Proof. (1) $\mathrm{Dih}(8)$ is not quasi-Hamiltonian. The pronorm $N$ of $G=\mathrm{Sym}(4)$ is obviously contained in the Wielandt subgroup of $G$ , which coincides with the Klein $4$ -subgroup $V$ of $G$ . However, the only element of $V$ pronormalizing $H$ is $(1,2)(3,4)$ , so $N=\{1\}$ because it is a proper $G$ -invariant subgroup of $V$ .

(2) It is not quasi-Hamiltonian, while the computation of the norm is easy.

(3) The pronorm is easily seen to coincide with the Sylow $2$ -subgroup. However, every cyclic subgroup of order $6$ is not pronormal.

(4) Every diagonal subgroup of order $3$ is neither pronormal nor even contained in the pronorm, which is easily seen to coincide with $Q_8\times \mathcal C_3$ .

(5) Write our group as $G=\langle a,b\rangle \times \langle c\rangle $ , where $\langle a,b\rangle \simeq Q_8$ and $o(c)=4$ . Clearly, $Z(G)\leq N_{\mathrm{pro}}(G)$ . Let $x=ac$ . Then, $\langle x\rangle $ is not pronormal in $G$ (conjugate by $b$ ). However, $x$ is not contained in the pronorm of $G$ because it does not pronormalize $\langle bc\rangle $ . Now, we easily see that the pronorm must coincide with the centre.

Proof. (1) Write our group as follows: $G=(\langle a_1\rangle \ltimes \langle b_1\rangle )\times (\langle a_2\rangle \ltimes \langle b_2\rangle )$ , where $a_i$ has order $2$ and $b_i$ has order $3$ for every $i\in \{1,2\}$ . Clearly, $a_i$ does not pronormalize $\langle b_1b_2\rangle $ , while $b_1,b_2$ and $a_1a_2$ are contained in the pronorm. Thus, a subgroup $H$ that is not contained in the pronorm must contain a conjugate of $a_1$ or $a_2$ . Suppose $H$ contains $a_1$ . If it contains $b_1$ , then we can factor $\langle a_1,b_1\rangle $ out and note that all subgroups of $\mathrm{Sym}(3)$ are pronormal. If it does not contain $b_1$ , but contains $b_2$ , then we factor by $\langle b_2\rangle $ and note that, again, all subgroups of the quotient are pronormal. If $H$ contains neither $b_1$ nor $b_2$ , then we may assume $H\leq \langle a_1\rangle \times \langle a_2\rangle $ and that actually either $H=\langle a_i\rangle $ for some $i$ or $H=\langle a_1,a_2\rangle $ . In both cases, $H$ is trivially pronormal in $G$ .

(2) Use GAP (or see Theorem 5.4).

(3)–(4) Easy.

(5) All subgroups are pronormal in this case (see the main theorem of [Reference Kuzennyi and Subbotin15]).

(6) Our group $G$ is polycyclic, so the pronorm is a subgroup of $G$ . Write ${G=(\langle a\rangle \ltimes \langle b\rangle )\times \langle c\rangle }$ , where $o(a)=2$ , $o(b)=3$ and $o(c)=\infty $ . To prove that ${N_{\mathrm{pro}}(G)=\langle b,c\rangle }$ , we only need to show that $b$ is contained in the pronorm ( $c$ being central and $G$ not having only pronormal subgroups). Clearly, $b$ centralizes $\langle b,c\rangle $ . Let $H$ be any subgroup of $G$ that is not contained in $\langle b,c\rangle $ and let $g$ be an element of $H$ that is not contained in $\langle b,c\rangle $ . Then, $g^2\in \langle c\rangle $ , and to show that $b$ pronormalizes $H$ , we may assume that $c^2=g^2=1$ and so, in particular, that $G$ is finite and $|H|\leq 4$ . However, now it is very easy to realize that $b$ pronormalizes $H$ .

Finally, if $H$ is any subgroup of $G$ which is not contained in $N_{\mathrm{pro}}(G)$ , then $H$ is easily seen to be pronormal in $G$ (essentially because one can assume that $o(c)=2$ , which in turn implies that all subgroups of $G$ are pronormal).

(7) Again, our group $G$ is polycyclic, so the pronorm is a subgroup of $G$ . Write

$$ \begin{align*}G=\langle a\rangle\ltimes(\langle b\rangle\times\langle c\rangle)\times\langle d\rangle,\end{align*} $$

where $o(a)=3$ , $o(b)=o(c)=2$ and $o(d)=\infty $ . As in item (6), to prove that $N_{\mathrm{pro}}(G)=\langle b,c,d\rangle $ , we may assume $o(d)=3$ , which immediately implies the claim (see also Lemma 3.8).

Finally, let $H$ be any subgroup of $G$ that is not contained in $N_{\mathrm{pro}}(G)$ . To show that $H$ is pronormal, we may assume $o(d)=3$ . Clearly, if $H$ contains a $3$ -element, then $H\trianglelefteq G$ . If not, then $H$ is a $2$ -subgroup, so either $|H|=2$ or $|H|=4$ . In both cases, $H$ is clearly pronormal in $G$ .

(8) This is an easy exercise (see also Lemma 3.8).

Proof. Suppose first that all subgroups of $G$ are normal or contained in the norm. Every finite subgroup of $G$ is quasi-Hamiltonian, so $G$ is quasi-Hamiltonian too. Thus, if $Q_8$ is involved in $G$ , then $G$ is of type (1). Assume $Q_8$ is not involved in $G$ . Then, there is a finite-exponent abelian subgroup $A$ of $G$ and an element $x\in G$ such that $G=\langle x\rangle A$ with $x$ acting on $A$ as a power automorphism which is $\equiv _41$ in the case where $p=2$ (see [Reference Schmidt20, Theorem 2.3.1]).

From now on, since the proof is exactly the same and for reader’s convenience, we prove the statement assuming that $G$ is finite.

Write $A=\langle a_1\rangle \times \langle a_2\rangle \times \cdots \times \langle a_n\rangle $ , where $o(a_1)\geq o(a_2)\geq \cdots \geq o(a_n)$ . We claim that $\langle a_2,\ldots ,a_n\rangle \leq Z(G)$ . Suppose not. Clearly, we may assume $n=2$ . If ${o(a_1)>o(a_2)}$ , then by induction on the order, $\langle a_2\rangle (\langle a_1a_2\rangle [p])/(\langle a_1a_2\rangle [p])$ and $\langle a_2\rangle /(\langle a_2\rangle [p])$ are centralized by $x$ , so $a_2$ is centralized by $x$ . Thus, $o(a_1)=o(a_2)$ . Assume that ${o(x)> o(a_1)}$ . Since $\langle x \rangle \cap A$ is a cyclic subgroup of $A$ , we may assume without loss of generality that $\langle x \rangle \cap A \leq \langle a_1 \rangle $ . It follows that $\langle xa_2 \rangle $ is not normalized by $a_1$ . However, $xa_2$ does not belong to the norm because it fails to normalize $\langle x \rangle $ , which yields a contradiction. Thus, $o(x)\leq o(a_1)$ and we may even assume that $\langle x\rangle \cap A=\{1\}$ . Now, $\langle a_1x\rangle $ is not normal in $G$ because it is not normalized by $a_2$ , so it must be contained in the norm of $G$ . However, $a_1x$ does not normalize $\langle x\rangle $ , so we obtain a contradiction. The claim is proved. In particular, $o(a_1)\geq o(a_2)|G'|$ .

We claim that $x$ and $a_1$ can be chosen in such a way that $o(x)<o(a_1)$ . Let ${L=\langle x,a_1\rangle }$ . If $|L'|=p$ , then $L$ is minimal nonabelian and metacyclic, so we may write $L=\langle x'\rangle \ltimes \langle a_1'\rangle $ for suitable elements $x',a_1'$ . In this case, $o(x')<o(a_1')$ ; otherwise, $\langle x'a_1'\rangle $ is not normal and does not even normalize $\langle x'\rangle $ . Similarly, if $M=\langle x\rangle \cap \langle a_1\rangle =\{1\}$ , then $o(x)<o(a_1)$ since otherwise, $\langle xa_1\rangle $ is not normal and does not even normalize $\langle x\rangle $ (put $x'=x$ and $a_1'=a_1$ in this case). Thus, we may assume $|L'|\geq p^2$ and $M\neq \{1\}$ . Let $S$ be the socle of $M$ . By induction, we may find $x'$ and $a_1'$ such that $L/S=\langle x'S\rangle \ltimes \langle a_1'S\rangle $ and $o(x'S)<o(a_1'S)$ . If $S\not \leq \langle a_1'\rangle $ , then $G'$ is not cyclic, which is a contradiction. Thus, $S\leq \langle a_1'\rangle $ , so $\langle a_1'\rangle $ is normalized by $x'$ and $o(a_1')>o(x')$ . In any case, $o(a_1')\geq o(a_1)$ .

Now, $A_1=\langle a_1',a_2,\ldots ,a_n\rangle $ is abelian and $x'$ acts on $A_1$ as a power automorphism because $o(a_1')\geq o(a_1)\geq o(a_2)|G'|$ . Finally, by suitably replacing $x'$ , we may further require that $\langle x'\rangle \cap A_1=\{1\}$ (it is in fact enough choosing $x'$ of smallest possible order).

It remains to prove the bound in point (2). Set $o(x) = p^t$ , $o(a_1) = p^n$ and ${a_1^x = a_1^{1+p^s}}$ for some $s\ge 1$ . In particular, $\langle a_1^{p^{n-s}}\rangle = Z(G)$ and $|\langle a_1 \rangle : Z(G)| = |G'|$ . Assume ${o(x)> |Z(G)|}$ . If $y$ is an element of $\langle a_1\rangle $ with $o(y)=o(x)$ , then $\langle xy\rangle $ is not normalized by $a_1$ and does not normalize $\langle x\rangle $ , which is a contradiction. Therefore, $o(x)\le |Z(G)|=o(a_1)/|G'|$ . Now, assume that $o(x)> o(a_1)p/|G'|^2$ . Let $y=a_1^{p^{(n-t)-(n-s)}}$ . Note that $o(a^{p^{n-t}})=p^t$ and that $\langle y\rangle>Z(G)$ . It follows that $\langle xy \rangle $ does not normalize $\langle x \rangle $ and is not normalized by $a_1$ because the assumption yields that $\langle xy\rangle \not \geq G'$ . If we replace $x$ by $xa_i$ , $i>1$ , in this argument, then we see that all the required conditions are satisfied.

Conversely, suppose $G$ is a group of type (2). Then, the conditions imply that every cyclic subgroup of $G$ that is not centralized by $x$ must contain $G'$ . Thus, $x\in N(G)$ . However, every cyclic subgroup that lies outside ${\langle x\rangle (\langle a_1,\ldots ,a_n\rangle \cap Z(G))\leq N(G)}$ contains $G'$ , so $G$ satisfies the required condition.

Proof. Since $P$ is not Dedekind, it must satisfy Condition (2) in the statement of Theorem 3.1. Write $P=\langle x\rangle \ltimes (\langle a_1\rangle \times \cdots \times \langle a_n\rangle )$ , where $o(a_2)\geq \cdots \geq o(a_n)$ , ${P'\leq \langle a_1\rangle }$ , $\langle a_2,\ldots ,a_n\rangle \leq Z(P)$ and $o(a_1)\geq |P'|^2/p\cdot \max \{o(x),o(a_2)\}$ . Clearly, $a_1$ does not normalize $\langle x\rangle $ , so $\langle a_1\rangle $ is pronormal in $G$ and, hence, (being subnormal) also normal. Similarly, $\langle xa_1\rangle $ and $\langle a_1a_i\rangle $ , $i>1$ , are normal in $G$ . Moreover, if $H$ centralizes $\langle a_1\rangle $ , it must also centralize $\langle xa_1\rangle $ and $\langle a_1a_i\rangle $ , $i>1$ , because the socles are the same, and $(|H|,p)=1$ , and hence, $H$ centralizes $P$ . Thus, we may assume that $C_H(a_1)=\{1\}<H$ and that $p$ is odd (because the automorphism group of a cyclic $2$ -group is a $2$ -group).

If $|P'|>p$ , then induction yields that $[H,P]\leq \Omega (P')$ , so $H$ centralizes $\langle a_1\rangle $ , which is a contradiction. Thus, $|P'|=p$ .

Now, $U=\Omega (\langle x,a_1\rangle )$ is elementary abelian of order $p^2$ , so $U=U_1\ltimes \langle a_1\rangle $ , where $U_1$ is $H$ -invariant. If $|U_1|\neq |\langle x\rangle |$ , then $U_1\trianglelefteq P$ , so $U_1\trianglelefteq G$ and, hence, $[H,P]\leq U_1$ by induction; but this yields again the contradiction that $H$ centralizes $a_1$ . Thus, we may assume $U_1=\langle x\rangle $ . Similarly, we can get rid of the $a_i$ with $i\geq 2$ and so assume $n=1$ .

Next, observe that every subgroup of $\langle x,a_1\rangle /\langle a_1^p\rangle $ is $H$ -invariant, so $H$ acts as a group of power automorphisms on it. However, $\langle a_1\rangle =C_G(a_1)$ , so $G/\langle a_1\rangle $ is abelian and, hence, $[H,U_1]=\{1\}$ . Thus, $H$ must centralize $a_1$ , which is the final contradiction.

Proof. It is enough to show that the critical-Sylow-tower groups $G$ do not satisfy the hypotheses. Suppose first that $G$ is critical-Sylow-tower of type $2$ . Then, there are primes $p_1,\ldots ,p_n,p_{n+1}=p_1$ such that

$$ \begin{align*}G=(\langle x_1\rangle\ltimes A_1)\times\cdots\times(\langle x_n\rangle\ltimes A_n),\end{align*} $$

$|x_i|=p_i$ , $|A_i|=p_{i+1}^{r_{i+1}}$ , $[x_i,A_i]\neq \{1\}$ and the $A_i$ are minimal normal subgroups of $G$ . Clearly, not every $A_i$ can be cyclic; otherwise, we would have $p_1<p_2<\cdots <p_n<p_1$ . Let $j$ be such that $A_j$ is not cyclic. Let $a\in A_n\setminus \{1\}$ and define $g$ as $x_1x_ja$ if $j\neq 1$ , or as $x_1a$ if $j=1$ .

Put $H=\langle g\rangle $ . Clearly, $H$ is not pronormal in $G$ since it is subnormal and nonnormal in $\langle x_2,x_1,a\rangle $ . However, $H$ is not even contained in the pronorm because otherwise, $A_j$ would have been cyclic.

Assume $G$ is critical-Sylow-tower of type $1$ . Thus, $|G|=p^\alpha q^\beta $ (here, $p$ and $q$ are distinct primes), $G$ has a unique minimal normal subgroup $F(G)$ and $G/F(G)$ is minimal nonnilpotent. Suppose $F(G)$ is a (necessarily elementary abelian) $p$ -group, and let $P$ and $Q$ be respectively a Sylow $p$ -subgroup and a Sylow $q$ -subgroup of $G$ ; in particular, $G/F(G)=P/F(G)\ltimes QF(G)/F(G)$ and $P/F(G)$ is cyclic.

If $P$ is Dedekind, then $F(G)\leq Z(P)$ , so $C_G(F(G))=G$ and, hence, $Q$ is characteristic in $QF(G)$ , which is a contradiction. Thus, $P$ is not Dedekind and so $P$ satisfies the structure of Theorem 3.2.

Since $F(G)$ is a minimal normal subgroup of $G$ and $Z(P)\neq F(G)$ , we have that $Z(P)\cap F(G)$ cannot be normalized by $Q$ . Now, let $a\in Z(P)\cap F(G)$ and $x\in Q$ . Then, $a^x=a'$ for some $a'\in F(G)$ . Set $P/F(G)=\langle yF(G)\rangle $ . If we now apply conjugation by $y$ , we find that $a^{x^y}=(a')^y$ , where $x^yF(G)=x^\ell F(G)$ for some positive integer ${1<\ell <q}$ . Thus, $a^{x^{y}}=a^x(\mathrm{mod}\ P')$ and so $a^{x^{y-1}}\in P'$ for every $a\in Z(P)$ . However, $P'\leq Z(P)$ and $[Q,\langle y\rangle ]F(G)=QF(G)$ , so $Z(P)$ is normal in $G$ . Therefore, $Z(P)=P$ , which is the final contradiction.

Proof. A standard inverse limit argument (using Theorem 3.3) shows that $G$ has a normal series with coprime factors. If $H/K$ is a factor of this series such that $H$ is not contained in the pronorm of $G$ , then $G/H$ has only pronormal subgroups; in particular,

$$ \begin{align*}\pi(G/\gamma_3(G)H)\cap\pi(\gamma_3(G)H/H)=\emptyset.\end{align*} $$

Let $W$ be the intersection of all the factors of the previous type. Clearly, $\pi (W)\cap \pi (G/W)=\emptyset $ and $W\leq N_{\mathrm{pro}}(G)$ . However, also $\pi (\gamma _3(G)W/W)\cap \pi (G/\gamma _3(G)W)=\emptyset $ and so we are done.

Proof. By Theorem 3.3, $G$ has a Sylow tower. Let $P$ be a normal Sylow $p$ -subgroup of $G$ for some prime $p$ . Clearly, we may assume by induction that $G$ has only one minimal normal subgroup $M$ , which is necessarily a $p$ -group.

Let $Q$ be a Sylow $p'$ -subgroup of $G$ , so $G=Q\ltimes P$ . Let $X$ be any subgroup of $Q$ that is not pronormal in $Q$ (and so in $G$ ). Then, $X$ is contained in the pronorm $N$ of $G$ . However, $Z(N)N'\leq P$ , so $N/(P\cap N)$ is cyclic and $X(P\cap N)/(P\cap N)$ is normal in $G/(P\cap N)$ . Then, $XP/P\trianglelefteq G/P$ and so $X\trianglelefteq Q$ , which is a contradiction. Thus, $Q$ is metabelian.

Now, suppose that $P$ is nonabelian. Then, Theorem 3.2 shows that $P$ is a Dedekind $2$ -group. By Maschke’s theorem, the socle of $P$ is cyclic, so $P$ is a quaternion group of order $8$ . Then, $C_G(P)=Z(P)$ because of the uniqueness of $M$ and, hence, $G/P$ is a proper subgroup of $\mathrm{Sym}(4)$ (which does not satisfy our hypotheses), so $G/P$ is metabelian and $G^{(3)}=\{1\}$ .

Proof. Let $G$ be a minimal counterexample. Then, $G$ has only one minimal normal subgroup $N$ , which has prime order $p$ . Clearly, $p>2$ and $G$ is not a $p$ -group. Let $P=O_p(G)$ and let $C$ be the intersection of all the centralizers of the chief factors of $G$ contained in $P$ . Then, $P\leq C$ and $G/C$ is abelian of order prime with $p$ . If $C\neq P$ , then there is a minimal normal subgroup $M/P$ of $G/P$ contained in $C/P$ which is a $q$ -group for some prime $q\neq p$ , so $M=P\times L$ , where $L$ is a nontrivial normal $q$ -subgroup of $G$ , which is a contradiction. Thus, $C=P$ and so $P$ is nonabelian. However, then Theorem 3.2 shows that $G=Q\times P$ , so $G$ is metabelian, which is the final contradiction.

Proof. Since $g \in P_G(H)\cap N_G(K)$ , there exists $u \in \langle H, H^g \rangle \leq N_G(K)$ such that ${H^u = H^g}$ . Thus, $gu^{-1} \in N_G(K)$ , so $(HK)^{gu^{-1}} = HK$ and, hence, $(HK)^g = (HK)^u$ . Therefore, $g\in P_G(HK)$ and the statement is proved.

Proof. Let $H$ be a subgroup of $G$ . Then, $H = H_{\pi '}H_\pi $ , where $H_\pi '$ is a Sylow $\pi '$ -subgroup of $H$ and $H_\pi $ is a Sylow $\pi $ -subgroup of $H$ . Now, $G$ is soluble, so $O_\pi (G)$ pronormalizes $H_{\pi '}$ . Consequently, if $x\in O_\pi (G)$ , then there exists ${u \in \langle H, H^x\rangle \le HO_\pi (G)}$ such that $H_{\pi '}^x = H_{\pi '}^u$ ; clearly, we may assume $u\in O_\pi (G)$ . Then, ${H^x = H_{\pi '}^xH_\pi ^x =H_{\pi '}^xH_\pi =H_{\pi '}^uH_\pi ^u = H^u}$ because $O_\pi (G)$ normalizes $H_\pi $ . The statement is proved.

Proof. Suppose first that $H=\langle h\rangle $ is cyclic. Write $K=Q\times W$ , where $Q=\langle a,b\rangle $ is a copy of the quaternion group of order $8$ and $W$ is elementary abelian. Set ${Z=Z(Q)\times W}$ . Since $K/Z$ is a Klein $4$ -group and $G'$ is abelian, we have that $[H,K]\leq Z$ . Write $Z=C_K(H)\times [H,Z]$ , and put $a^h=aa_1a_2$ and $b^h=bb_1b_2$ , where $a_1,b_1\in C_K(H)$ and $a_2,b_2\in [H,Z]$ . Replacing $h$ by $h^2$ , we have that $a_1=b_1=1$ . Thus, $a^h=aa_2$ and $b^h=bb_2$ .

Now, note that $[H,Z]=\{[h,z]\,:\, z\in Z\}$ , so there exist $z_1,z_2\in Z$ such that ${[h,z_1]=a_2}$ and $[h,z_2]=b_2$ . Let $Q_h=\langle az_1,bz_2\rangle $ be centralized by $H$ ; in fact, for example,

$$ \begin{align*}(az_1)^h=aa_2z_1^h=aa_2z_1[h,z_1]^{-1}=az_1.\end{align*} $$

This proves the statement when $H$ is cyclic.

Finally, we turn to the general case. Write $H=H_1\times \langle h\rangle $ . Then, $h$ centralizes not only $Q_1$ , but also $Q_1^H$ . Let $G_1=H_1\ltimes Q_1^H$ . By induction, we see that $Q_1^H$ contains a copy $Q_2$ of $Q_8$ that is centralized by $H_1$ . Consequently, $Q_2$ is centralized by $H$ and the statement is proved.

Proof. Assume first that every subgroup of $G$ is pronormal in $G$ or contained in the pronorm of $G$ . Obviously, if $K$ is nonpronormal in $Q$ , then $K$ is nonpronormal in $G$ . Consequently, $K \le \mathrm{N}_{\mathrm{pro}}(G)$ . Now, assume that $K \not \le \mathrm{N}_{\mathrm{pro}}(G)$ . Let $U$ be a subgroup of $V$ that is $K$ -invariant. Therefore, $K \ltimes U$ is pronormal in $G$ (because $K$ is not contained in the pronorm of $G$ ). If $g\in N_G(K)$ , then there exists ${u \in \langle KU, KU^g \rangle = K \ltimes \langle U, U^g \rangle }$ such that $(KU)^u = KU^g$ . Write $u=xy$ , where $x\in K$ and $y\in \langle U,U^g\rangle $ . Then, $KU^g=(KU)^u=(KU)^y=K^yU$ and, hence, $U^g=U$ , as desired. Thus, Conditions (1) and (2) are proved.

Set $P=\mathrm{PAut}_Q(V)$ . The fact that $Q/P$ has only pronormal subgroups follows from the observation that any nonpronormal subgroup of $Q$ must be contained in the pronorm of $G$ and so it must act as power automorphisms on $\gamma _3(V)$ . This proves Condition (3). Moreover, $P/C_Q(V)$ is clearly central in $Q/C_Q(V)$ , so $[P,Q]\leq C_Q(V)$ . This proves Condition (4).

To prove Condition (5), suppose that $Q/P$ contains a copy of the quaternion group of order $8$ . Since the Sylow subgroups of $Q$ are either Dedekind or of type described in Theorem 3.1, there exists a subgroup $W$ of $Q$ such that $W\simeq Q_8$ and $W\cap P=\{1\}$ . The latter condition implies that $Z(W)$ is not contained in the pronorm of $G$ , so $W$ normalizes every subgroup of $V$ which is normalized by $Z(W)$ . In turn, this shows that every cyclic subgroup of $V$ normalized by $W$ must be centralized by $Z(W)$ and so that $Z(W)$ centralizes $V$ , which is a contradiction. Therefore, the Sylow subgroups of $Q/P$ are abelian. This proves Condition (5).

Finally, Condition (6) follows at once from Theorem 3.1 and the fact that the centre of an irreducible subgroup of $\mathrm{GL}(n,p)$ is cyclic.

Conversely, assume that $G$ satisfies Conditions (1) and (2). Since $\pi (V) \cap \pi (Q) = \emptyset $ and $V$ is abelian, so $V \le N_{\mathrm{pro}}(G)$ by Lemma 3.8. Let $M$ be the subgroup generated by the nonpronormal subgroups of $Q$ . By Condition (1), $M\leq N_{\mathrm{pro}}(G)$ .

Now, let $H$ be a subgroup of $G$ that is not contained in ${N}_{\mathrm{pro}}(G)$ . Write ${H = H_Q \ltimes U}$ , where $U \le V$ and $H_Q$ is the Sylow $\pi (Q)$ -subgroup of $H$ . We claim that $H$ is pronormal in $G$ . Replacing $H$ with a conjugate (if necessary), we may assume that $H_Q \le Q$ . If $H_Q$ is not pronormal in $Q$ , then $H_Q \le M$ , so $H \le {N}_{\mathrm{pro}}(G)$ , which is a contradiction. Thus, $H_Q$ is pronormal in $Q$ . However, $H_Q $ is pronormal in ${H_Q \ltimes V}$ , since $V$ is contained in the pronorm of $G$ . Therefore, $H_Q$ is pronormal in $G=Q(H_QV)$ (see [Reference Brescia, Ingrosso and Trombetti1, Theorem 3.45]). Finally, if $g \in G$ , then there exists ${u \in \langle H_Q, H_Q^g \rangle \le \langle H, H^g \rangle }$ such that $H_Q^u = H_Q^g$ ; but $gu^{-1} \in N_G(H_Q)$ and so ${H^g = H^u}$ by Condition (2). This shows that $H$ is pronormal in $G$ and we are done.

Proof. By Theorem 3.3, $G$ has a Sylow tower $\{1\} =P_0 \trianglelefteq P_1 \trianglelefteq \cdots \trianglelefteq P_t = G,$ where $P_{i+1}/P_{i}$ is a Sylow $p_i$ -subgroup of $G/P_i$ (here, $p_i$ is a prime) for $i\in \{0, \ldots , t-1\}$ . Let $X$ be the largest term of this series that is contained in $N_{\mathrm{pro}}(G)$ . Then, we can write $G=T\ltimes (S\ltimes X)$ for some subgroup $T$ of $G$ and for some Sylow $p$ -subgroup $S$ of $G$ (here, $p$ is a prime) that is not contained in $N_{\mathrm{pro}}(G)$ . Since $S\ltimes X$ is not contained in the pronorm of $G$ , all subgroups of $T\simeq G/(S\ltimes X)$ are pronormal. Since $\pi (\gamma _3(X))\cap \pi (G/\gamma _3(X))=\pi (X)\cap \pi (G/X)=\emptyset $ , we may actually choose $S$ and $T$ such that $G= W\ltimes \gamma _3(X)$ , where $W=(T\ltimes S)\ltimes M$ and $X=M\ltimes \gamma _3(X)$ .

The fact that $[M,W]\leq C_M(\gamma _3(X))$ and $W"\leq \mathrm{PAut}_W(\gamma _3(X))$ follows at once from Lemma 3.10. In particular, if $\pi ([M,W])\cap \pi (C_M(W))=\emptyset $ , then we may re-arrange the Sylow subgroups of $G$ as follows: $G=((T\times C_M(W))\ltimes S)\ltimes ([M,W]\times \gamma _3(X))$ , where $[M,W]\times \gamma _3(X)$ is Dedekind. This proves the first part of the statement.

Assume $G$ is metabelian. Then, $S\ltimes X$ has an ascending $G$ -invariant series with elementary abelian factors. By [Reference Gorenstein10, Lemma 2.3], each of these factors is centralized by $T'$ because $\pi (T)\cap \pi (SX)=\emptyset $ and $G'$ is abelian. Consequently, $T'$ is subnormal in $G$ , so it centralizes $SX$ . In particular, $T'$ is normal in $G$ . Since $T'$ is contained in $T$ , so $T'\leq N_{\mathrm{pro}}(G)$ by Lemma 3.8. Now, $T=U\ltimes \gamma _3(T)$ , where $\pi (U)\cap \pi (\gamma _3(T))=\emptyset $ and $U$ is Dedekind, so replacing $T$ by $U$ and $X$ by $X\gamma _3(T)$ , we may assume that $T$ is Dedekind. Finally, the Sylow $q$ -subgroups of $G$ are Dedekind for every prime $q$ and, hence, ${G'\leq N_{\mathrm{pro}}(G)}$ by Lemma 3.8.

Now, we prove that $[T,S]=\{1\}$ . If $S$ is not Dedekind, then $[S,T]=\{1\}$ by Theorem 3.2. If $S$ is Dedekind nonabelian, then $T$ centralizes a copy of $Q_8$ contained in $S$ by Lemma 3.9. This implies that we can always write $S=C_S(T)\times [S,T]$ . If $x\in C_S(T)$ and $1\neq y\in [S,T]$ , then $\langle xy\rangle $ is not pronormal in $G$ , so $xy\in N_{\mathrm{pro}}(G)$ . Since we already know that $[S,T]\leq G'\leq N_{\mathrm{pro}}(G)$ , the arbitrariness of $x$ and $y$ yields that $S\leq N_{\mathrm{pro}}(G)$ , which is a contradiction. Thus, $[T,S]=\{1\}$ .

Case ( $\star $ ): $[S,M]=\{1\}$ . Then, $S\gamma _3(X)\trianglelefteq G$ and $G/S\gamma _3(X)\simeq T\ltimes M$ has only pronormal subgroups. Re-arranging the Sylow subgroups of $T\ltimes X$ (using also Lemma 3.9), we can write

$$ \begin{align*}TX=(T\times C_M(T))\ltimes ([M,T]\times \gamma_3(X)),\end{align*} $$

which is a group with only pronormal subgroups. Note that $[S,MT]=\{1\}$ and we are done.

Finally, if $S_1$ is a pronormal subgroup of $S$ that is not contained in the pronorm of $S$ , then it is not contained in the pronorm of $S\ltimes \gamma _3(X)$ and Lemma 3.10 yields that every $S_1$ -invariant subgroup of $\gamma _3(X)$ is also $N_S(S_1)$ -invariant.

Assume conversely that $G$ has the structure described in ( $\star $ ). Let $H$ be a nonpronormal subgroup of $S\times R$ . Then, $H$ acts as power automorphisms on $\gamma _3(X)$ . We claim that $H$ is contained in the pronorm of $G$ . Let $K$ be any other subgroup of $G$ . To show that $H$ pronormalizes $K$ , we may assume $K\cap [M,T]=\{1\}$ and so even $[M,T]=\{1\}$ by [Reference Brescia, Ingrosso and Trombetti1, Corollary 3.10]. Write $K=K_1\ltimes K_2$ , where $K_2=\gamma _3(X)\cap X$ . Also, we may reduce to $\langle H,K\rangle R\gamma _3(G)$ , so now $K_2$ is normal and $K_1=K$ is pronormalized by $H$ because it is pronormalized modulo $\gamma _3(G)$ (see [Reference Brescia, Ingrosso and Trombetti1, Corollary 3.10]). Therefore, the conditions of Lemma 3.10 are satisfied and we are done.

Proof. By Theorem 3.1, we may assume that $G$ is not a $2$ -group. Since $G$ is soluble, there are a Sylow $2'$ -subgroup $Q$ of $G$ and a Sylow $2$ -subgroup $P$ of $G$ (containing $H$ ) such that $G=PQ$ . Now, $P$ is a quasi-Hamiltonian $2$ -group involving $Q_8$ , so $P=H\times R$ , where $R$ is an elementary abelian $2$ -subgroup. Moreover, since $H\leq \overline {Z}(G)$ , we have that $[H,Q]=\{1\}$ . Thus, $H$ is normalized by both $P$ and $Q$ , and so by $G=PQ$ .

Now, let $K$ be any subgroup of $O_2(\overline {Z}(G))$ containing $H$ . We claim that $K$ is a direct factor of $G$ . First, note that $K$ is normal in $G$ by the previous paragraph because it is generated by copies of $Q_8$ . Let $N$ be a minimal normal subgroup of $G$ , so $N$ is an elementary abelian $p$ -group for some prime $p$ . If $p>2$ , then there is $X/N$ such that $G/N=KN/N\times X/N$ . Thus, $G=K\times X$ because $K$ is normal in $G$ . Suppose that every minimal normal subgroup of $G$ is a $2$ -group, put $L=O_2(G)$ and let $S$ be the socle of $L$ . Moreover, let $M/L$ be a minimal normal subgroup of $G/L$ , which necessarily is a $2'$ -subgroup. If every subgroup of $S$ is normal in $G$ , then $S\leq Z(G)$ , so $L\leq \overline Z(G)$ and, hence, $M=L\times U$ for some normal $2'$ -subgroup of $G$ , which is a contradiction. Thus, not every subgroup of $S$ is normal in $G$ and so not every subgroup of $S$ is normalized by $Q$ . Thus, there is a minimal normal $2$ -subgroup $W$ of order at least $4$ such that $W\cap K=\{1\}$ (by Maschke’s theorem). Finally, by induction, $G/W=KW/W\times D/W$ for some subgroup $D/W$ of $G/W$ , so $G=K\times D$ and we are done.

Proof. Clearly, we may assume that $G$ is finitely generated, so we need to prove that $G$ is finite. Let $R$ be the finite residual of $G$ . If every finite quotient of $G$ is soluble, then every finite quotient of $G$ is soluble of derived length at most $3$ by Corollary 3.5, so $G/R$ is soluble (of derived length at most $3$ ) and, hence, even finite. However, then $R=\{1\}$ and we are done.

It is therefore safe to assume that $G$ contains an insoluble finite quotient $G/N$ . Now, $N$ is finitely generated as well and no finite quotient of $N$ can be insoluble, otherwise, every subgroup of $G/N$ would be pronormal and $G/N$ would be soluble, which is a contradiction. By the previous paragraph, $N$ is finite and so also $G$ is finite. The statement is proved.

Proof. It follows from [Reference Brescia, Ingrosso and Trombetti1, Theorem 3.1] that for every subgroup $H$ of $G$ , the set $P_G(H)$ of all elements of $G$ pronormalizing $H$ coincides with the normalizer $N_G(H)$ . Thus, if $H$ is a pronormal subgroup of $G$ , then $H$ is normal in $G$ . However, if $H$ is contained in the pronorm, then $H$ normalizes every subgroup of $G$ , so $H$ is actually contained in the norm of $G$ . Thus, $HK=KH$ for all subgroups $H,K$ of $G$ .

Proof. This follows from Theorem 4.3 and the well-known fact that every subgroup of a locally nilpotent group is serial.

Proof. Clearly, if $H$ is pronormal in $G$ , then $H$ is normal in $G$ . Assume $H$ is contained in the pronorm of $G$ . Then, all subgroups of $H$ are pronormal in $G$ (and so in $H$ ), so $H$ is metabelian, being locally graded.

Now, $H'$ is an abelian subnormal subgroup of $G$ and so $L=(H')^G$ is locally nilpotent. By Theorem 4.3, every subgroup of $L$ is normal or contained in the norm of $L$ ; in particular, every subgroup of $L$ is subnormal of defect at most $2$ .

Let $S$ be a normal subgroup of $L$ such that $H'$ is normal in $S$ . Since $L$ and $H'$ are $H$ -invariant, we may further require that $S$ is $H$ -invariant. We claim that $H$ is subnormal in $HS$ of defect at most $2$ . To prove this, we may assume $H'=\{1\}$ , so $H$ is abelian and subnormal in $HS$ , and by the above, $H$ is subnormal of defect at most $2$ in $HS$ . In a similar fashion, we show that $HS$ is subnormal of defect at most $2$ in $HL$ , so $H$ is subnormal of defect at most $4$ in $HL$ . Finally, a further application of the same argument shows that $HL$ is subnormal in $(HL)^G$ of defect at most $2$ , so $H$ is subnormal in $G$ of defect at most $7$ .

Question 4.6. What is the smallest possible defect of subnormality in the statement of Theorem 4.5?

Proof. Clearly, we may always assume that $G$ is finitely generated. Then, a combination of Theorem 4.5 and [Reference Robinson18, Theorem 10.54] yields that $G$ is finite-by-nilpotent, and so even finite-by-(torsion-free nilpotent), being $G$ finitely generated.

Assume $G$ is nilpotent and nonperiodic. Then, $G$ is quasi-Hamiltonian by Corollary 4.4, so it satisfies the conditions in [Reference Schmidt20, Theorem 2.4.11]. If $G$ is nonabelian, then $G$ has a nonabelian section of the form $\mathcal C_{p^n}\ltimes \mathcal C_{p^m}$ , where $m$ is fixed and $n$ can be chosen arbitrarily large. This contradicts Theorem 3.1, and proves Statement (2).

Going back to the general case, we see that $G$ is finite-by-abelian, which proves the first half of Statement (1), and consequently even abelian in case $G$ is torsion-free, thus proving Statement (3).

To prove the second half of Statement (1), we may assume that the torsion subgroup $T$ of $G$ is a $p$ -group for some prime $p$ (see Theorem 3.3). Since $G$ is (central torsion-free)-by-finite, every finite quotient of $G$ is a quotient of a finite quotient of $G$ in which the centre has $p$ -elements of arbitrarily large orders. Thus, Theorem 3.1 yields that the $p$ -subgroups of every quotient of $G$ are abelian. This implies that there exists a torsion-free central subgroup $Z$ of $G$ such that $G/Z=U/Z\ltimes TZ/Z$ for some $p'$ -subgroup $U/Z$ . Clearly, $U$ is torsion-free and $G=U\ltimes T$ , as desired.

(4) To prove the statement, we may assume $G$ is finitely generated, so $G$ is finite-by-(abelian torsion free) by Statement (1). Since $G$ has a torsion-free abelian central subgroup of finite index, we have that $G$ is residually finite-by-(infinite cyclic), so we may actually assume that $G$ is finite-by-(infinite cyclic). Write $G=\langle w\rangle \ltimes H$ , where $H$ is finite and $o(w)=\infty $ , and set $C=C_{\langle w\rangle }(H)$ . It easily follows from Theorem 3.1 (and Example 2.2(5)) that the Sylow $p$ -subgroups of every quotient of $G$ are abelian (essentially because we can find a larger quotient containing a central $p$ -element of large order).

Suppose first that all subgroups of $H$ are pronormal. If $H$ is abelian, then $G$ is metabelian and we are done. Assume $H$ is not abelian and write $H=K\ltimes \gamma _3(H)$ , where as usual we have $\pi (K)\cap \pi (\gamma _3(H))=\emptyset $ , and both $K$ and $\gamma _3(H)$ are abelian. Since the Sylow subgroups of the quotients are abelian, we can write $K=C_{\langle w\rangle }(K)\times [\langle w\rangle ,K]$ modulo $\gamma _3(H)$ . Again, modulo $\gamma _3(H)$ , we have that the nontrivial Sylow subgroups of $[\langle w\rangle ,K]\times C$ are not pronormalized by $w$ , so $K$ is contained in the pronorm of $G$ . However, $K$ does not pronormalize a cyclic subgroup of $\gamma _3(H)\times \langle w\rangle $ , which hence is a contradiction. This implies that $[\langle w\rangle ,K]=\{1\}$ , so $G$ is metabelian.

Now, assume that $H$ does not have only pronormal subgroups. Then, $H$ has the structure outlined in Theorem 3.12, so we may write $H=(T\ltimes S)\ltimes X$ , where $T$ , $S$ and $X$ satisfy the conditions in that statement.

We start by deriving some additional information on $H$ . Set $L=H\times C$ . First, we claim that $T$ , $S$ and $X$ are abelian. By Statement (2), all Sylow subgroups of $G$ are abelian, so $S$ is abelian. Moreover, $T\times \langle a\rangle $ has only pronormal subgroups, so again $T$ is abelian. Let $n$ be the product of all primes in $\pi (H)$ and replace $G$ by $G/C^n$ , so $C$ is now finite of order $n$ . If $\gamma _3(X)\neq \{1\}$ , then $U=\gamma _3(X)\times O_{\pi (\gamma _3(X))}(\langle a\rangle )$ is not contained in the pronorm, so all subgroups of $G/U$ are pronormal; but $G/U$ has central $p$ -subgroups for every prime $p$ in $\pi (G/U)$ , so $G/U$ is abelian. If $w$ does not centralize a cyclic subgroup of $H/U$ , then $w$ does not normalize a cyclic subgroup of $H/U\times CU/U$ , so $S\leq N_{\mathrm{pro}}(G)$ , which is a contradiction.

Now, if $[T,S]\neq \{1\}$ , then $T$ does not pronormalize a cyclic subgroup of $S\times C$ , so $S$ is contained in the pronorm of $G$ , which is a contradiction. Thus, $[T,S]=\{1\}$ . If $w$ does not centralize a cyclic subgroup of $T\times S$ , then $w$ does not normalize a cyclic subgroup of $T\times S\times C$ , so $T\times S\leq N_{\mathrm{pro}}(G)$ , which is a contradiction. Consequently, $w$ centralizes $T\times S$ and $G$ is metabelian.

(5) Let $u$ be an element of $T$ that is not centralized by $x$ . By Statement (1), there is a power of $x$ , say $y=x^n$ , centralizing $u$ . However, then $\langle uy\rangle $ is not pronormal in $\langle uy,x\rangle $ , so $uy$ is contained in the pronorm of $G$ and must consequently act as power automorphisms on $T\times \langle y\rangle $ , and so it must centralize $T\times \langle y\rangle $ . Thus, $[y,T]=\{1\}$ , as desired.

Proof. Clearly, we may assume $G$ is finitely generated. By Corollary 4.7, $G$ is polycyclic and every finite quotient of $G$ is soluble of derived length at most $3$ (see Corollary 3.5). Since $G$ is residually finite, $G$ has derived length at most $3$ .

Proof. Since the property is inherited by subgroups, we only need to show that $G$ is a $T$ -group. Let $H$ be a subnormal subgroup of $G$ . If $H$ is pronormal, then $H$ is normal in $G$ . Assume $H$ is contained in the pronorm of $G$ , so $H$ is abelian (being locally graded). Then, $H^G$ is locally nilpotent and so abelian by Corollary 4.7(2). Consequently, $H^G\langle x\rangle $ is abelian for every $x\in G$ (by Corollary 4.7(3)), so $H^G\leq Z(G)$ and $H$ is again normal in $G$ .

Proof. Let $\{p_i\}_{i\in \mathbb N}$ and $\{q_i\}_{i\in \mathbb N}$ be families of prime numbers such that $q_i$ divides ${p_i-1}$ and $|\{p_i,q_i,p_j,q_j\}|=4$ for $i\neq j$ . For each $i\in \mathbb N$ , let $\langle a_i\rangle $ and $\langle b_i\rangle $ be groups of order $p_i$ and $q_i$ , respectively. For each $i\in \mathbb N$ , let $H_i=\langle b_i\rangle \ltimes \langle a_i\rangle $ be the natural semidirect product. Let $\langle x\rangle $ be an infinite cyclic group and define $G=\langle x\rangle \ltimes \mathrm{Dr}_{i\in I} H_i$ by setting $b_i^x=b_ia_i$ . Put $H=\mathrm{Dr}_{i\in I} H_i$ . Let $X$ be any subgroup of $G$ .

Assume $X\not \leq \langle x\rangle H'$ . We claim that $X$ is pronormal in $G$ . Without loss of generality, we may assume $X\cap H'=\{1\}$ and then even $X\cap H=\{1\}$ , so actually, $X$ is infinite cyclic. This obviously allows us to assume that $X \le G = \langle x \rangle N$ , where $N = \mathrm{ Dr}_{i=1}^nH_i$ for $n$ sufficiently large and $\pi (\langle x\rangle )=\pi (N)$ . Now, $Z(G) \le \langle x \rangle $ and $G/Z(G)$ is finite, so we may assume that $Z(G) = \{1\}$ and, hence, that $G$ is finite. Setting $z= xa_1^{-1} \cdots a_n^{-1}$ , we have that $G = \langle z \rangle \times N$ . By Lemma 3.10, every subgroup of $G$ is pronormal or contained in the pronorm. Moreover, by Lemma 3.8, $N_{\mathrm{pro}}(G) = \langle z \rangle \times N'=\langle x\rangle N'$ (note that the pronorm is maximal in $G$ and $G$ has not only pronormal subgroups), so $X$ is pronormal in $G$ .

Now, suppose $X\leq \langle x\rangle H'$ . Let $W$ be a subgroup of $G$ . We need to show that $X$ pronormalizes $W$ . By the previous paragraph, we may assume that $W\leq \langle x\rangle H'$ . However, $\langle x\rangle H'$ is abelian and so we are done.

Proof. Let $H$ be a nonpronormal subgroup of $G$ . We need to show that $H$ is contained in the pronorm of $G$ . The main result of [Reference Nogata16] shows that the only nonpronormal subgroups of $J_1$ are isomorphic with $\mathcal C_2$ and $\mathcal C_2\times \mathcal C_2$ , and so without loss of generality, we may assume that $H=\langle g\rangle $ is cyclic of order $2$ .

Now, let $K$ be any subgroup of $G$ and put $L=\langle K,K^g\rangle $ . The proof is complete if we can prove that $K$ is conjugate to $K^g$ in $L$ . To this aim, we may obviously assume that $L$ is a proper subgroup of $G$ . Furthermore, $x$ normalizes $L$ because $o(x)=2$ , so $\langle g\rangle L$ is in fact a proper subgroup of $G$ and, hence, it is contained in a maximal subgroup $M$ of $G$ . Then, $M$ is isomorphic to one of the following groups:

1. (1) $\mathrm{PSL}(2,11)$ ; (2) $(\mathcal C_3\ltimes \mathcal C_7)\ltimes \mathcal C_2^3$ ; (3) $\mathcal C_2\times \mathrm{Alt}(5)$ ; (4) $\mathcal C_{6}\ltimes \mathcal C_{19}$ ;

2. (5) $\mathcal C_{10}\ltimes \mathcal C_{11}$ ; (6) $\mathrm{Sym}(3)\times \mathrm{Dih}(10)$ ; (7) $\mathcal C_6\ltimes \mathcal C_7$ .

It is easy to see that groups of types (3)–(7) have all their subgroups that are pronormal or contained in the pronorm. Similarly, groups of type (1) and (2) satisfy the same property (use GAP [21], for example).

If $M$ is of type (4), (5) or (7), then all its subgroups are pronormal and there is nothing more to add. If $M$ is of type (2) or (6), then all its cyclic subgroups of order $2$ are not pronormal, so they must be contained in the pronorm of $M$ . If $M$ is of type (3), then the subgroups of order $2$ are either central or nonpronormal, so again we are done. If $M$ is of type (1), then all its involutions are conjugate so the involutions are nonpronormal (see for example [Reference Brescia, Ingrosso and Trombetti1, Corollary 3.52]) and, hence, they pronormalize $K$ . The statement is proved.

Proof. It follows from [Reference Nogata16] that the only nonpronormal subgroups are elementary abelian $2$ -groups. Let $H=\langle x\rangle $ be any cyclic subgroup of order $2$ and let $K$ be any subgroup of $G$ . As in the proof of Theorem 5.1, we may assume that $H$ normalizes $\langle K,K^x\rangle $ , so $\langle H,K\rangle $ is a proper subgroup of $G$ . Now, note that the subgroups of every maximal subgroup $M$ of $G$ are pronormal or contained in the pronorm of $M$ , which in turn implies that the subgroups of order $2$ of $M$ are contained in the pronorm of $M$ . This shows that $x$ pronormalizes $K$ and proves the result.

Proof. Since $t$ is an involution in $\mathrm{PSL}(2,q)$ and $q\equiv 5\ (\mathrm{mod}\ 8) $ , every lift $A\in \mathrm{SL}(2,q)$ of $t$ has trace $\mathrm{Tr}(A)=0$ ; hence, we may write

$$ \begin{align*} A=\begin{pmatrix}a & b\\[2pt] c & -a\end{pmatrix}, \quad a^2+bc=-1, \end{align*} $$

where the last identity comes from $\det (A)=1$ . Clearly,

$$ \begin{align*} A^{y_x}=y_x^{-1}Ay_x= \begin{pmatrix} a-cx & b+2ax-cx^2\\[2pt] c & -a+cx \end{pmatrix}. \end{align*} $$

Denote $P=AA^{y_x}$ and $Q=A^{y_x}A$ . A direct matrix multiplication gives

$$ \begin{align*} P&=\begin{pmatrix} a^2-acx+bc & 2a^2x-acx^2+bcx\\[2pt] -c^2x & a^2+acx+bc-c^2x^2 \end{pmatrix},\\Q&=\begin{pmatrix} a^2+acx+bc-c^2x^2 & -2a^2x+acx^2-bcx\\[2pt] c^2x & a^2-acx+bc \end{pmatrix}. \end{align*} $$

We now analyse when the classes of $A$ and $A^{y_x}$ commute in $\mathrm{PSL}(2,q)$ . Since the centre of $\mathrm{SL}(2,q)$ is $\{\pm I_2\}$ , commuting in $\mathrm{PSL}(2,q)$ means that

$$ \begin{align*} AA^{y_x}=\varepsilon\, A^{y_x}A, \quad \varepsilon\in\{\pm 1\}. \end{align*} $$

Case $\varepsilon =+1.$ The equation $P=Q$ implies, by comparing the $(2,1)$ -entries, that