$\newcommand{\sym}[1]{{\rm Sym}\,#1}\newcommand{\fix}[2]{{\rm Fix}_{#1}(#2)}\newcommand{\wr}{\,{\rm wr}\,}\newcommand{\Z}{\mathbb Z}$

The problem was reported by Joris Buloron.

Unfortunately, there is an oversight in the proof of Lemma 2.19, and so the assertion of the lemma may not be true for certain infinite permutation groups. A correct version of the lemma is the following.

**Lemma 2.19. **If $G$ is transitive on $\Omega$, $\omega\in\Omega$, and $N$ is a normal subgroup of $G$, then $G_\omega\leq N_G(N_\omega)$ and $\omega N_G(N_\omega)\subseteq\fix\Omega{N_\omega}$. Moreover, $\omega N_G(N_\omega)$ is a block of imprimitivity for $G$ of cardinality $|\omega N_G(N_\omega)| = |N_G(N_\omega):G_\omega|$. If $N_\omega$ has no infinite strictly ascending chain

\begin{equation}\label{eq:chain}

N_\omega < N_{\omega_1} < N_{\omega_2} < \cdots

\end{equation}

of stabilisers (with $\omega,\omega_i\in\Omega$ for all $i\geq 1)$, then $\omega N_G(N_\omega)=\fix\Omega{N_\omega}$. In particular, no such chain exists if either $\fix\Omega{N_\omega}$ is finite or $G$ is a torsion group.

**Proof. **Since $N_\omega=N\cap G_\omega$ and $N$ is normal in $G$, we obtain that $N_\omega$ is normal in $G_\omega$, and so $G_\omega\leq N_G (N_\omega)$. It follows from Lemma 2.14.that $\omega N_G(N_\omega)$ is a block of imprimitivity for $G$ of cardinaltiy $|\omega N_G(N_\omega)| = |N_G(N_\omega):G_\omega|$. The claim that $\omega N_G(N_\omega)\subseteq\fix\Omega{N_\omega}$ can be proved following the argument in the book.

Unfortunately, the proof that $\fix\Omega{N_\omega}\subseteq \omega N_G(N_\omega)$ is not correct in general when $G$ is infinite. Let us review our argument. If $\omega_1\in\fix\Omega{N_\omega}$, then there is some $g\in G$ such that $\omega_1=\omega g$. Furthermore, we have, for all $n\in N_\omega$, that $\omega gn=\omega g$, and, as $N$ is normal in $G$, $gng^{-1}\in G_\omega\cap N=N_\omega$. As this holds for all $n\in N_\omega$, this implies that $g N_\omega g^{-1}\leq N_\omega$, and so $N_\omega\leq g^{-1} N_\omega g$ (but it may not imply, as we claimed, that $N_\omega =g^{-1} N_\omega g$).

Assume that the proper containment $N_\omega < g^{-1}N_\omega g$ holds. Then we obtain that

$$

N_\omega <g^{-1}N_\omega g<g^{-1}(g^{-1}N_\omega g)g=g^{-2} N_\omega g^{2}.

$$

Calculating similarly the conjugate $g^{-k} N_\omega g^{k}$ for $k\geq 1$, we obtain the infinite strictly ascending chain of subgroups

$$

N_\omega< g^{-1}N_\omega g< g^{-2} N_\omega g^{2}<\cdots< g^{-k} N_\omega g^k<\cdots.

$$

Since $g^{-k} N_\omega g^{k}=N_{\omega_k}$ where $\omega_k=\omega g^k$, we obtain the chain of stabilisers as in \eqref{eq:chain}. Assuming that no such chain exists, the equality $N_\omega =g^{-1} N_\omega g$ must hold and $g\in N_G(N_\omega)$ and $\omega_1=\omega g\in\omega N_G(N_\omega)$. Hence, in this case, $\fix\Omega{N_\omega}\subseteq \omega N_G(N_\omega)$ and the equality $\fix\Omega{N_\omega}= \omega N_G(N_\omega)$ follows.

Let us finally show the last statement. Set $F:= \fix\Omega{N_\omega}$. Since $N_\omega \leq N_{\omega_i}$, we have that $\omega_i\in F$ for all $i\geq 1$. So if $F$ is finite, then we cannot have such an infinite chain. If $G$ is a torsion group, then $g^i=1$ for some $i\geq 1$, and so $\omega_i=\omega$, and we have a contradiction to the proper inclusion in \eqref{eq:chain}. $\Box$

Note that if $N_\omega$ admits the strictly ascending chain \eqref{eq:chain}, then $N_\omega$ also admits the strictly descending chain

$$

N_\omega > N_{\omega_{-1}}>\cdots> N_{\omega_{-k}}>\cdots

$$

where $\omega_{-k}=\omega g^{-k}$ for all $k\geq 1$. Therefore, in this case, $N_\omega$ admits the following chain of stabilisers which is infinite in both directions:

$$

\cdots < N_{\omega_{-k}}<\cdots<N_{\omega_{-1}}<N_\omega <N_{\omega_1}< N_{\omega_2}<\cdots<N_{\omega_k}<\cdots.

$$

**Example. **The following example is related to the construction given in the proof of Lemma 5.3 of the paper Permutation group subdegrees and the common divisor graph by C. E. Praeger and I. M. Isaacs.

Let $H$ be a group and let $G = H \wr \Z$ where $\Z$ is considered as an additive infinite cyclic group. The base group of $G$ is the normal subgroup $N = \prod_{z \in \Z} H_z$ where $H_0=H$ and $H_z = H^{z}$ for all $z\in\Z$. Consider the coset action of $G$ on the right coset space $\Omega=[G:M]$ where $M=\prod_{z< 0} H_z$. Let $K=\prod_{z\geq 0} H_z$. Note that every element of $\Omega$ can be written uniquely in the form $Mxz$ where $x\in K$ and $z\in\Z$. The action of $G$ under this identification can be worked out as follows. Let $x\in K$, $z\in\Z$ and $n\in N$. For $i\in \Z$ and $y\in N$, let $y_i$ denote the $i$-th coordinate of $y$. In our case, $x_i=1$ for $i<0$. Then

$$

(Mxz)n=M(x n^{-z})z=Myz

$$

where $y\in K$ and $y_i=1$ for all $i<0$ while $y_i=x_in_{i+z}$ for $i\geq 0$. Moreover, if $u\in\Z$, then

$$

(Mxz)u=Mx(zu).

$$

Choosing $\omega=M$, we have that $G_\omega=N_\omega=M$. The subgroup $M$ is corefree in $G$, and so $G$ can be viewed as a permutation group on $\Omega$.

The stabilizer $N_\omega=G_\omega=M$ fixes precisely the points of the set $$\{ Mxz \mid x\in K \mbox{ and } z\geq 0\}.$$ Note that $N_G(G_{\omega})=N_G(N_{\omega})=N$, and so $$\omega N_G(G_\omega) = \omega N_G(N_\omega)=\omega N = \{Mx\mid x\in K\}.$$

Thus $\omega N_G(N_\omega)\subset \fix\Omega{N_\omega}$ with proper containment.

For each $z>0$, $M^{z} = \prod_{i< z}H_i$ and this is the stabiliser (in $G$ and in $N$) of the point $\omega_i:= \omega^{z} = Mz$. W__e have an infinite strictly ascending chain $N_\omega < N_{\omega_1} < \cdots$ and a descending chain $N_\omega > N_{\omega_{-1}} > \cdots$.__

__Further examples of groups in which a conjugate of a subgroup $X$ is strictly contained in $X$ are described in this discussion of MathOverFlow.__

The first sentence of Section 2.6 should read as follows.

*Lemma 2.19 shows us that, for $H \leq G_\omega$ , the normaliser $N_G(H)$ is transitive on the set $\fix \Omega H$ when $H = N_\omega$ , for some normal subgroup $N$ of $G$ that has no infinite strictly ascending chain of stabilisers.*

In consequence of the modifications in Lemma 2.12, Corollary 2.21 also needs to be adjusted.

**Corollary 2.21.** Let $G$ be a group acting primitively on $\Omega$ and let $N$ be a normal subgroup of $G$ which is not contained in the kernel of the action. Then $N$ is transitive and either $N$ induces a regular permutation group on $\Omega$ or $N_G(N_\omega)=G_\omega$ for each $\omega\in\Omega$. In particular, a non-trivial normal subgroup of a primitive permutation group is transitive. Furthermore, if $N_\omega$ has no infinite strictly ascending chain of stabilisers as in \eqref{eq:chain}, then either $N$ induces a regular permutation group on $\Omega$ or $\fix \Omega{N_\omega}=\{\omega\}$ for each $\omega\in\Omega$.

Also, in Theorem 3.2, statement (ii) will need to be changed:

**Theorem 3.2(ii).** $\omega C_{\sym\Omega}(G)=\omega N_G(G_\omega)\subseteq \fix\Omega{G_\omega}$.

Parts (i) and (iii) of Theorem 3.2 can be proved as in the book. The new part (ii) follows from part (i) and from the new Lemma 2.19 above.

In the proof of part (iv), more care needs to be taken. Recall that $G$ is a transitive permutation group on $\Omega$. Assume that $C_{\sym\Omega}(G)$ is transitive. Then, $\omega C_{\sym\Omega}(G)=\Omega$, and hence part (ii) implies that $\fix\Omega {G_\omega}=\Omega$. That is, $G$ is regular. Suppose now that $G$ is regular. Then $G_\alpha=1$ for each $\alpha\in\Omega$, and so there is no infinite strictly ascending chain of stabilisers starting with $G_\omega$. Applying Lemma 2.19 with $N=G$, we obtain that $\fix\Omega{G_\omega}=\omega N_G(G_\omega)=\omega C_{\sym\Omega}(G)$. Since $G$ is regular, $\fix\Omega{G_\omega}=\Omega$, and hence $C_{\sym\Omega}(G)$ is transitive.