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\title{On the factorised subgroups of products of cyclic and non-cyclic finite $p$-groups}
\alttitle{Sur les sous-groupes factoris\'{e}s des produits de $p$-groupes finis cycliques et non cycliques}

\author{\firstname{Brendan} \lastname{McCann}}
\address{Department of Computing and Mathematics, South-East Technological University Waterford, Cork Road, Waterford, Ireland}
\email{brendan.mccann@setu.ie}

\subjclass{20D40, 20D15}

\keywords{\kwd{factorised groups}
\kwd{products of groups}
\kwd{finite $p$-groups}}

\altkeywords{\kwd{groupes factoris\'{e}s}
\kwd{produit de groupes}
\kwd{$p$-groupes finis}}

\begin{abstract}
Let $p$ be an odd prime and let $ G = AB $ be a finite $p$-group that is the product of a cyclic subgroup $A$ and a non-cyclic subgroup $B$. Suppose in addition that the nilpotency class of $B$ is less than $\frac{p}{2}$. We denote by $\uOmega_i(B) $ the subgroup of $B$ generated by the $p^i$-th powers of elements of $B$, that is $ \uOmega_i(B) = \langle b^{p^i} \mid b \in B \rangle $. In this article we show that, for all values of $i$, the set $ A \uOmega_i(B) $ is a subgroup of $G$. We also present some applications of this result. 
\end{abstract}

\begin{altabstract}
Soient $p$ un nombre premier impair et $ G = AB $ un $p$-groupe fini qui est le produit des sous-groupes $A$ et $B$, tels que $A$ soit un sous-groupe cyclique et $B$ soit un sous-groupe non cyclique. Supposons \'{e}galement que la classe de nilpotence de $B$ soit inf\'{e}rieure \`{a} $\frac{p}{2}$. On note $ \uOmega_i(B) $ le sous-groupe de $B$ engendr\'{e} par les puissances $p^i$ des \'{e}l\'{e}ments de $B$, alors $ \uOmega_i(B) = \langle b^{p^i} \mid b \in B \rangle $. Dans cet article nous montrons que, pour chaque valeur du nombre $i$, l'ensemble $ A \uOmega_i(B) $ est sous-groupe du groupe $G$. Nous pr\'{e}sentons \'{e}galement quelques applications de ce r\'{e}sultat. 
\end{altabstract}

\dateposted{2024-05-02}
\begin{document}
\maketitle

\vspace{-1em}

\section{Introduction}

We say that the group $G$ is the product of the subgroups $A$ and $B$ if $G$ is equal to the set product of $A$ and $B$, that is $ G = \{ ab \mid a \in A, b\in B \} $. Such groups are also referred to as factorised groups. For a factorised group $ G = AB $, is seems natural to ask whether we can identify subgroups of the ``factors'' $A$ and $B$ whose set product with each other forms a subgroup of $G$. By a result from elementary Group Theory (see, for instance, \cite[I 2.12 Hilfssatz]{Huppert}), this is equivalent to determining subgroups $ A_1 \leqslant A $ and $B_1 \leqslant B$ such that their set products satisfy $ A_1 B_1 = B_1 A_1 $. Despite an extensive literature on factorised groups, as documented in Amberg, Franciosi and de Giovanni~\cite{Ambergetal} and Ballester-Bolinches, Esteban-Romero and Assad~\cite{Ballester-Bolinchesetal}, results in this direction remain scarce, even when both factors are abelian. The most notable result goes back to that of Huppert (see~\cite[Satz~3]{Huppertzyklisch} or~\cite[Corollary~3.1.9]{Ballester-Bolinchesetal}), which states that if the finite $p$-group $ G = AB $ is the product of two cyclic subgroups $A$ and $B$, then $G$ is the \emph{totally permutable} product of $A$ and $B$, that is $ A_1 B_1 \leqslant G$ for each $A_1 \leqslant A $ and $B_1 \leqslant B$. Since $A$ and $B$ are cyclic $p$-groups, this can be restated as $ \Omega_s(A) \Omega_t(B) \leqslant G $ for all values of $s$ and $t$, where the characteristic subgroup $ \Omega_i(W) $ of the finite $p$-group $W$ is defined by $ \Omega_i(W) = \langle w \in W \mid w^{p^i} = 1 \rangle $. In general, we cannot expect that $G = AB $ will be a totally permutable product if either of the subgroups $A$ or $B$ is non-cyclic. However, if the prime $p$ is odd, then in the case where $ A \cap B = 1 $, $A$ is cyclic and the nilpotency class $ \bmcl(B) $ of B satisfies $ \bmcl(B) < \frac{p}{2} $, it has been shown in~\cite[Theorem~2.6$\MK$(i)]{McCann4} that $ A \Omega_i(B) \leqslant G $ for all values of $i$. Since $A$ is cyclic, it is then straightforward to deduce that $ \Omega_s(A) \Omega_t(B) \leqslant G $, for all values of $s$ and $t$. This can be viewed as a partial analogue to Huppert's result for products of cyclic subgroups. It is notable that Example~2.8 of~\cite{McCann4} further shows that $ A \Omega_i(B) $ is not always a subgroup of $G$ when $ A \cap B \neq 1 $. This places a limit on the extent to which Huppert's result can be directly generalised. 

In this paper we present some results that can be considered as dual to those contained in~\cite{McCann4}. Our main result is Theorem~\ref{e7}, which states that if $G = AB $ is a finite $p$-group for subgroups $A$ and $B$ such that $A$ is cyclic and $ \bmcl(B) < \frac{p}{2} $ then, for all~$i$, $ A \uOmega_i(B) \leqslant G $. Here $ \uOmega_i(B) $ denotes the characteristic subgroup of $B$ generated by the $p^i$-th powers of elements of $B$, that is $ \uOmega_i(B) = \langle b^{p^i} \mid b \in B \rangle $. We apply Theorem~\ref{e7} in Theorem~\ref{e8} to provide results that are dual to~\cite[Theorems~2.6 and 2.9]{McCann4}. This leads to a new derivation of established results concerning the structure of products of cyclic and non-cyclic finite $p$-groups (see~\cite[Theorems~2.9 and 4.1]{McCann4}). Namely we show in Corollary~\ref{e9} that if $p$ is an odd prime and $G = AB $ is a finite $p$-group, where $A$ is a cyclic subgroup and $B$ is a subgroup such that $ \bmcl(B) < \frac{p}{2}$ and $ \bmexp(B) = p^k$, then $ \Omega_k(A) B \trianglelefteqslant G $ and $ \bmdl(G) \le 1 + k + \bmdl(B)$. 

We denote the $n^{ \text{th} }$ term of the derived series of a group $G$ by $G^{(n)}$. Thus $ G^{(0)} = G $, $G^{(1)} = G^{'} $ and $ G^{(n+1)} = [G^{(n)}, G^{(n)}] $ for $ n \ge 1 $. We denote the derived length of a soluble group $G$ by $ \bmdl(G) $. The $ i^{\text{th}}$ term of the lower (or descending) central series of $G$ will be denoted by $ K_i(G) $, that is $ K_1(G) =G $, $ K_2(G) = G^{'} $ and $ K_{i+1}(G) = [K_i(G), G] $ for $ i \ge 2 $. If $G$ is nilpotent then $ \bmcl(G) $ will denote the class of $G$. We note in particular that if $\bmcl(G) < s $, then $ K_s(G) = 1 $. The normal closure of the subgroup $U$ in $G$ is denoted by $U^G$, so that $ U^G = \langle U^g \mid g \in G \rangle $. We remark that if $ N \trianglelefteqslant G $, then $ (UN/N)^{G/N} = U^G N / N $. We finally denote the cyclic group of order $p^n$ by $ C_{p^n} $. 


\section{Results and Proofs}

We begin with some elementary lemmas that will find application in the proofs of Theorems~\ref{e7} and~\ref{e8}. 

\begin{lemm}\label{e1}
Let $G$ be a finite $p$-group and let $ N \trianglelefteqslant G $. Let $U$ be a subgroup of $G$. Then $ \uOmega_i(UN/N) = \uOmega_i(U) N/N $. 
\end{lemm}

\begin{proof}
By definition we have
\[
\uOmega_i(UN/N) = \langle (u N)^{p^i} \mid u \in U \rangle N/ N = \langle u^{p^i}N \mid u \in U \rangle N / N = \langle u^{p^i} \mid u \in U \rangle N / N. 
\]
It then follows that $ \uOmega_i(UN/N) = \uOmega_i(U) N / N $. 
\end{proof}

\begin{lemm}\label{e2}
Let $G$ be a group such that $G = AB $, where $A$ and $B$ are subgroups of $G$. Suppose that $ B_1 \trianglelefteqslant B $ is such that $ A B_1 \leqslant G$. Then $ B_1^G = B_1^{ AB_1 } = A_1 B_1 $, where $ A_1 = A \cap B_1^G $. 
\end{lemm}

\begin{proof}
Since $ B_1 \trianglelefteqslant B $, we see that
\[
B_1^G = \langle B_1^g \mid g \in G \rangle = \langle B_1^{ ba } \mid b \in B, \ a \in A \rangle = \langle B_1^a \mid a \in A \rangle \leqslant B_1^{ A B_1 } \leqslant A B_1. 
\]
But $ B_1 \leqslant B_1^{ AB_1 } \leqslant B_1^G $, so we have
\[
B_1^G = B_1^{ AB_1 } = (A \cap B_1^G) B_1 = A_1 B_1,
\]
where $ A_1 = A \cap B_1^G $. 
\end{proof}


\begin{lemm}\label{e3}
Let $G$ be a group and let $ N \trianglelefteqslant G $. Suppose that $H$ and $K$ are subgroups of $G$ with $ N \leqslant H \cap K $ such that $ (H/N) (K/N) \leqslant G/N $. Then $ HK \leqslant G $. Moreover, if $ (H/N) (K/N) \trianglelefteqslant G/N $, then $ HK \trianglelefteqslant G $. 
\end{lemm}

\begin{proof}
We let $ h \in H $ and $ k \in K $. Since $ (H / N) (K / N) \leqslant G / N $, we have $ (H/N) (K/N) = (K/N) (H/N) $. Hence there exist $ h_1 \in H $ and $ k_1 \in K $ such that $ (hN) (kN) = (k_1N) (h_1 N) $. Equivalently, we see that $ hkN = k_1 h_1 N $. Thus there exists $ x \in N $ such that $ h k = k_1 h_1 x $. But $ N \leqslant H \cap K $, so $ h_1 x \in H $. It follows that
\[
hk = k_1 (h_1 x) \in K H. 
\]
Hence $ HK \subseteq KH $. Since $ (K/N) (H/N) = (H/N) (K/N) $, we similarly see that $ KH \subseteq HK $. We conclude that $ HK = KH $, so $ HK \leqslant G $. 

We now make the additional assumption that $ (H/N) (K/N) \trianglelefteqslant G/N $. We let $ h \in H $ and $ k \in K $. If $ g \in G $ then, by normality, we have $ ((hN)(kN))^{gN} = (hkN)^{gN} = (hk)^{g}N \in (H/N) (K/N) $. Hence there exist $ h_1 \in H $ and $k_1 \in K $ such that $ (hk)^g N = (h_1N) (k_1N) = h_1k_1 N $. It follows that there exists $ x \in N \leqslant H \cap K $ such that $ (hk)^g = h_1 k_1 x $. Since $ x \in K $, we have $ k_1 x \in K $, so $ (hk)^g = h_1 (k_1 x) \in HK $. We thus conclude that $ HK \trianglelefteqslant G $. 
\end{proof}


Our next lemma is a consequence of the Hall-Petrescu identity (see~\cite[III 9.4 Satz]{Huppert}). 

\begin{lemm}\label{e4}
Let $p$ be a prime and let $G$ be a finite $p$-group such that $ \bmcl(G) < p $. Suppose, in addition, that $ \bmexp(G^{'}) \le p $. Then for $ g_1, g_2 \in G $, we have $ (g_1 g_2)^p = g_1^p g_2^p $. 
\end{lemm}

\begin{proof}
Since $ \bmcl(G) < p $, we have $ K_p(G) = 1 $. We further have $ K_i(G) \leqslant G^{'} $ for $ i \ge 2 $. Hence $ \bmexp(K_i(G)) \le p $ for $ i \ge 2 $. By the Hall-Petrescu identity, there exist $ c_2, \dots, c_p $, with $ c_s \in K_s(G) $ for $ s = 2, \dots, p $, such that
\[
(g_1 g_2)^p = g_1^p g_2^p c_2^{ \binom{p}{2} } \cdots c_{p-1}^{ \binom{p}{p-1} } c_p. 
\]
We note that $ c_p \in K_p(G) = 1 $. Since $ \bmexp(K_s(G)) \le p $ for $ s \ge 2 $, we further see that $ c_s^p = 1 $ for $ s \ge 2 $. But $ p $ is a divisor of $ \binom{p}{s} $ for $ s = 2, \dots, p-1$. Hence $ c_s^{ \binom{p}{s} } = 1 $ for $ s = 2, \dots, p-1 $. It follows that $ (g_1 g_2)^p = g_1^p g_2^p $. 
\end{proof}


\begin{coro}\label{e4a}
Let $p$ be an odd prime and let $G$ be a finite $p$-group such that $ \bmcl(G) < p $ and $ \bmexp(G) = p^2 $. Let $ y \in G $ be such that $ \bmo(y) = p ^ 2 $ and suppose that there exists $ W \leqslant G $ such that $ | G : W | = p $ and $ \bmexp(W) = p $. Then $ \uOmega_1(G) = \langle y^p \rangle $. 
\end{coro}

\begin{proof}
By definition, we have $ \langle y^p \rangle \leqslant \uOmega_1(G)$. Since $ \bmexp(W) = p $, we see that $ y \notin W $. Now $ | G : W | = p $ so, by comparison of orders, we have $ G = W \langle y \rangle $. In addition, we have $ W \trianglelefteqslant G $ and $ G / W \cong C_p $. Hence $ G^{'} \leqslant W $, so $ \bmexp(G^{'}) \le p $. We let $ g \in G $. Then there exist a suitable value $ \alpha$ and an element $ w \in W $ such that $ g = w y^{\alpha} $. By Lemma~\ref{e4}, we have $ g^p = (w y^{\alpha})^p = w^p (y^{\alpha})^p = w^p (y^p)^{\alpha} $. But $ \bmexp(W) = p $, so $ w^p = 1 $. Hence $ g^p = (y^p)^{\alpha} \in \langle y^p \rangle $, so $ \uOmega_1(G) \leqslant \langle y^p \rangle $. 
\end{proof}


The following theorem is a restatement of results of Philip Hall concerning regular $p$-groups (see~\cite[III 10.2 Satz and 10.5 Hauptsatz]{Huppert}). 

\begin{theo}\label{e5}
Let $p$ be an odd prime and let $G$ be a finite $p$-group such that $ \bmcl(G) < p $. Then, for all~$i$,
\begin{enumerate}\romanenumi %[(i)]
\item \label{6i} $ \Omega_i(G) = \{ g \in G \mid g^{p^i} = 1 \} $. In particular $ \bmexp(\Omega_i(G)) \le p^i $;
\item \label{6ii} $ \uOmega_i(G) = \{ g^{p^i} | g \in G \} $. 
\end{enumerate}
\end{theo}


We next prove a consequence of Theorem~\ref{e5}. 

\begin{coro}\label{e5a}
Let $p$ be an odd prime and let $G$ be a finite $p$-group such that $ \bmcl(G) < p $. Then, for all~$i$, we have $ \uOmega_{i+1}(G) = \uOmega_1(\uOmega_i(G)) $. 
\end{coro}

\begin{proof}
By Theorem~\ref{e5}$\MK$\eqref{6ii}, we see that $ \uOmega_{i+1}(G) = \{ g^{p^{i+1}} \mid g \in G \} = \{ (g^{p^i})^p \mid g \in G \} = \{ u^p \mid u \in \{ g^{p^i} \mid g \in G \} \} = \{ u^p \mid u \in \uOmega_i(G) \} = \uOmega_1(\uOmega_i(G)) $. 
\end{proof}


We make extensive use of the next result, which deals with $p$-groups that are the product of a cyclic subgroup and a subgroup that has exponent $p$ and class less than $ \frac{p}{2} $. The first part is a restatement of~\cite[Lemma~2.2]{McCann4}. 

\begin{lemm}\label{e6}
Let $p$ be an odd prime and let $ G = AB $ be a finite $p$-group for subgroups $A$ and $B$ such that $A$ is cyclic, $ \bmexp(B) = p $ and $ \bmcl(B) < \frac{p}{2} $. Then:
\begin{enumerate}\romanenumi%[(i)]
\item \label{8i} $ \Omega_1(A) B \trianglelefteqslant G $;
\item \label{8ii} $ \bmexp(B^G) = p $. 
\end{enumerate}
\end{lemm}

\begin{proof}
We note that the proof of~\eqref{8i} is given in~\cite[Lemma~2.2]{McCann4}. For \eqref{8ii}, we see that the result is trivial if $ B \trianglelefteqslant G$. Letting $ A = \langle x \rangle $, we can thus assume that $ B^x \neq B $. In particular $A$ is non-trivial, so $ \Omega_1(A) \cong C_p $. By \eqref{8i}, we have $ \Omega_1(A) B \trianglelefteqslant G $. Hence, by comparison of orders, we have $ B^G = \Omega_1(A) B = B B^x $. We further see that $ | B^G : B | = | B^G : B^x | = | \Omega_1(A) | = p $. In particular, $ B$ and $B^x$ are normal in $B^G$. It follows that $ \bmcl(B^G) \le \bmcl(B) + \bmcl(B^x) < \frac{p}{2} + \frac{p}{2} = p $. Now $ B^G = B B^x $, so $ B^G = \Omega_1(B^G) $. By Theorem~\ref{e5}, we then have $ \bmexp(B^G) \le p $. But $ \bmexp(B) = p $, so we conclude that $ \bmexp(B^G) = p $. 
\end{proof}

We now proceed to our main result. 

\begin{theo}\label{e7}
Let $p$ be an odd prime and let $ G = AB $ be a finite $p$-group for subgroups $A$ and $B$ such that $A$ is cyclic and $ \bmcl(B) < \frac{p}{2} $. Then, for all~$i$, we have $ A \uOmega_i(B) \leqslant G $. 
\end{theo}

\begin{proof}
The bulk of our proof is devoted to showing that $ A \uOmega_1(B) \leqslant G$. For this, we use induction on $|G|$. If $ G = A $, then the result is trivial. Hence we can assume that $A$ is a proper subgroup of $G$. We can further assume that $ B \not\trianglelefteqslant G $, as otherwise $ \uOmega_1(B) \trianglelefteqslant G $ and we trivially have $ A \uOmega_1(B) \leqslant G $. Since $ | G : B | \le p $ is then excluded, we have $ |A| \ge p^2 $. In particular, we see that $ \Omega_1(A) $ is a proper subgroup of $A$ with $ \Omega_1(A) \cong C_p $. 

By a result of Morigi (see~\cite[Lemma~1]{Morigi} or~\cite[Lemma~3.3.8]{Ballester-Bolinchesetal}), there exists a non-trivial normal subgroup $ W \trianglelefteqslant G $ such that either $ W \leqslant A $ or $ W \leqslant B$. Since $G$ is a finite $p$-group, we then have either $ A \cap Z(G) \neq 1 $ or $ B \cap Z(G) \neq 1 $. We assume first that $ A \cap Z(G) \neq 1 $. By minimality, it follows that $ \Omega_1(A) \leqslant Z(G)$. By induction, we see that $ (A /\Omega_1(A)) \uOmega_1(B \Omega_1(A) / \Omega_1(A)) \leqslant G / \Omega_1(A) $. By Lemma~\ref{e1}, we further have $ \uOmega_1(B \Omega_1(A) / \Omega_1(A)) = \uOmega_1(B) \Omega_1(A) / \Omega_1(A) $. Hence
\[
A /\Omega_1(A) (\uOmega_1 (B) \Omega_1(A) / \Omega_1(A)) \leqslant G / \Omega_1(A). 
\]
We now apply Lemma~\ref{e3} to see that $ A \uOmega_1(B) \Omega_1(A) \leqslant G $. Since $ \Omega_1(A) \leqslant Z(G) $, we have
\[
A \uOmega_1(B) \Omega_1(A) = A \Omega_1(A) \uOmega_1(B) = A \uOmega_1(B),
\]
so $ A \uOmega_1(B) \leqslant G $. We can therefore assume that $ A \cap Z(G) = 1 $ and thus $ B \cap Z(G) \neq 1 $. 

We let $ z \in B \cap Z(G) $ be such that $ \langle z \rangle \cong C_p $. By induction, we have $ (A \langle z \rangle / \langle z \rangle) (\uOmega_1(B /\langle z \rangle) \leqslant G / \langle z \rangle $. By Lemma~\ref{e1}, we see that $ \uOmega_1(B /\langle z \rangle) = \uOmega_1(B \langle z \rangle / \langle z \rangle) = \uOmega_1(B) \langle z \rangle / \langle z \rangle $. Thus $ (A \langle z \rangle / \langle z \rangle) (\uOmega_1(B) \langle z \rangle / \langle z \rangle) \leqslant G / \langle z \rangle $. By Lemma~\ref{e3}, we then have $ A \langle z \rangle \uOmega_1(B) \langle z \rangle \leqslant G $. But $ A \langle z \rangle \uOmega_1(B) \langle z \rangle = A \uOmega_1(B) \langle z \rangle $. Hence
\[
A \uOmega_1(B) \langle z \rangle \leqslant G. 
\]
Since $A$ is a proper subgroup of $G$, we let $M$ be a maximal subgroup of $G$ such that $ A \leqslant M $. Then $ M = A (B \cap M) $. We let $ B_1 = B \cap M $, so that $ M = A B_1$. Since $ A \leqslant M $, we have $ A \cap B = A \cap B \cap M = A \cap B_1 $. Hence
\[
| G | = \dfrac{ |A| |B| }{ | A \cap B | } = p |M| = p \dfrac{ |A| |B_1| }{ | A \cap B_1 | } = p \dfrac{ |A| |B_1| }{ | A \cap B | }. 
\]
It follows that
\[
| B : B_1 | = \dfrac{ |B| }{ |B_1| } = p. 
\]

We consider the case where $ \uOmega_1(B_1) \neq 1 $. By induction, we have $ A \uOmega_1(B_1) \leqslant AB_1 \leqslant G $. Now $ \uOmega_1(B_1) $ is characteristic in $B_1$ and $ | B : B_1 | = p $, so $ B_1 \trianglelefteqslant B$. Hence $ \uOmega_1(B_1) \trianglelefteqslant B $. Applying Lemma~\ref{e2}, we have $ \uOmega_1(B_1)^G = (A \cap \uOmega_1(B_1)^G) \uOmega_1(B_1) $. We let $ N = \uOmega_1(B_1)^G $. Since $ \uOmega_1(B_1) \neq 1 $, we see that $N$ is a non-trivial normal subgroup of $G$. By our induction hypothesis and Lemma~\ref{e1}, we then have
\[
(AN/N) \uOmega_1(BN/N) = (AN/N) (\uOmega_1(B) N / N) \leqslant G / N. 
\]
Hence, by Lemma~\ref{e3}, we have $ A N \uOmega_1(B) N = A N \uOmega_1(B) \leqslant G $. But $ N = \uOmega_1(B_1)^G = (A \cap \uOmega_1(B_1)^G) \uOmega_1(B_1) $, so
\[
A N \uOmega_1(B) = A (A \cap \uOmega_1(B_1)^G) \uOmega_1(B_1) \uOmega_1(B) = A \uOmega_1(B). 
\]
Thus, in the case where $ \uOmega_1(B_1) \neq 1 $, we conclude that $ A \uOmega_1(B) \leqslant G $. 

We now assume that $ \uOmega_1(B_1) = 1 $. Thus $ \bmexp(B_1) \le p $. Since $ | B : B_1 | = p $, it follows that $ \bmexp(B) \le p^2 $. If $ \bmexp(B) = p $, then $ \uOmega_1(B) = 1 $ and the result is trivial. Hence we can assume that there exists $ y \in B $ such that $ \bmo(y) = p^2 $. But $ | B : B_1 | = p $ and $ \bmcl(B) < \frac{p}{2} < p $, so we can apply Corollary~\ref{e4a} to see that $ \uOmega_1(B) = \langle y^p \rangle \cong C_p $. In particular, we have $ \uOmega_1(B) \leqslant Z(B) $. From the above, there exists $ z \in B \cap Z(G) $ with $ \langle z \rangle \cong C_p $ such that $ A \uOmega_1(B) \langle z \rangle \leqslant G $. But $ \uOmega_1(B) \leqslant Z(B) $, so $ \uOmega_1(B) \langle z \rangle \leqslant Z(B) $. Hence we can apply Lemma~\ref{e2} to see that $ (\uOmega_1(B) \langle z \rangle)^G = (\uOmega_1(B) \langle z \rangle)^{ A \uOmegas_1(B) \langle z \rangle } $. Since $ \uOmega_1(B) \cong \langle z \rangle \cong C_p $, we see that $ \uOmega_1(B) \langle z \rangle $ is elementary abelian. In particular, we have $ \bmexp(\uOmega_1(B) \langle z \rangle) = p $ and $ \bmcl(\uOmega_1(B) \langle z \rangle) = 1 < \frac{p}{2} $. We can thus apply Lemma~\ref{e6}$\MK$\eqref{8i} to see that $ \Omega_1(A) \uOmega_1(B) \langle z \rangle \trianglelefteqslant A \uOmega_1(B) \langle z \rangle $. Hence
\[
(\uOmega_1(B) \langle z \rangle)^G = (\uOmega_1(B) \langle z \rangle)^{ A \uOmegas_1(B) \langle z \rangle } \leqslant \Omega_1(A) \uOmega_1(B) \langle z \rangle. 
\]

We can assume that $ B $ is a proper subgroup of $G$, since otherwise $ \uOmega_1(B) \trianglelefteqslant G $ and it trivially follows that $ A \uOmega_1(B) \leqslant G $. We let $M_1$ be a maximal subgroup of $G$ such that $ B \leqslant M_1 $. Then $ | G : M_1 | = p $ and $ M_1 = (A \cap M_1) B $. We let $ A_1 = A \cap M_1 $. As above, we have $ | A : A_1 | = p $. Since $ |A| \ge p^2 $, we see that $ \Omega_1(A) \leqslant A_1 $. Hence $ \Omega_1(A) = \Omega_1(A_1) \cong C_p $. By induction, we have $ A_1 \uOmega_1(B) \leqslant A_1 B \leqslant G $. We can thus apply Lemma~\ref{e2} and Lemma~\ref{e6}$\MK$\eqref{8i} to see that
\[
\uOmega_1(B)^{ A_1 B } = \uOmega_1(B)^{ A_1 \uOmegas_1(B) } \leqslant \Omega_1(A) \uOmega_1(B) \trianglelefteqslant A_1 \uOmega_1(B). 
\]
Bearing in mind that $ \Omega_1(A) \cong C_p $, we see by comparison of orders that either $ \uOmega_1(B)^{ A_1 B } = \Omega_1(A) \uOmega_1(B) $ or $ \uOmega_1(B)^{ A_1 B } = \uOmega_1(B) $. 

If $ \uOmega_1(B)^{ A_1B } = \Omega_1(A) \uOmega_1(B) $, then $ \Omega_1(A) \uOmega_1(B) \trianglelefteqslant A_1 B $. In particular $ \Omega_1(A) \uOmega_1(B) B = \Omega_1(A) B \leqslant A_1 B $, so $ \Omega_1(A) B $ is a subgroup of $G$. We can then apply Lemma~\ref{e2} to see that
\[
\Omega_1(A)^G = \Omega_1(A)^{ \Omega_1(A) B } \leqslant \Omega_1(A) \uOmega_1(B) \trianglelefteqslant A_1 B. 
\]
But $ \uOmega_1(B) \cong C_p $, so either $ \Omega_1(A)^G = \Omega_1(A) $ or $ \Omega_1(A)^G = \Omega_1(A) \uOmega_1(B) $. In the former case, we have $ \Omega_1(A) \trianglelefteqslant G $. But $ \Omega_1(A) \cong C_p $, so $ 1 \neq \Omega_1(A) \leqslant A \cap Z(G) $, which has been excluded. If $ \Omega_1(A)^G = \Omega_1(A) \uOmega_1(B) $ then $ \Omega_1(A) \uOmega_1(B) \trianglelefteqslant G $, so $ A \uOmega_1(B) = A \Omega_1(A) \uOmega_1(B) \leqslant G $, and we are done. We can thus assume that
\[
\uOmega_1(B)^{ A_1 B } = \uOmega_1(B) \trianglelefteqslant A_1 B. 
\]
Since $ \uOmega_1(B) \cong C_p $, it follows that $ \uOmega_1(B) \leqslant Z(A_1 B) $. But $ A_1 B = M_1 \trianglelefteqslant G $, so we further have $ \uOmega_1(B)^G \leqslant Z(A_1 B) \trianglelefteqslant G $. Now, for $z$ as above, we have $ \langle z \rangle \leqslant B \cap Z(G) \leqslant Z(A_1 B) $. We therefore see that
\[
(\uOmega_1(B) \langle z \rangle)^G = \uOmega_1(B)^G \langle z \rangle \leqslant Z(A_1 B). 
\]

We recall from the above that $ (\uOmega_1(B) \langle z \rangle)^G \leqslant \Omega_1(A) \uOmega_1(B) \langle z \rangle $. Since $ \Omega_1(A) \cong C_p $, we have either $ \Omega_1(A) \leqslant (\uOmega_1(B) \langle z \rangle)^G \leqslant Z(A_1 B) $ or $ \Omega_1(A) \cap (\uOmega_1(B) \langle z \rangle)^G = 1 $. In the former case we see that $B$ centralises $ \Omega_1(A) $. But then $ G = AB \leqslant C_G(\Omega_1(A)) $, so $ \Omega_1(A) \leqslant Z(G) $. Since this is excluded, we have $ \Omega_1(A) \cap (\uOmega_1(B) \langle z \rangle)^G = 1 $. By comparison of orders, it follows that
\[
(\uOmega_1(B) \langle z \rangle)^G = \uOmega_1(B) \langle z \rangle \trianglelefteqslant G. 
\]
We let $ N_1 = \uOmega_1(B) \langle z \rangle $ and have $ N_1 \leqslant B $. Hence $ G/N_1 = (AN_1/N_1) (B/N_1) $. If $ A \cap N_1 \neq 1 $ then, by minimality, we have $ \Omega_1(A) \leqslant N_1 = (\uOmega_1(B) \langle z \rangle)^G $. But this has been excluded, so $ A \cap N_1 = 1 $. Hence $ \Omega_1(AN_1 / N_1) = \Omega_1(A) N_1 / N_1 \cong \Omega_1(A) \cong C_p $. We further note that $ \bmcl(B/N_1) \le \bmcl(B) < \frac{p}{2} $. Since $ \bmexp(B) = p^2 $ and $ \uOmega_1(B) \leqslant N_1 $, we also see that $ \bmexp(B/N_1) = p $. We can thus apply Lemma~\ref{e6}$\MK$\eqref{8i} to see that
\[
\Omega_1(AN_1 / N_1) (B/N_1) = (\Omega_1(A) N_1 / N_1) (B/ N_1) \trianglelefteqslant G / N_1. 
\]
By Lemma~\ref{e3}, we then have $ \Omega_1(A) N_1 B = \Omega_1(A) B \trianglelefteqslant G $. 

We let $ A = \langle x \rangle $. We can assume that $ B^x \neq B $ as otherwise $ B \trianglelefteqslant G $, which has been excluded. Since $ \Omega_1(A) \cong C_p $, we then have $ \Omega_1(A) \cap B = 1 $ and see that $ | \Omega_1(A) B : B | = | \Omega_1(A) B : B^x | = | \Omega_1(A) | = p $. Thus both $ B $ and $ B^x $ are normal in $ \Omega_1(A) B $. By comparison of orders, we further see that $ \Omega_1(A) B = B B^x $. Since $ \bmcl(B^x) = \bmcl(B) < \frac{p}{2} $, it follows that
\[
\bmcl(\Omega_1(A) B) = \bmcl(B B^x) \le \bmcl(B) + \bmcl(B^x) < \frac{p}{2} + \frac{p}{2} = p. 
\]

From the above there exists $ B_1 \leqslant B $, with $ | B : B_1 | = p $ and $ \bmexp(B_1) = p $, such that $ A B_1 \leqslant G $. Since $ \Omega_1(A) \cap B = 1 $, we see by minimality that $ A \cap B = 1 $. In particular, we have $ A \cap B_1 = 1 $. Since $A$ is non-trivial, it follows that $B_1$ is a proper subgroup of $ A B_1 $. Hence $B_1$ is a proper subgroup of $ N_{AB_1}(B_1) $. But $ N_{ AB_1 }(B_1) = (A \cap N_{A B_1}(B_1)) B_1 $, so $ A \cap N_{A B_1}(B_1) $ is non-trivial. It follows that $ \Omega_1(A) \leqslant A \cap N_{A B_1}(B_1) $. In particular, we see that $ \Omega_1(A) B_1 \leqslant G $. Since $ | B : B_1 | = p $, we further see that
\[
| \Omega_1(A) B : \Omega_1(A) B_1 | = | B : B_1 | = p. 
\]
Now $ \bmexp(B_1) = p $ and $ \Omega_1(A) \cong C_p $, so $ \Omega_1(A) B_1 $ is generated by elements of order $p$. Thus $ \Omega_1(A) B_1 = \Omega_1(\Omega_1(A) B_1) $. But $ \bmcl(\Omega_1(A) B_1) \le \bmcl(\Omega_1(A) B) < p $. Hence we can apply Theorem~\ref{e5} to see that
\[
\bmexp(\Omega_1(A) B_1) = p. 
\]

As shown above, there exists $ y \in B $ such that $ \bmo(y) = p^2 $ and $ \uOmega_1(B) = \langle y^p \rangle$. It follows, in particular, that $ \bmexp(\Omega_1(A) B) \ge p^2 $. But $ | \Omega_1(A) B : \Omega_1(A) B_1 | = p $, so $ \Omega_1(A) B / \Omega_1(A) B_1 \cong C_p $. Since $ \bmexp(\Omega_1(A) B_1) = p $, we further see that $ \bmexp(\Omega_1(A) B) \le p^2 $. Hence
\[
\bmexp(\Omega_1(A) B) = p^2. 
\]
We can therefore apply Corollary~\ref{e4a} to see that
\[
\uOmega_1(\Omega_1(A) B) = \langle y^p \rangle = \uOmega_1(B). 
\]
Since $ \Omega_1(A) B \trianglelefteqslant G $, we then have $\uOmega_1(B) \trianglelefteqslant G$. We thus conclude that $ A \uOmega_1(B) \leqslant G $. 

Having established that $ A \uOmega_1(B) \leqslant G $, we use induction on $i$ to show that $ A \uOmega_i(B) \leqslant G $ for all~$i$. We assume that the result holds for $ i = s $, where $ s \ge 1 $. Since $ \bmcl(B) < \frac{p}{2} < p $, we see, by Corollary~\ref{e5a}, that $ \uOmega_{s+1}(B) = \uOmega_1(\uOmega_s(B)) $. By induction, we can assume that $ A \uOmega_s(B) \leqslant G $. Applying the result we have proven for $ i = 1 $, we then have $ A \uOmega_1(\uOmega_s(B)) \leqslant A \uOmega_s(B) \leqslant G $. Since $ \uOmega_1(\uOmega_s(B)) = \uOmega_{s+1}(B) $, it follows that $ A \uOmega_{s+1}(B) \leqslant G $. 
\end{proof}


We use Theorem~\ref{e7} to prove the following result, which can be considered as an analogue to~\cite[Theorem~2.6]{McCann4}. 

\begin{theo}\label{e8}
Let $p$ be an odd prime and let $ G = AB $ be a finite $p$-group for subgroups $A$ and $B$ such that $A$ is cyclic, $ \bmcl(B) < \frac{p}{2} $ and $ \bmexp(B) = p^k $, where $ k \ge 1 $. Then, for all~$i$ such that $ 1 \le i \le k $, we have:
\begin{enumerate}\romanenumi%[(i)]
\item \label{10i} $ A \uOmega_{k-i}(B) \leqslant G $;
\item \label{10ii} $ \uOmega_{k-i}(B)^G \leqslant \Omega_i(A) \uOmega_{k-i}(B) \leqslant G $;
\item \label{10iii} $ \bmexp(\uOmega_{k-i}(B)^G) = p^i $. 
\end{enumerate}
\end{theo}

\begin{proof}
We see that \eqref{10i} holds by Theorem~\ref{e7}. For \eqref{10ii} and \eqref{10iii}, we first deal with the case where $ i = 1 $. Since $ \bmcl(B) < \frac{p}{2} < p $, we see by Theorem~\ref{e5} that $ \uOmega_{k-1}(B) = \{ b^{p^{k-1}} \mid b \in B \} $. Since $ \bmexp(B) = p^k $, it follows that $ \bmexp(\uOmega_{k-1}(B)) = p $. By $(i)$, we have $ A \uOmega_{k-1}(B) \leqslant G $. Hence we can apply Lemma~\ref{e2} to see that
\[
\uOmega_{k-1}(B)^G = \uOmega_{k-1}(B)^{ A \uOmegas_{k-1}(B) }. 
\]
Now $ \bmcl(\uOmega_{k-1}(B)) \le \bmcl(B) < \frac{p}{2} $. Hence, by Lemma~\ref{e6}$\MK$\eqref{8i}, we have $ \Omega_1(A) \uOmega_{k-1}(B) \trianglelefteqslant A \uOmega_{k-1}(B) $. It follows that
\[
\uOmega_{k-1}(B)^G = \uOmega_{k-1}(B)^{ A \uOmegas_{k-1}(B) } \leqslant \Omega_1(A) \uOmega_{k-1}(B). 
\]
Thus \eqref{10ii} holds for $ i = 1 $. By Lemma~\ref{e6}$\MK$\eqref{8ii}, we further see that $ \bmexp(\uOmega_{k-1}(B)^G) = \bmexp(\uOmega_{k-1}(B)^{ A \uOmegas_{k-1}(B) }) = p $, so \eqref{10iii} also holds for $ i = 1 $. 

We now assume that $ k \ge 2 $ and further assume that \eqref{10ii} and \eqref{10iii} hold for $i = s$, where $ 1 \le s < k $. Thus $ \uOmega_{k-s}(B)^G \leqslant \Omega_s(A) \uOmega_{k-s}(B) $ and $ \bmexp(\uOmega_{k-s}(B)^G) = p^s $. By Lemma~\ref{e2}, we have $ \uOmega_{k-s}(B)^G = \uOmega_{k-s}(B)^{ A \uOmegas_{k-s}(B) } = A_1 \uOmega_{k-s}(B) $, where $ A _1 = A \cap \uOmega_{k-s}(B)^G $. Since $ \bmexp(A_1) \le \bmexp((\uOmega_{k-s}(B))^G) = p^s $, we see that $ A_1 = \Omega_t(A) $ for some $ t \le s $. 

We let $ N = \uOmega_{k-s}(B)^G $. Then $ A \cap N = \Omega_t(A) \leqslant \Omega_s(A) $. If $ \Omega_s(A) = A $, then $ \Omega_{s+1}(A) = A $ and $ \Omega_1(AN/N) \leqslant AN / N = \Omega_{s+1}(A) N / N $. If $ \Omega_s(A) $ is a proper subgroup of $A$, then $ \Omega_{s+1}(A) \cong C_{p^{s+1}} $. Since $ \bmexp(N) = p^s $, we see that $ \Omega_{s+1}(A) \not\leqslant N $. By minimality, we then have $ \Omega_1(AN/N) \leqslant \Omega_{s+1}(A) N/N $. Thus, in every case, we have
\[
\Omega_1(AN/N) \leqslant \Omega_{s+1}(A) N / N. 
\]

By Lemma~\ref{e1}, we have $ \uOmega_{k-s-1}(BN / N) = \uOmega_{k-s-1}(B) N / N $. Since $ \bmcl(B) \le \frac{p}{2} < p $ we see, by Corollary~\ref{e5a}, that $ \uOmega_1(\uOmega_{k-s-1}(B)) = \uOmega_{k-s}(B) \leqslant N $. Hence $ \bmexp(\uOmega_{k-s-1}(B) N / N) \le p $. But, by our inductive assumption, we have $ \bmexp(N) = \bmexp(\uOmega_{k-s}(B)^G) = p^s $. In addition, $ \bmexp(B) = p^k $, so $ \uOmega_{k-s-1}(B) $ contains elements of order $p^{s+1}$. Hence $ \bmexp(\uOmega_{k-s-1}(B) N / N) = p $. Since $ A \uOmega_{k-s-1}(B) \leqslant G $ and $ N = \Omega_t(A) \uOmega_{k-s}(B) \leqslant A \uOmega_{k-s-1}(B) $, we see that
\[
(AN/N) (\uOmega_{k-s-1}(B) N / N) = A \uOmega_{k-s-1}(B) / N \leqslant G / N. 
\]
But $ \bmexp(\uOmega_{k-s-1}(B) N / N) = p $ and $ \bmcl(\uOmega_{k-s-1}(B) N / N) \le \bmcl(B) < \frac{p}{2} $. Hence, we can apply Lemma~\ref{e6}$\MK$\eqref{8i} to see that
\[
\Omega_1(AN/N) (\uOmega_{k-s-1}(B) N / N) \trianglelefteqslant A \uOmega_{k-s-1}(B) / N. 
\]
By Lemma~\ref{e6}$\MK$\eqref{8ii}, we further have $ \bmexp((\uOmega_{k-s-1}(B) N/ N)^{ A \uOmegas_{k-s-1}(B) / N }) = p $. We note that $ (\uOmega_{k-s-1}(B) N/ N)^{ A \uOmegas_{k-s-1}(B) / N } = \uOmega_{k-s-1}(B)^{ A \uOmegas_{k-s-1}(B) } N / N $. In addition, we see by Lemma~\ref{e2} that $ \uOmega_{k-s-1}(B)^{ A \uOmegas_{k-s-1}(B) } = \uOmega_{k-s-1}(B)^G $. It then follows that
\[
\bmexp(\uOmega_{k-s-1}(B)^G N / N) = \bmexp((\uOmega_{k-s-1}(B) N/ N)^{ A \uOmegas_{k-s-1}(B) / N }) = p. 
\]

We let $ W / N = \Omega_1(AN/N) $. Then $ N \leqslant W \leqslant AN $, so $ W = (A \cap W) N $. From the above, we have $ W \leqslant \Omega_{s+1}(A) N $. Since $A$ is cyclic, we see that if $ A \cap W \not\leqslant \Omega_{s+1}(A) $ then $ \Omega_{s+1}(A) $ is a proper subgroup of $ A \cap W $. It follows that $ \Omega_{s+2}(A) \leqslant A \cap W $, where $ \Omega_{s+2}(A) \cong C_{p^{s+2}} $. Hence $ \Omega_{s+2}(A) \leqslant W \leqslant \Omega_{s+1}(A) N $, so $ \Omega_{s+2}(A) = \Omega_{s+1}(A) (\Omega_{s+2}(A) \cap N)$. But, for $t$ as above, we have $ \Omega_{s+2}(A) \cap N \leqslant A \cap N \leqslant \Omega_t(A)$. Since $ t \le s $, the contradiction $ \Omega_{s+2}(A) \leqslant \Omega_{s+1}(A) \Omega_t(A) = \Omega_{s+1}(A) \cong C_{p^{s+1}}$ then arises. We can thus assume that $ A \cap W = \Omega_m(A) $ for some $m$ such that $ t \le m \le s+1$. Hence
\[
\Omega_1(AN / N) (\uOmega_{k-s-1}(B) N / N) = (\Omega_m(A) N / N) (\uOmega_{k-s-1}(B) N / N) \trianglelefteqslant A \uOmega_{k-s-1}(B) / N. 
\]
By Lemma~\ref{e3}, it follows that $\Omega_m(A) N \uOmega_{k-s-1}(B) N \trianglelefteqslant A \uOmega_{k-s-1}(B) $. But $ N = \Omega_t(A) \uOmega_{k-s}(B) $ and $ \Omega_m(A) N \uOmega_{k-s-1}(B) N = \Omega_m(A) N \uOmega_{k-s-1}(B) $. Hence
\[
\Omega_m(A) \Omega_t(A) \uOmega_{k-s}(B) \uOmega_{k-s-1}(B) = \Omega_m(A) \uOmega_{k-s-1}(B) \trianglelefteqslant A \uOmega_{k-s-1}(B). 
\]
We can now apply Lemma~\ref{e2} to see that
\[
\uOmega_{k-s-1}(B)^G = \uOmega_{k-s-1}(B)^{ A \uOmegas_{k-s-1}(B) } \leqslant \Omega_m(A) \uOmega_{k-s-1}(B). 
\]
Since $A$ is cyclic and $ A \uOmega_{k-s-1}(B) \leqslant G $, we see that $ \Omega_j(A) \uOmega_{k-s-1}(B) \leqslant G $ for all $j$. Now $ m \le s+1 $, so
\[
\uOmega_{k-s-1}(B)^G \leqslant \Omega_m(A) \uOmega_{k-s-1}(B) \leqslant \Omega_{s+1}(A) \uOmega_{k-s-1}(B) \leqslant G. 
\]
We thus conclude that \eqref{10ii} holds for $ i = s+1 $. 

From the above, we have $ \bmexp(\uOmega_{k-s-1}(B)^G N / N)) = p $. But $ N = \uOmega_{k-s}(B)^G $ and, by our inductive assumption, we have $ \bmexp(\uOmega_{k-s}(B)^G) = p^s $. Hence $ \bmexp(\uOmega_{k-s-1}(B)^G) \le p^{s+1} $. Now $ \bmexp(B) = p^k $ so there exists $ b \in B $ such that $ \bmo(b) = p^k $. Since $ s+1 \le k $, we see that $ \bmo(b^{ p^{k - s-1} }) = p^{s+1} $. Thus $ b^{p^{k-s-1}} $ is an element of order $ p^{s+1} $ in $ \uOmega_{k-s-1}(B) $. Hence $ \bmexp(\uOmega_{k-s-1}(B)^G) \ge p^{s+1}. $We conclude that $ \bmexp(\uOmega_{k-s-1}(B)^G) = p^{s+1} $, so \eqref{10iii} also holds for $ i = s+1 $. 
\end{proof}

In our final result we use Theorem~\ref{e8} to provide an alternative derivation of two results concerning the structure of products of cyclic $p$-groups with $p$-groups of class less than $ \frac{p}{2} $ (see~\cite[ Theorems~2.9 and 4.1]{McCann4}). 

\begin{coro}\label{e9}
Let $p$ be an odd prime and let $ G = AB $ be a finite $p$-group for subgroups $A$ and $B$ such that $A$ is cyclic, $ \bmcl(B) < \frac{p}{2} $ and $ \bmexp(B) = p^k $, where $ k \ge 1 $. Then:
\begin{enumerate}\romanenumi%[(i)]
\item \label{11i} $ \Omega_k(A) B \trianglelefteqslant G $;
\item \label{11ii} $ \bmdl(G) \le 1 + k + \bmdl(B) $. 
\end{enumerate}
\end{coro}

\begin{proof}
We let $ i = k $ in Theorem~\ref{e8}$\MK$\eqref{10iii} and see that $ \bmexp(B^G) = p^k $. Now $ B \leqslant B^G $, so $ B^G = (A \cap B^G) B $. We have $ A \cap B^G = \Omega_t(A) $, for a suitable $t$. Since $ \bmexp(B^G) = p^k $, we can assume that $ t \le k $. Now $ G/B^G = G / \Omega_t(A) B $ is isomorphic to a subgroup of $A$, so $ G/B^G $ is cyclic. Since $ B^G = \Omega_t(A) B \leqslant \Omega_k(A)B $, we then see that $ \Omega_k(A) B / B^G \trianglelefteqslant G $. It follows that $ \Omega_k(A) B \trianglelefteqslant G$, so \eqref{11i} is established. 

For \eqref{11ii}, we note that $ G / \Omega_k(A) B $ is isomorphic to a factor group of the cyclic group $A$. Hence $ G^{'} \leqslant \Omega_k(A) B$. Since $A$ is cyclic, we see that $ \Omega_1(A) B \leqslant \cdots \leqslant \Omega_k(A) B \leqslant G $. For $ i = 1, \dots, k $, we have $ | \Omega_i(A)B : \Omega_{i-1}(A) B | \le | \Omega_i(A) : \Omega_{i-1}(A) | \le p $, so $ \Omega_{i-1}(A) B \trianglelefteqslant \Omega_i(A) B $. We further see that $ \Omega_i(A) B / \Omega_{i-1}(A) B $ is isomorphic to a factor group of the cyclic group $ \Omega_i(A) / \Omega_{i-1} (A) $. Hence $ (\Omega_i(A)B)^{'} \leqslant \Omega_{i-1}(A) B $ for $ i = 1, \dots, k $, so $ G^{(1 + k)} \leqslant B $. It then follows that $ G^{(1 + k + \bmdl(B))} = 1 $, in accordance with \eqref{11ii}. 
\end{proof}

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