Proof of Theorem 1.3. Let $\pi : \mathcal {E} \to X$ be the projection, and let $\varphi \in {\rm Aut}(\mathcal {E}\to X)$. Then, it follows from the definition that $\varphi$ induces a well-defined automorphism $\psi _{\varphi }$ of $X$ such that

\[ \pi \circ \varphi = \psi_{\varphi}\circ \pi. \]

That is, the following diagram commutes:

Now, let

\[ k: {\rm Aut}(\mathcal{E}\to X)\to {\rm Aut}(X) \]

be the homomorphism given by $\varphi \mapsto \psi _{\varphi }$. Then, clearly the homomorphism $k$ is surjective, and the kernel ${\rm Ker}(k)$ of $k$ is equal to the group $\mathcal {G}(\mathcal {E}\to X)$ of all automorphisms of $\mathcal {E} \to X$ whose induced automorphism of $X$ is the identity. Hence, we can obtain the following short exact sequence of groups

(2.1)\begin{equation} \begin{matrix} 1 & \to & \mathcal{G}(\mathcal{E}\to X) & \stackrel{j}{\hookrightarrow} & {\rm Aut}(\mathcal{E}\to X) & \stackrel{k}{\to} & {\rm Aut}(X) & \to & 1.\\ & & & & \varphi & \mapsto & \psi_{\varphi} & & \end{matrix} \end{equation}

Proof. (1) For the proof, let $H$ be any finite subgroup of $\mathcal {G}(\mathcal {E}\to X)$. For each $x\in X$, let $H_{x}$ be the subgroup of $GL(\mathcal {E}_x)\cong GL(r,\, k)$ obtained by restricting the automorphisms of $\mathcal {G}(\mathcal {E}\to X)$ to the fibre $\mathcal {E}_{x}$. Then, clearly $H_{x}$ is a finite subgroup of $GL(\mathcal {E}_{x})$. Note that there is a point $x\in X$ such that $|H|$ is equal to $|H_{x}|$. Moreover, since the linear automorphism group $GL(r,\, k)$ has bounded finite subgroups, $H_{x}$ should also have bounded finite subgroups with $N(H_{x}) \le N(GL(r,\, k))$. Hence, the order of $H$ is always less than or equal to $N(GL(r,\, k))$. Since $H$ is an arbitrary finite subgroup of $\mathcal {G}(\mathcal {E}\to X)$, this implies that $\mathcal {G}(\mathcal {E}\to X)$ has bounded finite subgroups with

\[ N(\mathcal{G}(\mathcal{E}\to X))= N(GL(r, k)), \]

as desired.

1. (1) The proof of Lemma 2.2 (2) is similar to that of Lemma 2.2 (1). Indeed, let $A$ be any finite abelian subgroup of $\mathcal {G}(\mathcal {E}\to X)$. For each $x\in X$, let $A_{x}$ be the subgroup of $GL(\mathcal {E}_x)\cong GL(r,\, k)$ obtained by restricting the automorphisms of $\mathcal {G}(\mathcal {E}\to X)$ to the fibre $\mathcal {E}_{x}$. Then, clearly $A_{x}$ is a finite abelian subgroup of $GL(\mathcal {E}_{x})$. Note that there is a point $x\in X$ such that $|A|$ is equal to $|A_{x}|$. Moreover, since the linear automorphism group $GL(r,\, k)$ has a finite ${\rm Rk}_f (GL(r,\, k))$, $A_{x}$ should also have a finite ${\rm Rk}_f(A_{x}) \le {\rm Rk}_f(GL(r,\, k))$. Hence, ${{\rm RK}}_f (A)$ is always less than or equal to ${\rm Rk}_f(GL(r,\, k))$. Since $H$ is an arbitrary finite abelian subgroup of $\mathcal {G}(\mathcal {E}\to X)$, this implies that $\mathcal {G}(\mathcal {E}\to X)$ has a finite

\[ {\rm Rk}_f (\mathcal{G}(\mathcal{E}\to X))\le {\rm Rk}_f(GL(r, k)). \]

This completes the proof of Lemma 2.2.

Proof. By Lemma 3.1, ${\rm Aut}(N_{Z/X}\to Z)$ contains an isomorphic copy of every finite subgroup of ${\rm Aut}(X,\, Z)$. On the other hand, since ${\rm Aut}(Z)$ is assumed to be strongly Jordan, it follows from Theorem 1.3 that ${\rm Aut}(N_{Z/X} \to Z)$ is strongly Jordan. This completes the proof.

Proof. (1) By assumption, there exits a modification $\mu : \tilde {X}\to X$ obtained as a composition of blowups with smooth centres such that $\tilde {X}$ is a compact Kähler manifold. For simplicity, we assume that $X$ and $\tilde {X}$ are connected. Note that $\dim \tilde {X}=\dim X$. Thus we have the Gysin map

\[ \mu_\ast : H^{1}(\tilde{X}, \Omega_{\tilde{X}}^{1}) \to H^{1}(X, \Omega_X^{1}) \]

defined by using the Serre duality. Let $\tilde {\omega }$ be a Kähler form on $\tilde {X}$. Since $\mu$ is a modification obtained by several blowups with smooth centres, there is a Hermitian form $\omega$ on $X$ such that $\mu ^{\ast }(\omega )=\tilde {\omega }$ on $\tilde {X}$ minus exceptional divisors of the modification $\mu$. Here, in order to obtain a Hermitian form $\omega$ defined everywhere on $X$, by Hartogs’ extension principle we have extended $\omega$ initially defined only on $X$ minus smooth centres to all of $X$ in a natural way. It then follows from the projection formula that we have

\[ \mu_\ast\mu^{{\ast}} ([\omega]) =[\omega]. \]

If we consider $H^{1}(X,\, \Omega _X^{1})$ as a Dolbeault cohomology group of currents, there is a smooth function $\varphi$ on $X$ such that

\[ \mu_\ast \mu^{{\ast}} (\omega) +\sqrt{-1}\partial \bar{\partial} \varphi =\omega \ge -\varepsilon \omega \]

for every $\varepsilon >0$. This implies that

\[ c=[\omega]=\mu_\ast\mu^{{\ast}}([\omega]) \]

is a nef class on $X$. On the other hand, since we have

\[ \int_{X} (\mu_\ast \mu^{{\ast}}(\omega))^{n} =\int_{\tilde{X}} (\mu^{{\ast}}(\omega))^{n} =\int_{\tilde{X}} \tilde{\omega}^{n} >0. \]

Hence, it follows from a result of Demailly and P$\breve {{\rm a}}$un in [Reference Demailly and Păun3] that $c$ is a big class on $X$.

1. (1) For the proof, we essentially follow the arguments of Lieberman in [Reference Lieberman6]. To do so, as in (1) above let $c$ denote the nef and big class defined by $\mu _\ast \mu ^{\ast } ([\omega ])=[\omega ]$. Then, we calculate the volume of the graph $\varGamma _f$ of an automorphism $f \in {\rm Aut}_c (X)$ with respect to $\omega$, as follows.

(4.1)\begin{equation} {\rm Vol}(\varGamma_f) =(\omega+f^{{\ast}}\omega)^{n} = (c+f^{{\ast}} c)^{n} =2^{n} c^{n}. \end{equation}

Since we have an uniform bound (4.1) on the volume of $\Gamma _f$, we may use the Bishop's theorem as in the proof of Fujiki and Lieberman in their papers [Reference Fujiki4, Reference Lieberman6], in order to complete the proof of Theorem 4.2 (2).

Proof. By the pull-back action of ${\rm Aut}(X)$ on $H^{2}(X,\, \mathbb {Q})$, we can obtain a faithful representation of ${\rm Aut}(X)/{\rm Aut}_r(X)$, as follows.

(4.2)\begin{equation} {\rm Aut}(X)/{\rm Aut}_r (X)\to GL(H^{2}(X, \mathbb{Q})). \end{equation}

Note that $GL(H^{2}(X,\, \mathbb {Q}))$ has bounded finite subgroups. Since

\[ {\rm Aut}_r(X)/{\rm Aut}_0(X) \]

is finite by Corollary 4.3, it follows from (4.2) that ${\rm Aut}(X)/{\rm Aut}_0(X)$ also has bounded finite subgroups.

On the other hand, according to [Reference Popov10, Theorem 5], ${\rm Aut}_0(Y)$ is strongly Jordan for any compact complex space $Y$. Hence, it follows from Lemma 2.1 that ${\rm Aut}(X)/{\rm Aut}_0(X)$ is strongly Jordan, as required.

Proof of Theorem 5.1. For the proof, we divide it into several cases, as follows.

1. (1) Assume that $Z$ is not connected. Then, we first choose a finite-index subgroup $\varGamma _0$ of $\varGamma$ such that $\varGamma _0$ fixes all finitely many components of $Z$. As a consequence, from now on we may assume that $Z$ is connected.

2. (2) Assume that $Z$ is a compact connected Kähler manifold of dimension less than or equal to $3$. Since $\varGamma$ is a non-abelian free subgroup of ${\rm Aut}(Z)$ acting freely on $Z$, it follows from Theorem 5.2 that $\dim M=3$ and $Z$ is a compact complex torus, as desired.

3. (3) Assume that $Z$ is singular. Note that the singular locus ${\rm Sing}(Z)$ of $Z$ is always $\varGamma$-invariant, non-empty, closed analytic subset of $Z$ such that $\dim {\rm Sing}(Z)$ is less than $\dim Z\le 3$. Hence, we may replace $Z$ by the singular locus ${\rm Sing}(Z)$ with a strictly lower dimension, and so we assume that $\dim Z$ is less than or equal to $2$. Since the Fujiki's class $\mathcal {C}$ is closed under taking closed subspaces, $Z$ is still in Fujiki's class $\mathcal {C}$. Note that $\dim Z$ cannot be equal to zero, since $\varGamma$ acts freely on $Z$.

   1. (i) Assume that again $Z$ is singular. Then, there is a singular locus ${\rm Sing}(Z)$ with $\dim {\rm Sing}(Z) =1$ which is $\varGamma$-invariant, non-empty, closed analytic subset of $Z$. Thus, by [Reference Cantat, Paris-Romaskevich and Xie1, Lemma 3.3] $\varGamma$ should be cyclic, which is a contradiction.

   2. (ii) Assume that $Z$ is smooth and $\dim Z=1$. Then, $\varGamma$ should be cyclic by [Reference Cantat, Paris-Romaskevich and Xie1, Lemma 3.3], which is a contradiction.

   3. (iii) Assume that $Z$ is a compact Kähler manifold and $\dim Z=2$. Then, once again $\varGamma$ should be cyclic by [Reference Cantat, Paris-Romaskevich and Xie1, Lemma 3.4], which is a contradiction.

   4. (iv) Assume that $Z$ is smooth and connected in Fujiki's class $\mathcal {C}$, but not Kähler and that $\dim Z=2$. Then, it follows from [Reference Demailly and Păun3, Theorem 0.7] and [Reference Tosatti13, Section 2.4] that there is a big real $(1,\,1)$-class $[\alpha ]\in H^{1,1}_{\partial \bar {\partial }}(Z,\, \mathbb {R})$ whose non-Kähler class $E_{nK}(\alpha )$ is $\varGamma$-invariant, non-empty, closed analytic subset of $Z$. Since $\dim E_{nK}(\alpha )$ is equal to $1$, once again by [Reference Cantat, Paris-Romaskevich and Xie1, Lemma 3.3] $\varGamma$ should be cyclic, which is a contradiction.

4. (4) Assume that $Z$ is compact connected in Fujiki's class $\mathcal {C}$, but not Kähler and that $\dim Z$ is less than or equal to $3$. Then, as in (2)-(iv) there is a big real $(1,\,1)$-class $[\alpha ]\in H^{1,1}_{\partial \bar {\partial }}(Z,\, \mathbb {R})$ whose non-Kähler class $E_{nK}(\alpha )$ is $\Gamma$-invariant, non-empty, closed analytic subset of $Z$. Note that $\dim E_{nK}(\alpha )$ is less than or equal to $2$. Now, by applying the arguments as in (3) to the singular locus $E_{nK}(\alpha )$ we can derive a contradiction.

This completes the proof of Theorem 5.1.

Proof. Suppose that $M$ is not Kähler. Then, it follows from the proof of Theorem 5.1 (more precisely, case (4)) that we can derive a contradiction. This implies that $M$ should be a compact connected Kähler manifold of dimension $\le 3$, equipped with a free action of a non-abelian free subgroup $\varGamma$ of ${\rm Aut}(M)$. Hence, by Theorem 5.2 we have $\dim M=3$, and $M$ is a compact complex torus. This completes the proof.