\end{enumerate}
\end{exams}
Let $\mathbb{P}$ denote the set of all positive rational primes. Given a finite group $G$ and a system $\set_G$ of nilpotent subgroups of $G$, we set
\[
m_G\coloneqq \gcd\left\{p-1: p\in\ \mathbb{P}, \,p\mid (G:M) \text{ for some }M\in\mathcal{M}_G\right\}.
\]
If
\[
\left\{p\in \mathbb{P}: p \mid (G:S) \text{ for some } S\in\mathcal{M}_G\right\} = \emptyset,
\]
then $\vert S\vert = \vert G\vert$ for $S\in\mathcal{M}_G$, so $G = S\in\set_G$, contradicting the second part of Condition~\eqref{(a)}. Hence, assuming $G\not\in\set_G$, the constant $m_G$ is well defined.
We note the following.
\begin{lemm}\label{Lem:mGdiv}
Let $G$ be a finite group and let $\set_G$ be a system of nilpotent subgroups of $G$ satisfying Conditions~\eqref{(a)} and \eqref{(e)}. If $K \leq G$ and $K\not\in\set_G$, then $m_G \mid m_K$.
\end{lemm}
\begin{proof}
Since $K\not\in\set_G$, we have $K\not\in\set_K$, so that $m_K$ is defined. Suppose that $p\in \mathbb{P}$ is such that $p \mid (K:S)$ for some $S\in\mathcal{M}_K$. Then $p \mid (G:T)$ for $T \in \mathcal{M}_G$ by Condition~\eqref{(e)}, so that $m_G \mid p-1$. Hence, $m_G \mid m_K$, as claimed.
\end{proof}
Our main result is now as follows.
\begin{theo}\label{Thm:Main}
Let $G$ be a finite group, and let $\set_G$ be a system of nilpotent subgroups of $G$ satisfying Conditions~\eqref{(a)}{\rm--}\eqref{(e)}. For $S\in \set_G,$ set
\[
\set_G(S) \coloneqq \left\{T \in\mathcal{M}_G: S \leq T \right\}.
\]
Then $\left\lvert \set_G(S) \right\rvert \equiv 1\pmod{m_G}$ for all $S\in \set_G$.
\end{theo}
Combining Theorem~\ref{Thm:Main} with Example~\eqref{(I)}, we derive an arithmetic property of the collection of Sylow $p$-subgroups of a finite group $G$ containing a fixed $p$-subgroup.
\begin{coro}\label{Cor:Sylowp}
Let $G$ be a finite group such that $\left\lvert\pi(G)\right\rvert>1,$ let $p$ be a prime number dividing the order of $G,$ and let $H$ be a fixed $p$-subgroup of $G$. Then the number $n_p(G,H)$ of Sylow $p$-subgroups of $G$ containing $H$ satisfies the congruence
\begin{equation}\label{Eq:SylowpCong}
n_p(G,H) \equiv 1\pmod{m_G},
\end{equation}
where
\[
m_G = \gcd\left\{p-1: p\in\pi(G)\setminus\{p\}\right\}.
\]
\end{coro}
\begin{rema}
Hirsch's original Claim~\ref{claim} follows from the special case of Corollary~\ref{Cor:Sylowp}, where $G$ has odd order, and $H$ is cyclic.
\end{rema}
Similarly, by combining Theorem~\ref{Thm:Main} with Example~\eqref{(II)}, we find the following.
\begin{coro}\label{Cor:HallPi}
Let $G$ be a finite soluble group, let $\pi$ be a non-empty set of prime numbers such that $\left\lvert G\right\rvert$ is not a $\pi$-number, and let $H$ be a fixed $\pi$-subgroup of $G$. If $G$ contains a nilpotent Hall $\pi$-subgroup, then the number $n_\pi(G,H)$ of Hall $\pi$-subgroups of $G$ containing $H$ satisfies the congruence
\begin{equation}\label{Eq:HallPiCong}
n_\pi(G,H) \equiv 1\pmod{m_G},
\end{equation}
where
\[
m_G = \gcd\left\{p-1: p\in\pi(G) - \pi\right\}.
\]
\end{coro}
\begin{rema}
The condition in Corollary~\ref{Cor:HallPi} that the group $G$ should contain a nilpotent Hall $\pi$-subgroup can in fact be disposed of, leading to the following general result for finite $\pi$-separable groups.
\end{rema}
\begin{theo}\label{Thm:separable}
Let $\pi$ be a non-empty set of primes, let $G$ be a finite $\pi$-separable group such that $\left\lvert G\right\rvert$ is not a $\pi$-number, and let $H$ be a fixed $\pi$-subgroup of $G$. Then the number $n_\pi(G,H)$ of Hall $\pi$-subgroups of $G$ containing $H$ satisfies the congruence
\[
n_\pi(G,H) \equiv 1\pmod{m_G},
\]
where
\[
m_G = \gcd\left\{p-1: p\in \pi(G) - \pi\right\}.
\]
\end{theo}
The proof of Theorem~\ref{Thm:separable}, whose proper setting is the theory of projectors relative to Schunck classes, will be discussed in a separate publication, together with certain related results.
\section{Proof of Theorem~\ref{Thm:Main}}\label{Sec:MainProof}
We begin with two easy reductions.
First, suppose that $S\in\set_G$ is such that $S \unlhd G$. Then $\set_G(S) = \mathcal{M}_G$ by~\eqref{(d)} with $G=K$, and so
\[
\left\lvert \set_G(S) \right\rvert = \left\lvert\mathcal{M}_G \right\rvert = (G : \N_G(T)),
\]
where $T\in \mathcal{M}_G$. Consequently, if $p$ is a prime dividing the cardinality of the set $\set_G(S)$, then $p \mid (G:T)$, hence $p\equiv 1\pmod{m_G}$ by definition of $m_G$, implying $\left\lvert \set_G(S)\right\rvert \equiv 1\pmod{m_G}$. Therefore, we may suppose that $\N_G(S) < G$.
Second, if $S\in\mathcal{M}_G$, then $\set_G(S) = \{S\}$, so $\left\lvert\set_G(S)\right\rvert = 1 \equiv 1\pmod{m_G}$. Thus, we may assume that $S\not\in \mathcal{M}_G$.
Suppose for a contradiction that Theorem~\ref{Thm:Main} is false, and let $G$ be a counterexample of least possible order. Fix a system $\set_G$ of nilpotent subgroups satisfying Conditions~\eqref{(a)}{\rm--}\eqref{(e)} for which $G$ fails, and, among the elements $S\in\set_G$ with $\left\lvert \set_G(S)\right\rvert \not\equiv 1\pmod{m_G}$, let $S_0$ be one of largest possible order. By the observations above, we have $K \coloneqq \N_G(S_0) < G$ and $S_0\not\in\mathcal{M}_G$.
Let $K_1, K_2, \ldots, K_t$ be the pairwise distinct elements of $\set_K$ occurring as intersections $S\cap K$ for $S\in\set_G(S_0)$, and set
\[
\set_j(S_0) \coloneqq \left\{S\in\set_G(S_0): \N_S(S_0) = K_j\right\},\quad 1\leq j\leq t,
\]
so that
\[
\set_G(S_0) = \bigsqcup_{j=1}^t \set_j(S_0)
\]
for some $t>0$, where $\sqcup$ denotes disjoint union. We note that, for $j\in[t]$ and $S\in\set_j(S_0)$,
\[
S_0 < K_j = \N_S(S_0) \leq S
\]
since $S_0 < S$ by our second reduction, so that its normaliser in the nilpotent group $S$ grows.
Each of these elements $K_j$ may or may not lie in $\mathcal{M}_K$, i.e., be a maximal element of $\set_K$. After permuting indices if necessary, we may suppose that
\[
K_j\in\mathcal{M}_K\;\Longleftrightarrow\: j\in [s]
\]
for some $s$ with $0\leq s\leq t$. We claim that
\[\label{Eq:sdivide}
\mathcal{M}_K = \left\{K_1, K_2, \ldots, K_s\right\}.
\]
To see this, we first note that $\mathcal{M}_K = \set_K(S_0)$ by Condition~\eqref{(d)}, since $S_0\unlhd K$. Hence, a given element $S\in\mathcal{M}_K$ is contained in some $T \in\set_G(S_0)$, and we have $S \leq T \cap K\in \set_K$ by Condition~\eqref{(c)}, so that
\[
T \cap K = S = K_j
\]
for some $j\in [s]$. It follows that $\mathcal{M}_K \subseteq \{K_1, \ldots, K_s\}$, and the reverse inclusion holds by definition of~$s$.
It follows now that $s\equiv 1\pmod{m_G}$. Indeed, by Condition~\eqref{(d)}, we have
\[
s = \left\lvert \mathcal{M}_K \right\rvert = (K : \N_K(S))\, \big \vert\, (K:S)
\]
for some $S\in\mathcal{M}_K$. If $K\in \set_G$, then $\mathcal{M}_K = \{K\}$, so $s = 1 \equiv 1\pmod{m_G}$. Suppose, on the other hand, that $K\not\in\set_G$ (so that $K\not\in \set_K$ and $m_K$ is defined), and let $p\in\mathbb{P}$ be such that $p\mid s$. Then $p \mid (K:S)$, so $p \equiv 1\pmod{m_K}$ by definition of $m_K$, implying $s \equiv 1\pmod{m_K}$. Furthermore, by Lemma~\ref{Lem:mGdiv}, $m_G \mid m_K$ in this case, and we again deduce that $s\equiv 1\pmod{m_G}$.
For $j\in[s]$, we have $\set_j(S_0) = \set_G(K_j)$, thus $\left\lvert \set_j(S_0)\right\rvert \equiv 1\pmod{m_G}$ by choice of $S_0$, since $K_j > S_0$. Hence,
\begin{equation}\label{Eq:sPart}
\left\lvert \bigsqcup_{j=1}^s \set_j(S_0) \right\rvert = \sum_{j=1}^s \left\lvert \set_j(S_0) \right\rvert \equiv s \equiv 1\pmod{m_G}.
\end{equation}
In order to obtain the desired contradiction (to the existence of a counterexample $G$), it remains to show that
\begin{equation}\label{Eq:>sPart}
\left\lvert \bigsqcup_{j=s+1}^t \set_j(S_0)\right\rvert \equiv 0\pmod{m_G}.
\end{equation}
In fact, we shall establish the stronger property that
\begin{equation}\label{Eq:>sPart2}
\left\lvert \set_j(S_0)\right\rvert \equiv 0\pmod{m_G},\quad ssPart2} is false. Then, among the indices $j$ with $s < j \leq t$ and $\left\lvert \set_j(S_0) \right\rvert \not\equiv 0 \pmod{m_G}$, we may choose one, $j_0$ say, such that $\N_S(S_0) = K_{j_0}$ is of largest order among the subgroups $K_j$ of $K$ corresponding to these indices $j$. Consider the complex $\set_G(K_{j_0})$. Since $K_{j_0} > S_0$, we have $\left\lvert \set_G(K_{j_0}) \right\rvert \equiv 1\pmod{m_G}$ by our choice of $S_0$. We now split the set $\set_G(K_{j_0})$ according to the intersections of its elements with the normaliser $K = \N_G(S_0)$. Let $K_{j_0}, L_1, L_2, \ldots, L_r$ be the distinct elements of $\set_K$ arising in this way, so that
\begin{equation}\label{Eq:SGKj0Decomp}
\set_G(K_{j_0}) = \set_{j_0}(S_0) \,\sqcup\, \bigsqcup_{\rho=1}^r \left\{S\in\mathcal{M}_G: \N_S(S_0) = L_\rho \right\},
\end{equation}
where $L_\rho > K_{j_0}$ for all $\rho\in [r]$, with some $r$ such that $0\leq r < t$. Since $\set_G(K_{j_0}) \subseteq \set_G(S_0)$, the groups $L_\rho$ we encounter in this way are among the subgroups $K_1, \ldots, K_t$ of $K$ which occurred earlier on in the proof, so that $L_\rho = K_{\psi(\rho)}$ for a suitable injective map $\psi: [r] \rightarrow [t]$. We thus have
\[
\left\{S\in\mathcal{M}_G: \N_S(S_0) = L_\rho \right\} = \set_{\psi(\rho)}(S_0),\quad 1\leq \rho\leq r,
\]
and~\eqref{Eq:SGKj0Decomp} may be written more briefly in the form
\begin{equation}\label{Eq:SGKj0Decomp2}
\set_G(K_{j_0}) = \set_{j_0}(S_0)\,\sqcup\,\bigsqcup_{\rho=1}^r \set_{\psi(\rho)}(S_0).
\end{equation}
Now, as before, a given $L_\rho$ may or may not be a maximal element of $\set_K$, while $K_{j_0}\not\in\mathcal{M}_K$ since $j_0 > s$. After permuting indices if necessary, we may suppose that
\[
L_\rho \in \mathcal{M}_K\;\Longleftrightarrow\: \rho\in [q]
\]
for some integer $q$ with $0\leq q\leq r$. We have
\[
\set_{\psi(\rho)}(S_0) = \left\{S\in\mathcal{M}_G: \N_S(S_0) = L_\rho \right\} = \set_G(L_\rho),\quad 1\leq \rho\leq q,
\]
so $\left\lvert \set_{\psi(\rho)}(S_0) \right\rvert \equiv 1\pmod{m_G}$ for $1\leq \rho\leq q$ by our choice of $S_0$, since $L_\rho > K_{j_0} > S_0$. Hence,
\begin{equation}\label{Eq:ContraPrepare}
\left\lvert \bigsqcup_{\rho=1}^q \set_{\psi(\rho)}(S_0) \right\rvert = \sum_{\rho=1}^q \left\lvert \set_{\psi(\rho)}(S_0) \right\rvert \equiv q \pmod{m_G}.
\end{equation}
Next, we claim that
\begin{equation}\label{Eq:qdiv}
\set_K(K_{j_0}) = \left\{L_1, L_2, \ldots, L_q\right\}.
\end{equation}
Indeed, let $S\in \set_K(K_{j_0})$. Then there exists some $S'\in\mathcal{M}_G$ such that $S' \geq S$, a fortiori $S'\in \set_G(K_{j_0})$, and
\[
S' \cap K = S = L_\rho
\]
for some $\rho\in [q]$. This proves the forward inclusion of~\eqref{Eq:qdiv}, while the reverse inclusion holds by definition of $q$.
It now follows that $q\equiv 1\pmod{m_G}$. Indeed, if $K\in\set_G$, then $\mathcal{M}_K = \{K\}$, and
\[
q = \left\lvert \set_K(K_{j_0})\right\rvert = 1 \equiv 1\pmod{m_G}.
\]
If, on the other hand, $K\not\in\set_G$, then $K\not\in\set_K$, so that $m_K$ is defined, and the pair $(K, \set_K)$ satisfies Conditions~\eqref{(a)}{\rm--}\eqref{(e)} by Lemma~\ref{Lem:Inherit}. Consequently, we have
\[
q = \left\lvert \set_K(K_{j_0})\right\rvert \equiv 1\pmod{m_K}
\]
by choice of $G$, since $K < G$. Furthermore, $m_G \mid m_K$ by Lemma~\ref{Lem:mGdiv}, so that $q \equiv 1\pmod{m_G}$ follows again.
Combining~\eqref{Eq:ContraPrepare} with our last observation, we now conclude that
\[
\left\lvert \bigsqcup_{\rho=1}^q \set_{\psi(\rho)}(S_0) \right\rvert \equiv 1\pmod{m_G}.
\]
Moreover, for $\rho$ in the range $q < \rho \leq r$, we have $\psi(\rho) > s$ and $K_{\psi(\rho)} > K_{j_0}$. Hence, by our choice of the index $j_0$, it follows that
\[
\left\lvert \set_{\psi(\rho)}(S_0)\right\rvert \equiv 0\pmod{m_G},\quad q < \rho \leq r.
\]
We deduce that
\[
1 \equiv \left\lvert \set_G(K_{j_0})\right\rvert \equiv \left\lvert \set_{j_0}(S_0)\right\rvert + 1 \not\equiv 1 \pmod{m_G},
\]
since $\left\lvert \set_{j_0}(S_0)\right\rvert \not\equiv 0\pmod{m_G}$ by our choice of $j_0$. This contradiction shows that~\eqref{Eq:>sPart2} holds; thus also~\eqref{Eq:>sPart} holds true. Combining~\eqref{Eq:sPart} with~\eqref{Eq:>sPart}, we now deduce that
\[
\left\lvert \set_G(S_0)\right\rvert \equiv 1\pmod{m_G},
\]
contradicting our choice of the data $G$, $\set_G$, and $S_0$. This final contradiction shows that no counterexample to our theorem exists, and the result is proven. \qed
\section{Some remarks concerning the proof of Theorem~\ref{Thm:HirschClass}}\label{Sec:Hirsch}
Let $G$ be a group of odd order $N$. If $x\in G$ lies in a conjugacy class of size $h$, then the number of elements $y\in G$ commuting with $x$ is $N/h$. Thus, the total number of solutions $(x,y)\in G^2$ of the equation
\begin{equation}\label{Eq:Comm}
x^{-1} y^{-1} xy = 1
\end{equation}
is
\[
\sum_{\rho=1}^{k(G)} h_\rho N/h_\rho = N k(G).
\]
Hirsch's principal idea is to recount the non-trivial solutions $(x,y)$ of~\eqref{Eq:Comm} in batches according to the subgroup $\langle x, y\rangle$ they generate. Hence, we need to compute $\varphi_2(\langle x, y\rangle)$. Here, for a finite group $G$ and a positive integer $n$, $\varphi_n(G)$, the $n$th Eulerian number of $G$, is the number of $n$-tuples $(x_1, x_2, \ldots, x_n)\in G^n$ such that $G = \langle x_1, x_2, \ldots, x_n\rangle$. These numbers were introduced by Philip Hall~\cite{HallP}, who shows among other things that
\begin{equation}\label{Eq:HallEuler}
\varphi_n(G) = \sum_{H\leq G} \mu_G(H, G) \vert H\vert^n,
\end{equation}
where $\mu_G$ is the M\"obius function for the lattice of subgroups of $G$; cf.~\cite[Eqn.~(3.12)]{HallP}. Let $x = x_1 x_2\cdots x_k$ and $y=y_1y_2\cdots y_k$, where $x_i$ and $y_i$ are $p_i$-elements. Then
\[
\varphi_2(\langle x, y\rangle) = \varphi_2(\langle x_1, y_1\rangle) \varphi_2(\langle x_2, y_2\rangle) \cdots \varphi_2(\langle x_k, y_k\rangle).
\]
Now $\langle x_i, y_i\rangle$, if non-trivial, is either cyclic of order $p_i^m$, say, or an abelian group of the form $C_{p_i^m} \times C_{p_i^n}$, where $m\geq n$. For a prime $p$ and a positive integer $m$, we have
\begin{equation}\label{Eq:phi2Cyclic}
\varphi_2(C_{p^m}) = p^{2m} -p^{2m-2} = p^{2m-2} (p^2-1),
\end{equation}
since we only need to rule out those pairs in which the orders of both components are less than $p^m$. In dealing with the case of an abelian $p$-group of rank $2$, Hirsch offers, essentially without proof, the formulae
\begin{align}
\varphi_2(C_{p^m} \times C_{p^m}) &= (p^{2m} - p^{2m-2})[(p^{2m} - p^{2m-2}) - (p^m - p^{m-1})],\label{Eq:HirschForm1}
\\
\varphi_2(C_{p^m} \times C_{p^n}) &= \varphi(p^m) p^n \varphi(p^n)(p^m + p^{m-1}),\quad m>n,\label{Eq:HirschForm2}
\end{align}
where $\varphi$ is Euler's totient function. Of these, \eqref{Eq:HirschForm1} is false (the right-hand side overcounts whenever $m>1$), while~\eqref{Eq:HirschForm2} turns out to be correct. The fact that Hirsch distinguishes the cases $m=n$ and $m>n$ suggests that what he had in mind here is a direct enumeration of pairs of generators in these groups. Viewed in this way, the proof of~\eqref{Eq:HirschForm2} is considerably harder than that of~\eqref{Eq:HirschForm1}, so that the correctness of~\eqref{Eq:HirschForm2}, against the backdrop of~\eqref{Eq:HirschForm1} being erroneous, comes as somewhat of a surprise. In any case, Hall's formula~\eqref{Eq:HallEuler} suggests a more elegant and uniform approach to the computation of the Eulerian numbers of a finite abelian $p$-group, which we explain next.
If $G$ is a finite abelian $p$-group, then, by duality,
\[
\mu_G(H, G) = \mu_G(1, G/H),\quad H\leq G.
\]
Moreover, Delsarte~\cite{Delsarte} shows in this case that, for $H\leq G$,
\[
\mu_G(1, H) =
\begin{cases}
(-1)^s p^{\binom{s}{2}},& H\cong C_p^s,\\
0,& \text{otherwise.}
\end{cases}
\]
Hence, for
\[
G = C_{p^{\lambda_1}} \times C_{p^{\lambda_2}} \times \cdots \times C_{p^{\lambda_\ell}}
\]
of type $\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_\ell)$ with $\lambda_1 \geq \lambda_2\geq \cdots \geq \lambda_\ell \geq1$, we have
\begin{align*}
\varphi_n(G) &= \sum_{s=0}^\ell \sum_{\substack{H\leq G\\G/H \cong C_p^s}} \mu_G(1, G/H) \vert H\vert^n\\
&= \sum_{s=0}^\ell \sum_{\substack{H\leq G\\G/H \cong C_p^s}} (-1)^s p^{\binom{s}{2}} p^{n(\lambda_1 + \cdots + \lambda_\ell - s)}.
\end{align*}
Again by duality, the number of subgroups $H\leq G$ with $G/H \cong C_p^s$ equals the number of subgroups $K\leq G$ with $K \cong C_p^s$. By a well-known result, the number of subgroups of type $\nu = (\nu_1,\ldots,\nu_\ell)$ in an abelian $p$-group of type $\lambda = (\lambda_1, \ldots, \lambda_\ell)$ is given by the formula
\begin{equation}\label{Eq:psubCount}
\prod_{i\geq 1} p^{\nu_{i+1}'(\lambda_i' - \nu_i')} \kbinom{\lambda_i' - \nu_{i+1}'}{\nu_i' - \nu_{i+1}'}_p,
\end{equation}
where $\lambda', \nu'$ are the conjugates of the partitions $\lambda$ and $\nu$, respectively, and
\begin{equation}
\kbinom{n}{k}_p = \prod_{i=0}^{k-1} \frac{1 - p^{n-i}}{1-p^{i+1}}
\end{equation}
is the number of $k$-dimensional subspaces of an $n$-dimensional vector space over the field $\mathbb{Z}/p\mathbb{Z}$; see, for instance,~\cite[Eqn.~(1)]{Butler}. If $H\leq G$ is such that $H\cong C_p^s$, then its type $\nu$ takes the form
\[
\nu = (\underbrace{1, 1, \ldots, 1}_{s}, \underbrace{0, 0, \ldots, 0}_{\ell - s}),
\]
and, consequently,
\[
\nu' = (s, \underbrace{0, 0, \ldots, 0}_{\ell-1}).
\]
It thus follows from~\eqref{Eq:psubCount} that the number of subgroups $H$ in a group $G$ of type $\lambda$ with $H \cong C_p^s$ is simply given by the Gau{\ss} coefficient $\kbinom{\ell}{s}_p$. We thus obtain
\[
\varphi_n(G) = \sum_{s=0}^\ell (-1)^s p^{\binom{s}{2}} \kbinom{\ell}{s}_p p^{n(\lambda_1+\cdots +\lambda_\ell - s)}.
\]
Applying the $q$-binomial theorem
\[
(1+z)(1+qz) \cdots (1+q^{\ell-1}z) = \sum_{s=0}^\ell \kbinom{\ell}{s}_q q^{\binom{s}{2}} z^s
\]
with $q=p$ and $z = -p^{-n}$, we find the following.
\begin{prop}\label{Prop:EulerNumb}
Let
\[
G = C_{p^{\lambda_1}} \times C_{p^{\lambda_2}} \times \cdots \times C_{p^{\lambda_\ell}}
\]
be a finite abelian $p$-group of type $\lambda = (\lambda_1, \ldots, \lambda_\ell),$ where $\ell\geq 1$ and $\lambda_1 \geq \lambda_2\geq \cdots \geq \lambda_\ell \geq 1$, and let $n$ be a positive integer. Then the $n$th Eulerian number $\varphi_n(G)$ of $G$ is given by the formula
\begin{equation}\label{Eq:Eulernp}
\varphi_n(G) = p^{n(\lambda_1+\cdots+\lambda_\ell) - \binom{n+1}{2} + \binom{n-\ell+1}{2}} \prod_{i=0}^{\ell-1}(p^{n-i}-1).
\end{equation}
\end{prop}
In particular, for $\ell = n = 2$, Equation~\eqref{Eq:Eulernp} yields
\begin{equation}\label{Eq:phi2p}
\varphi_2(C_{p^m} \times C_{p^r}) = p^{2m+2r-3} (p^2-1)(p-1),\quad m\geq r,
\end{equation}
which is what Hirsch needs at this point.
Since $\langle x, y\rangle$ is non-trivial by assumption, at least one of its Sylow subgroups $\langle x_i, y_i\rangle$ must be non-trivial, thus $d\mid\varphi_2(\langle x, y\rangle)$ by~\eqref{Eq:phi2Cyclic} and~\eqref{Eq:phi2p}, so $N k(G) \equiv 1 \pmod{d}$. Since $N^2\equiv 1\pmod{d}$ by definition of $d$, we find that $k(G) \equiv N \pmod{d}$. The second (and more subtle) part of Hirsch's argument then deals with the task of strengthening the modulus $d$ by a factor $2$, and is correct, apart from relying on the unsubstantiated Claim~\ref{claim}.
%\subsection*{Acknowledgements}
%This research was supported through the program ``Research in Pairs'' by the Mathematisches Forschungsinstitut Oberwolfach in 2019.
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