$p$-nilpotency criteria for some verbal subgroups

Suppose $G$ is $p$-nilpotent. Then it has a normal $p$-complement $H$. Let $x,y\in G$ be such that $p$ does not divide $o(x)$ and $p$ divides $o(y)$, with $y\neq 1$. Then $x\in H$ and $xy\not\in H$ as $y\not\in H$. Hence $p$ divides $o(xy)$. The other implication follows immediately from Theorem A. ∎

We have $O_{p^{\prime}}(Fit_{p}(G))\leq O_{p^{\prime}}(G)=1$. Since $Fit_{p}(G)$ is $p$-nilpotent, we also have $Fit_{p}(G)=TO_{p^{\prime}}(Fit_{p}(G))$ for some Sylow $p$-subgroup $T$ of $Fit_{p}(G)$. Hence $Fit_{p}(G)=T$ is a normal $p$-subgroup of $G$. Thus $Fit_{p}(G)\leq O_{p}(G)$. On the other hand, $O_{p}(G)$ is a normal $p$-nilpotent subgroup of $G$, and so we conclude that $Fit_{p}(G)=O_{p}(G)$. ∎

By Lemma 2.3 there exists a commutator-closed and symmetric subset $X$ of $G$ such that $G=\langle X\rangle$ and every element of $X$ has prime power order. Hence by [10, Lemma 3.6 item (c)] we get $\gamma_{k}(G)=\langle[x_{1},\dots,x_{t}]\mid x_{i}\in X\cup X^{-1},t\geq k\rangle$. Note that $X\cup X^{-1}=X$, and $[x_{1},\dots,x_{t}]$ is a $\gamma_{k}$-value for every $t\geq k$. Also, since $X$ is commutator-closed, we get that $[x_{1},\dots,x_{t}]\in X$ has prime power order. Thus, $\gamma_{k}(G)$ is generated by $\gamma_{k}$-values of prime power order. ∎

If $H\leq G$ then for every $h\in H_{w}$ we have $h\in G_{w}$. Thus $H$ satisfies $P(w,p)$ if $G$ does.

Assume that $N$ is a normal subgroup of $G$ of $p^{\prime}$-order and consider $\overline{G}=G/N$. Let $xN\in\overline{G}_{w}$ be a $w$-value of $p^{\prime}$-order and let $1\neq yN\in\overline{G}_{w}$ be a $w$-value of order divisible by $p$. Then we can assume that $x,y\in G_{w}$. Also, $p$ divides the order of $y$ and, since $N$ has $p^{\prime}$-order, $p$ does not divide the order of $x$. Thus by $P(w,p)$ we deduce that $p$ divides the order of $xy$. In particular $xy\notin N$ and $p$ divides the order of $xyN=xN\cdot yN$. This shows that $\overline{G}$ satisfies property $P(w,p)$. ∎

Set $N=O_{p^{\prime}}(G)$. Aiming for a contradiction, suppose $N\neq 1$ and consider $\overline{G}=G/N$. Then $|\overline{G}|<|G|$ and by Lemma 3.2 the group $\overline{G}$ satisfies $P(w,p)$. Since $G$ is a minimal $P(w,p)$-exception we deduce that $w(\overline{G})=w(G)N/N$ is $p$-nilpotent. Then $w(G)N/N=S/N\cdot H/N$ where $S/N\in\hbox{\rm Syl}_{p}(w(G)N/N)$ and $H/N=O_{p^{\prime}}(w(G)N/N)$. Set $K=H\cap w(G)$. Since $H\mathrel{\unlhd}w(G)N$, we deduce that $K\mathrel{\unlhd}w(G)$. Also, $|K|$ divides $|H|=[H\colon N]|N|$ and so $|K|$ is prime to $p$. Finally, $[w(G)\colon K]=[w(G)H\colon H]$ divides $[w(G)N\colon H]$, that is a power of $p$. We deduce that $K$ is a normal $p$-complement of $w(G)$ and $w(G)$ is $p$-nilpotent, a contradiction. Therefore we must have $O_{p^{\prime}}(G)=1$. The second part of the statement now follows from Lemma 2.1.

∎

Note that $G^{\prime}$ satisfies $P(w,p)$ by Lemma 3.2. Since $G$ is a minimal $P(w,p)$-exception and $|G^{\prime}|<|G|$ we deduce that $w(G^{\prime})$ is $p$-nilpotent. Note that $w(G^{\prime})$ is a normal subgroup of $G^{\prime}$, so $O_{p^{\prime}}(w(G^{\prime}))\leq O_{p^{\prime}}(G^{\prime})\leq O_{p^{\prime}}(G)=1$ by Lemma 3.3. Hence $O_{p^{\prime}}(w(G^{\prime}))=1$ and since $w(G^{\prime})$ is $p$-nilpotent we deduce that $w(G^{\prime})$ is a $p$-group. In particular $w(G^{\prime})$ is nilpotent. By assumption $G^{\prime}/w(G^{\prime})$ is nilpotent. Hence the group $G^{\prime}$ is metanilpotent. Let $F(G^{\prime})$ denote the Fitting subgroup of $G^{\prime}$. Then $F(G^{\prime})\leq Fit_{p}(G^{\prime})\leq Fit_{p}(G)=O_{p}(G)$ (again by Lemma 3.3), so $F(G^{\prime})$ is a $p$-group. Let $P\in\hbox{\rm Syl}_{p}(G^{\prime})$ be a Sylow $p$-subgroup. Then $F(G^{\prime})\leq P$. On the other hand, $[P,O_{p^{\prime}}(F(G^{\prime}))]=[P,1]=1$ and by Lemma 2.2 we get $P\leq F(G^{\prime})$. Therefore $P=F(G^{\prime})$ is normal in $G^{\prime}$. ∎

By Lemma 3.3 we have $O_{p^{\prime}}(G)=1$. Since $H$ is normal in $G$ we have $O_{p^{\prime}}(H)=1$ and so $Fit_{p}(H)=O_{p}(H)\leq P$ (with the same proof used to show that $Fit_{p}(G)=O_{p}(G)$). Then $F(H)\leq Fit_{p}(H)\leq P$ and so $C_{H}(P)\leq C_{H}(F(H))$. Since $G$ is soluble, $H$ is soluble as well and by [9, Theorem 6.1.3] we get

Therefore $C_{H}(P)\leq P$. ∎

It is enough to prove that $G_{\gamma_{2}}=G_{\gamma_{3}}$, then the result will follow by induction on $k$. Clearly every $\gamma_{3}$-value is a commutator, so $G_{\gamma_{3}}\subset G_{\gamma_{2}}$.

Now, let $g=[a,b]\in G_{\gamma_{2}}$ be a commutator. Using Ore’s conjecture and the fact that $G$ is quasisimple, we can write $a=xz$ where $x$ is a commutator and $z\in Z(G)$. So $g=[xz,b]=[x,b]\in G_{\gamma_{3}}$. Therefore $G_{\gamma_{2}}\subset G_{\gamma_{3}}$. This completes the proof. ∎

Note that for every $g\in G$ we have

Aiming for a contradiction, suppose there exists $g\in G$ such that $p$ divides the order of $[g,_{k-1}x]$. Then $[g,_{k-1}x]\neq 1$ and by $P(\gamma_{k},p)$ we deduce that $p$ divides the order of $[g,_{k-1}x]\cdot x^{-1}$. Thus $p$ divides the order of $x^{-[g,_{k-2}x]}$, that coincides with the order of $x$, a contradiction. This proves the statement. ∎

Let $g\in P$ be an element. By Lemma 4.2 the element $[g,_{k-1}x]$ has $p^{\prime}$-order. On the other hand, since $x\in N_{G}(P)$ we have that $[g,_{k-1}x]\in P$. Therefore the only possibility is $[g,_{k-1}x]=1$. This shows that $[P,_{k-1}x]=1$. Therefore by [9, Theorem 5.3.6] we get $[P,x]=[P,_{k-1}x]=1$. ∎

Note that $G$ is a minimal $P(\gamma_{k},p)$-exception. Hence by Lemma 3.3 we have $Fit_{p}(G)=O_{p}(G)$. By Lemma 2.5 the group $G$ is generated by $\gamma_{k}$-values of $p^{\prime}$-order. Since $Fit_{p}(G)=O_{p}(G)$, Lemma 4.2 implies that $Fit_{p}(G)\leq Z(G)$. On the other hand, $Z(G)$ is abelian and so $p$-nilpotent and it is normal in $G$, so $Z(G)\leq Fit_{p}(G)$. Thus $Fit_{p}(G)=Z(G)$.

It remains to prove that $G$ is quasisimple. Since $G=G^{\prime}$, it is enough to show that $G/Z(G)$ is simple. Let $N/Z(G)\mathrel{\unlhd}G/Z(G)$ be a proper normal subgroup of $G/Z(G)$. Then $N\mathrel{\unlhd}G$ is a proper subgroup and $N$ has property $P(\gamma_{k},p)$ by Lemma 3.2. Hence $\gamma_{k}(N)$ is $p$-nilpotent by minimality of $G$. Therefore $\gamma_{k}(N)\leq Fit_{p}(G)=Z(G)$. In particular $\gamma_{k+1}(N)=[\gamma_{k}(N),N]=1$ and so $N$ is nilpotent and $N\leq Fit_{p}(G)=Z(G)$. Thus $N/Z(G)=1$. This shows that $G/Z(G)$ is simple and completes the proof. ∎

If every element of $G$ is a commutator, then by induction on $k$ we deduce that every element of $G$ is a $\gamma_{k}$-value. Suppose $G$ contains elements that are not commutators. Then by Proposition 2.8 one of the following holds:

1. (1)

   $p=3$, $G/Z(G)\cong\mathbf{A}_{6}$ and $Z(G)\cong C_{3}$;

2. (2)

   $p=2$, $G/Z(G)\cong\mathrm{PSL}(3,4)$ and $Z(G)\cong C_{2}\times C_{4}$;

3. (3)

   $p=2$, $G/Z(G)\cong\mathrm{PSL}(3,4)$ and $Z(G)\cong C_{4}\times C_{4}$.

Using GAP we can check that none of the groups in the above list satisfies property $P(\gamma_{2},p)$ (where $p=3$ in the first case and $p=2$ in the others). Hence by Lemma 4.1 the groups in the list do not satisfy property $P(\gamma_{k},p)$, a contradiction. ∎

We first show that $G>G^{\prime}$. Aiming for a contradiction, suppose $G=G^{\prime}$. Then by Lemma 4.4 we deduce that $G$ is quasisimple and $Z(G)=O_{p}(G)$ is a $p$-group. By Lemma 4.5 every element of $G$ is a $\gamma_{k}$-value. Let $P\leq G$ be a $p$-subgroup and let $y\in\mathrm{N}_{G}(P)$ be an element of $p^{\prime}$-order. Then $y$ is a $\gamma_{k}$-value of $p^{\prime}$-order and by Lemma 4.3 we deduce that $[P,y]=1$. Hence $N_{G}(P)/C_{G}(P)$ is a $p$-group. By Frobenius criterion we conclude that $G$ is $p$-nilpotent, a contradiction. Hence $G>G^{\prime}$. The fact that $G^{\prime}$ has a normal Sylow $p$-subgroup follows from Lemma 3.4.

It remains to show that $G$ is soluble. Since $G$ is a minimal $P(\gamma_{k},p)$-exception and $G>G^{\prime}$, $\gamma_{k}(G^{\prime})$ is a $p$-nilpotent normal subgroup of $G$. Hence $\gamma_{k}(G^{\prime})\leq Fit_{p}(G)$ is a $p$-group by Lemma 3.3. By Lemma 2.7 applied to $G^{\prime}$ we get that $G^{(k)}=(G^{\prime})^{(k-1)}\leq\gamma_{k}(G^{\prime})$. Thus $G^{(k)}$ is a $p$-group and $G$ is soluble. ∎

If $\gamma_{k}(G)$ is $p$-nilpotent then $G$ satisfies property $P(\gamma_{k},p)$ by Corollary B. Suppose that $G$ has property $P(\gamma_{k},p)$. Aiming for a contradiction, suppose $G$ is minimal (with respect to the order) such that $\gamma_{k}(G)$ is not $p$-nilpotent. Then by Lemma 4.6 we get that $G$ is soluble and $G^{\prime}$ has a normal Sylow $p$-subgroup $T$. Set $P=T\cap\gamma_{k}(G)\in\hbox{\rm Syl}_{p}(\gamma_{k}(G))$. Note that $G$ is a minimal $P(\gamma_{k},p)$-exception, so by Lemma 3.5 we deduce that $C_{\gamma_{k}(G)}(P)\leq P$. Since $P\mathrel{\unlhd}\gamma_{k}(G)$, Lemma 4.3 implies that every $\gamma_{k}$-value of $G$ has order divisible by $p$. Using Lemma 2.4 we conclude that every $\gamma_{k}$-value of $G$ has $p$-power order. Therefore $\gamma_{k}(G)\leq P$ is a $p$-group and so it is $p$-nilpotent, a contradiction. This completes the proof. ∎

Let $g\in G$ be an element and let $x\in G_{\delta_{k}}$ be a $\delta_{k}$-value of $p^{\prime}$-order. Note that

is a $\delta_{k}$-value of $G$. Also,

Now, $x^{-1}$ and $x^{-[g,x]}$ are $\delta_{k}$-values of $G$ of $p^{\prime}$-order (their order is equal to the one of $x$). Since $G$ satisfies $P(\delta_{k},p)$, we deduce that $[g,x,x]$ has $p^{\prime}$-order. ∎

Let $g\in P$ be an element. By Lemma 5.1 the element $[g,x,x]$ has $p^{\prime}$-order. On the other hand, since $x\in N_{G}(P)$ we have that $[g,x,x]\in P$. Therefore the only possibility is $[g,x,x]=1$. This shows that $[P,x,x]=1$. Hence by [9, Theorem 5.3.6] we get $[P,x]=[P,x,x]=1$. ∎

If $G^{(k)}$ is $p$-nilpotent then $G$ satisfies property $P(\delta_{k},p)$ by Corollary B.

Suppose $G$ is soluble and satisfies $P(\delta_{k},p)$ but is minimal such that $G^{(k)}$ is not $p$-nilpotent. Note that every proper subgroup and every proper quotient of $G$ is soluble, so $G$ is a minimal $P(\delta_{k},p)$-exception. By Lemma 3.2 the group $G^{\prime}$ satisfies $P(\delta_{k},p)$. Since $G$ is soluble, we have $G>G^{\prime}$ and so $G^{(k+1)}=\delta_{k}(G^{\prime})$ is $p$-nilpotent. In particular $G^{(k+1)}$ is contained in $Fit_{p}(G)$, that is a $p$-group by Lemma 3.3. Let $P\in Syl_{p}(G^{(k)})$. Then $G^{(k+1)}\leq P$ and