UDC 519.41

MATHEMATICS

Academician of the Academy of Sciences of the BSSR D. A. SUPRUNENKO

ON THE INDECOMPOSABILITY OF TRANSITIVE SUBGROUPS OF THE GROUP \(SF(X)\)

1. Let \(X\) be an arbitrary nonempty set; \(S(X)\) the symmetric group acting on \(X\), i.e., the group of all bijective mappings \(f: X \to X\). For infinite \(X\) we introduce the group \(SF(X)\), consisting of all such \(f \in S(X)\) that \(f(x) \ne x\) only for a finite set of points \(x \in X\). Obviously, \(SF(X)\) is a transitive locally finite invariant subgroup of the group \(S(X)\). If \(\mathfrak A\) is a group, then an arbitrary homomorphism \(g: \mathfrak A \to S(X)\) is called a representation of the group \(\mathfrak A\) by substitutions of \(S(X)\). A representation \(h: \mathfrak A \to S(Y)\) is called equivalent to the representation \(g\) if there exists a bijective mapping \(\varphi: X \to Y\) such that for every \(a \in \mathfrak A\) the equality \(h(a)\varphi=\varphi g(a)\) holds.

A group \(\Gamma\) from \(S(X)\) is called regular if it is transitive and its stationary subgroup coincides with the identity subgroup. As is known, every regular subgroup of \(S(X)\) is the image of the left regular representation of some group. Consequently, if \(X\) is an infinite set and \(\Gamma\) is a regular subgroup of \(S(X)\), then \(\Gamma \cap SF(X)=(e)\), where \(e\) is the identity of \(S(X)\).

We shall say that groups \(\mathfrak G_1\) and \(\mathfrak G_2\) from \(S(X)\) have one and the same orbital type if there is a mapping \(f\) in \(S(X)\) such that for every orbit \(X_\alpha\) of the group \(\mathfrak G_1\), \(f(X_\alpha)\) is an orbit of \(\mathfrak G_2\).

1. Lemma. Let a transitive group \(G\) from \(S(X)\) be representable in the form \(G=HF\), where \(H\) and \(F\) are elementwise permutable normal divisors of \(G\), with \(H\) intransitive, and

\[ X=\bigcup X_\alpha,\quad \alpha\in I, \tag{1} \]

is the decomposition of \(X\) into orbits of the group \(H\). Construct the representations \(r_\alpha: H \to S(X_\alpha)\), \(r_\alpha(h)=h_\alpha=h|X_\alpha\), \(h\in H\), where \(h|X_\alpha\) is the restriction of \(h\) to \(X_\alpha\), \(\alpha\in I\). Then the representations \(r_\alpha\) are pairwise equivalent.

Proof. By virtue of a well-known theorem, (1) is a decomposition of \(X\) into systems of imprimitivity of the group \(G\). Let \(\alpha\) and \(\beta\) be arbitrary indices from \(I\). Then there is a substitution \(f_\beta\) in \(F\) such that \(f_\beta(X_\alpha)=X_\beta\). Put \(\varphi=f_\beta|X_\alpha\). For \(x\in X_\alpha\), \(h\in H\), one can write
\[ r_\beta(h)f_\beta(x)=h_\beta f_\beta(x)=hf_\beta(x)=f_\beta h(x)=f_\beta h_\alpha(x)=f_\beta r_\alpha(h)(x). \]
Hence \(r_\beta(h)\varphi=\varphi r_\alpha(h)\).

1. Theorem. Let \(X\) be an infinite set. Then a transitive subgroup \(G\) of the group \(SF(X)\) cannot be represented in the form of a product of two proper elementwise permutable normal divisors. In particular, \(G\) is indecomposable into a direct product.

Proof. Let a transitive subgroup \(G\) of the group \(SF(X)\) be representable in the form of the product \(HF=G\) of its proper elementwise permutable normal divisors \(H\) and \(F\). Then two cases are possible: 1) both groups \(H\) and \(F\) are transitive; 2) at least one of them is intransitive. We shall show that both cases lead to a contradiction. By virtue of Lemma 1 of the paper \((^1)\), in the first case \(H\) and \(F\) are regular groups. The latter contradicts the inclusion \(H \subset SF(X)\). Consider the second case. Let, for example, the group \(H\) be intransitive and (1) be the decomposition of the set \(X\) into orbits of the group \(H\). Then, according to the lemma, the representations \(r_\alpha\), \(\alpha\in I\), of the group-

the groups \(H\) are pairwise equivalent. Hence, and from the inclusion \(H \subset SF(X)\), it follows that the set \(I\) is finite.

Consequently, the orbits of the group \(H\) are infinite, and the group \(H\) itself is infinite. The same arguments are applicable also to the group \(F\); consequently, \(F\) is an infinite group. Since \(H\) is invariant in \(G\), (1) is a decomposition of \(X\) into systems of imprimitivity of the group \(G\).

Consider the homomorphism determined by this decomposition:
\[ \gamma: G \to S_k, \tag{2} \]
where \(S_k\) is the symmetric group of degree \(k\), and \(k\) is the cardinality of the set \(I\). Let \(N=\ker\gamma\), \(\Phi=F\cap N\). Then \(F:\Phi \leq k!\). Consequently, the group \(\Phi\) is infinite. Put \(H_\alpha=H|X_\alpha,\ \Phi_\alpha=\Phi|X_\alpha\); since the representations \(r_\alpha\) are pairwise equivalent, the group \(H_\alpha\) is infinite for every \(\alpha\in I\). From the finiteness of \(I\) and the infinitude of the group \(\Phi\) there follows the existence of such an \(\alpha\in I\) that the group \(\Phi_\alpha\) is infinite. Choose \(\alpha\) so that \(\Phi_\alpha\) is infinite.

Obviously, \(H_\alpha\) is a transitive subgroup of \(S(X_\alpha)\), and \(\Phi_\alpha\) is contained in the centralizer of the group \(H_\alpha\) in \(S(X_\alpha)\). If \(\Phi_\alpha\) is transitive, then we arrive at a contradiction as in the first case. Let \(\Phi_\alpha\) be intransitive and
\[ X_\alpha=\bigcup Y_\beta,\quad \beta\in B \tag{3} \]
be the decomposition of \(X_\alpha\) into orbits of the group \(\Phi_\alpha\). According to Lemma 2 of article \((^1)\), the representations \(\rho_\beta:\Phi_\alpha\to S(Y_\beta)\), where \(\rho_\beta(x)=x|Y_\beta\), are pairwise equivalent. Consequently, the group \(F_{\alpha\beta}=\Phi_\alpha|Y_\beta\) is infinite, and the cardinality of \(B\) is finite.

Put \(H_\alpha^\beta=\{h\mid h\in H_\alpha,\ h(Y_\beta)=Y_\beta\}\), \(H_{\alpha\beta}=H_\alpha^\beta|Y_\beta\).

It is easy to see that (3) is a decomposition of \(X_\alpha\) into systems of imprimitivity of the group \(H_\alpha\). Consequently, \(H_{\alpha\beta}\) is a transitive subgroup of \(S(Y_\beta)\); \(F_{\alpha\beta}\) is also a transitive subgroup of \(S(Y_\beta)\). The groups \(H_{\alpha\beta}\) and \(F_{\alpha\beta}\) commute elementwise. According to Lemma 1 of article \((^1)\), \(F_{\alpha\beta}\), \(H_{\alpha\beta}\) are regular groups. But, on the other hand, \(H_{\alpha\beta}\subset SF(Y_\beta)\), and in \(SF(Y_\beta)\) there are no regular groups. The theorem is proved.

1. From the theorem just proved, in particular, there again follows the assertion of article \((^1)\) that every transitive locally nilpotent subgroup of \(SF(X)\) for infinite \(X\) is a \(p\)-group, and a maximal transitive locally nilpotent subgroup of \(SF(X)\) is a Sylow \(p\)-subgroup of \(SF(X)\).

I. D. Ivanyuta in \((^2)\) proved that the transitive Sylow \(p\)-subgroups of \(SF(X)\), for countable \(X\), are pairwise conjugate in \(SF(X)\), while for uncountable \(X\) there are no transitive Sylow \(p\)-subgroups in \(SF(X)\). Hence, and from the preceding, it follows that the orbits of a locally nilpotent subgroup of the group \(SF(X)\) are either countable or finite. In \((^3)\) it is proved that, for finite \(X\), the maximal transitive nilpotent subgroups of \(S(X)\) are pairwise conjugate. Thus, up to conjugacy in \(SF(X)\), a maximal locally nilpotent subgroup of \(SF(X)\) is determined by its orbital type.

We note that the proposition of article \((^1)\) concerning \(ZA\)-groups needs clarification. The correct formulation should read as follows. Every \(ZA\)-group contained in \(SF(X)\) is a subdirect product of nilpotent groups.

The following problem suggests itself naturally. Let \(X\) be an infinite set, and let \(\mathfrak{G}\) be an abstract group. Under what conditions does there exist an injective homomorphism \(h:\mathfrak{G}\to SF(X)\) such that \(\operatorname{Im} h\) is a transitive group?

Institute of Mathematics
Academy of Sciences of the BSSR

Received
21 I 1969

CITED LITERATURE

\(^1\) D. A. Suprunenko, DAN, 167, 302 (1966).
\(^2\) I. D. Ivanyuta, Dissertation, Institute of Mathematics, Academy of Sciences of the Ukrainian SSR, 1964.
\(^3\) D. A. Suprunenko, DAN, 99, 23 (1954).