A. L. Shmelkin

THE SEMIGROUP OF VARIETIES OF GROUPS

(Presented by Academician P. S. Aleksandrov, 18 X 1962)

A variety is any class $\mathfrak M$ of groups closed under the operations of taking subgroups, homomorphic images, and complete direct products of groups from $\mathfrak M$. On the other hand (see (${}^1$)), a variety is the class of all groups satisfying some system of identities; therefore to each variety there corresponds a certain verbal subgroup (see (${}^2$)).

H. Neumann (${}^3$) introduced an associative operation on the set of all varieties: the product $\mathfrak A\mathfrak B$ of varieties $\mathfrak A$ and $\mathfrak B$ is the variety consisting of all possible extensions of groups from $\mathfrak A$ by means of groups from $\mathfrak B$. Moreover, if the varieties $\mathfrak A$ and $\mathfrak B$ correspond to the verbal subgroups $U$ and $V$, then to their product $\mathfrak A\mathfrak B$ there corresponds the verbal subgroup $U(V)$, generated by all possible words of the form $u(v_1,\ldots,v_n)$ with $u(x_1,\ldots,x_n)\in U$, $v_i\in V_i$, $i=1,\ldots,n$.

In what follows, by the semigroup of varieties we shall mean the set of all varieties, except the variety of all groups and the variety consisting of the one-element group, with the operation defined above.

In (${}^3$) it was proved that the right cancellation law holds in the semigroup of varieties, and that the indecomposable (with respect to product) varieties are generators of this semigroup. In the same work H. Neumann posed the question whether the semigroup of varieties is free. The present note is devoted to a positive solution of this problem.*

The main tool for solving the problem will be the operations of the discrete and complete wreath product of groups. The discrete (complete) wreath product $A \operatorname{wr} B$ ($A \operatorname{Wr} B$) is the split extension of the direct product

\[ \bar A=\prod_{b\in B}^{\times} A(b) \]

(the complete direct product

\[ \tilde A=\prod_{b\in B}^{\sim\times} A(b) \]

) by means of the group $B$, where $A(b)\cong A$, $b\in B$, such that $b_1^{-1}a(b)b_1=a(bb_1)$; $b,b_1\in B$.

Lemma 1. For any extension $G$ of a group $A$ by means of a group $B$ there exists a monomorphism $\varphi:G\to A\operatorname{Wr}B$, and the commutative diagram

\[ \begin{array}{ccccc} E & \to & A & \to & G & \to & B & \to & E\\ & & \downarrow \varphi_A & & \downarrow \varphi & & \downarrow \varepsilon_B & & \\ E & \to & \tilde A & \to & A\operatorname{Wr}B & \to & B & \to & E, \end{array} \]

holds, where $\varepsilon_B$ is the identity automorphism of the group $B$; $\varphi_A$ is the restriction of $\varphi$ to the subgroup $A$.

Proof (cf. (${}^{4,5}$)). In each coset of $G$ modulo $A$ we choose a representative $s_b$. Then $\varphi$ is defined as follows:

\[ g_1\varphi = b_1\prod_{b\in B}s_{bb_1}^{-1}g_1s_b(b), \]

where $b_1$ is the image of $g_1$ in $B$. The lemma is verified without difficulty.

Lemma 2. All identities of the group $A\operatorname{wr}B$ also hold in the group $A\operatorname{Wr}B$.

Proof. Indeed, suppose that an identity $v(x_1,\ldots,x_n)$ of the group $A\operatorname{wr}B$ fails for some $g_1,\ldots,g_n\in A\operatorname{Wr}B$, i.e. $v=v(g_1,\ldots,g_n)\ne1$. Obviously, $v\in \tilde A$; hence some, say the $b_0$-th, component of the element $v$ is not equal to 1, but in the process of reducing the word $v(g_1,\ldots,g_n)$ to canonical form in the sense of the group $A\operatorname{Wr}B$, in forming each component only

* Note added in proof. After the manuscript had been submitted for publication, the author learned of the work (${}^6$), in which the problem is also solved.

finite number of components of the elements \(g_1,\ldots,g_n\). In that case, if the elements \(g_i\) are replaced by \(g_i' \in A \operatorname{wr} B \subseteq A \operatorname{Wr} B\), obtained from \(g_i\) by replacing by \(1\) all components except the component from \(B\) and also those which take part in the formation of the \(b_0\)-th component of the element \(v\), then we obtain that
\[ v(g_1',\ldots,g_n') \ne 1, \]
since this element has the same \(b_0\)-th component as \(v\). The contradiction obtained proves the lemma.

Thus, now, in order to prove that any extension of \(A\) by means of \(B\) belongs to some multivariety, it suffices, as follows from Lemmas 1 and 2, to prove that \(A \operatorname{wr} B\) lies in this multivariety.

Lemma 3. In the semigroup of multivarieties the law of left cancellation holds, i.e. from \(\mathfrak A \mathfrak B_1=\mathfrak A \mathfrak B_2\) it follows that \(\mathfrak B_1=\mathfrak B_2\).

Proof. Let the verbal subgroups \(U,V_1,V_2\) correspond to the multivarieties \(\mathfrak A,\mathfrak B_1,\mathfrak B_2\). Take the group
\[ F=A \operatorname{wr} B, \]
where \(A\) is the \(\mathfrak A\)-free group of countable rank with free generators \(a_1,a_2,\ldots\),
\[ B=\prod_{i=0}^{\infty}{}^{*} B_i, \]
and each group \(B_i\) is a \(\mathfrak B_1\)-free group of countable rank with free generators \(b_{i1},b_{i2},\ldots\). Take \(V_2(F)\) \(\bigl(V_2(F)\in\mathfrak A\), since \(U(V_2(F))=U(V_1(F))=E\bigr)\), and prove that \(V_2(F)\subseteq \bar A\). Suppose, on the contrary, that there is an element \(b\cdot \bar a\in V_2(F)\), \(b\ne1\). Suppose, for definiteness, that the projection \(b_0\) of the element \(b\) onto \(B_0\) is not equal to \(1\). It has the form \(b_0=f(b_{01},\ldots,b_{0n})\); obviously, \(b_0\in V_2(F)\) and in \(V_2(F)\) there are also contained the nonidentity elements
\[ b_i=f(b_{i1},\ldots,b_{in}). \]
The commutators contained in \(V_2(F)\),
\[ k_i=[b_0,a_i(1)]=a_i(1)\cdot a_i^{-1}(b_0), \]
obviously freely generate an \(\mathfrak A\)-free group
\[ K=\{k_1,k_2,\ldots\}. \]
Take in \(V_2(F)\) the subgroup \(\{K,B'\}\), where
\[ B'=\{b_1,b_2,\ldots\}=\prod_{i=1}^{\infty}{}^{*}\{b_i\}, \]
and prove that
\[ \{K,B'\}\cong K \operatorname{wr} B'. \]
Indeed, denote by
\[ K(1)=K,\qquad K(b')=b'^{-1}Kb',\quad b'\in B'. \]
It is checked directly that
\[ \{K(b'),\, b'\in B'\}=\prod_{b'\in B'}{}^{*} K(b'), \]
and if \(k(b')\in K(b')\), then
\[ b''^{-1}k(b')b''=k(b'b'') \]
for \(b''\in B'\), i.e.
\[ \{K,B'\}\cong K \operatorname{wr} B'. \]
Thus \(K\operatorname{wr}B'\in\mathfrak A\), since this group is embedded in \(V_2(F)\in\mathfrak A\). Hence, according to Lemmas 1 and 2, \(\mathfrak A\) contains every extension of \(K\) by means of \(B'\). But \(K\) is a free group of countable rank of the multivariety \(\mathfrak A\), and \(B'\) is also a free group of some nonunit multivariety \(\mathfrak B\) of abelian groups; therefore \(\mathfrak A\) contains the free group of countable rank of the multivariety \(\mathfrak A\mathfrak B\). Therefore \(\mathfrak A=\mathfrak A\mathfrak B\), but then \(\mathfrak A=\mathfrak A\mathfrak B^s\) for any \(s\). This is impossible, since (see (3)) the number of factors in a decomposition of the multivariety \(\mathfrak A\) cannot exceed the minimal length of the elements of \(U\). The lemma is proved.

Lemma 4. If \(\mathfrak A_1\mathfrak B_1=\mathfrak A_2\mathfrak B_2\), then either \(\mathfrak A_1\supseteq\mathfrak A_2,\ \mathfrak B_1\subseteq\mathfrak B_2\), or \(\mathfrak A_1\subseteq\mathfrak A_2,\ \mathfrak B_1\supseteq\mathfrak B_2\).

Proof. Denote by \(U_1,U_2,V_1,V_2\) the verbal subgroups corresponding to the multivarieties \(\mathfrak A_1,\mathfrak A_2,\mathfrak B_1,\mathfrak B_2\), respectively. Suppose \(\mathfrak B_1\nsubseteq\mathfrak B_2\). Take the group
\[ F=A\operatorname{wr}B, \]
where \(A\) is the \(\mathfrak A_1\)-free group of countable rank with free generators \(a_1,a_2,\ldots\), and \(B\) is the \(\mathfrak B_1\)-free group of countable rank. By assumption, \(V_2(B)\ne E\); let \(b\in V_2(B)\), \(b\ne1\). Then \(V_2(F)\) contains the elements
\[ k_i=[b,a_i(1)]=a_i(1)a_i^{-1}(b), \]
freely generating an \(\mathfrak A_1\)-free group \(K\). Since
\[ U_2(V_2(F))=E, \]
we have
\[ U_2(K)=E \]
and \(\mathfrak A_1\subseteq\mathfrak A_2\). If, in addition, \(\mathfrak B_2\nsubseteq\mathfrak B_1\), then in exactly the same way we show that \(\mathfrak A_2\subseteq\mathfrak A_1\) and, using the law of left cancellation, find \(\mathfrak B_1=\mathfrak B_2\). Therefore \(\mathfrak B_2\subseteq\mathfrak B_1\), as was required. There remains the case \(\mathfrak B_1=\mathfrak B_2\), but here we may use H. Neumann’s law of right cancellation, already proved, and obtain \(\mathfrak A_1=\mathfrak A_2\). The lemma is proved.

Theorem. The semigroup of multivarieties is a free semigroup with the indecomposable multivarieties as the system of free generators.

Proof. Since H. Neumann showed that every

a variety is represented as a product of indecomposable ones, it remains to prove the uniqueness of such a representation. Let \(\mathfrak P=\mathfrak A_1\mathfrak B_1=\mathfrak A_2\mathfrak B_2\), where \(\mathfrak A_1\) and \(\mathfrak A_2\) are indecomposable. We shall prove that \(\mathfrak A_1=\mathfrak A_2\). Suppose that \(\mathfrak A_1\ne \mathfrak A_2\); then also \(\mathfrak B_1\ne \mathfrak B_2\). By Lemma 4 the factors are connected by certain inclusions; let, for example, \(\mathfrak A_1\subset \mathfrak A_2\), \(\mathfrak B_1\supset \mathfrak B_2\). Take the group \(F=A\operatorname{Wr}B\), where \(A\) is an \(\mathfrak A_1\)-free group and \(B\) is a \(\mathfrak B_1\)-free group of countable rank. By Lemma 1, \(F\) contains a \(\mathfrak P\)-free group \(\Phi\) of countable rank. The group \(V_2(\Phi)\) is an \(\mathfrak A_2\)-free group of countable rank, and \(V_2(\Phi)\supset V_1(\Phi)\) (since \(\mathfrak B_1\supset \mathfrak B_2\)); therefore \(V_2(\Phi)\) is an extension of the group \(V_1(\Phi)\in\mathfrak A_1\) by means of \(V_2(\Phi)/V_1(\Phi)\cong V_2(B)\ne E\) (the isomorphism follows from Lemma 1); denote \(V_2(B)=G\). Let \(\mathfrak M=\mathfrak M(G)\) be the variety generated by the group \(G\), i.e. the class of all groups in which all identical relations holding in \(G\) are satisfied. We shall show that any \(\mathfrak M\)-free group is embedded in the complete direct product of a certain number of copies of the group \(G\). Indeed, let \(X\) be a set of arbitrary cardinality. Take the set of all single-valued mappings \(\varphi_i:X\to G\), \(i\in I\), and the group
\[ S=\widetilde{\prod_{i\in I}^{X}}G_i,\qquad G_i\cong G. \]
Then, if by \(\varphi\) we denote the mapping
\[ x\varphi=\prod_{i\in I}x\varphi_i\in S,\qquad x\in X, \]
then, obviously, \(\varphi\) is one-to-one, and \(X\varphi\) freely generates an \(\mathfrak M\)-free group. Now take the group \(F'=A'\operatorname{wr}B'\), where \(A'\) is an \(\mathfrak A_1\)-free group of arbitrary rank with free generators \(a_j,\ j\in J\),
\[ B'=B\times \widetilde{\prod_{i\in I}^{X}}B_i,\qquad B_i\cong B. \]
Obviously,
\[ V_2(B')=V_2(B)\times \widetilde{\prod_{i\in I}^{X}}V_2(B_i). \]
We may assume that \(V_2(B_i)=G_i\). Take in \(G=V_2(B)\) an element \(g\ne 1\). In \(V_2(F')\) lie the commutators \(k_j=[g,a_j(1)]\), \(j\in J\), which freely generate a group \(K\cong A'\). Just as in the proof of Lemma 3, it is easy to show that
\[ \{K,S\}\cong K\operatorname{wr}S,\qquad \text{where } S=\widetilde{\prod_{i\in I}^{X}}G_i. \]
Take in \(S\) an \(\mathfrak M\)-free subgroup \(M\), generated by the set \(X\varphi\); obviously, \(\{K,M\}\cong K\operatorname{wr}M\). The latter group is embedded in \(V_2(F')\in\mathfrak A_2\), and hence itself lies in \(\mathfrak A_2\). Further, since for any normal divisor \(N\) of the group \(K\) one has
\[ (K\operatorname{wr}M)/\bar N\cong (K/N)\operatorname{wr}M, \]
where \(\bar N\) is the normal divisor generated by \(N\) in \(K\operatorname{wr}M\), it follows, by the closure of varieties with respect to homomorphisms, that (using Lemmas 1 and 2) \(\mathfrak A_2\) contains all extensions of any group \(H\in\mathfrak A_1\) by means of any \(\mathfrak M\)-free group \(M\). Let \(R=M\psi\) be an arbitrary group from \(\mathfrak M\). Take a split extension \(\Gamma\) of the group
\[ \bar H=\widetilde{\prod_{r\in R}^{X}}H(r) \]
by means of \(M\), where \(H(r)\cong H\), in which the elements of \(M\) induce the following automorphisms in \(\bar H\):
\[ m^{-1}h(r)m=h(r\cdot m\psi). \]
Obviously, \([\operatorname{Ker}\psi,\bar H]=E\) and
\[ \Gamma/\operatorname{Ker}\psi\cong H\operatorname{wr}R. \]
Thus, finally, we obtain that \(\mathfrak A_2\) contains any extension of any group \(H\in\mathfrak A_1\) by means of any group \(R\in\mathfrak M\), i.e. \(\mathfrak A_1\mathfrak M\subseteq\mathfrak A_2\). But, on the other hand, as was shown above, an \(\mathfrak A_2\)-free group of countable rank can be represented as an extension of \(V_1(\Phi)\in\mathfrak A_1\) by means of \(G\in\mathfrak M\), and therefore \(\mathfrak A_1\mathfrak M=\mathfrak A_2\), and the variety \(\mathfrak A_2\) is decomposable, contrary to the supposition. Hence \(\mathfrak A_1=\mathfrak A_2\) and (by Lemma 3) \(\mathfrak B_1=\mathfrak B_2\). The assertion of the theorem is evident.

Moscow State University
named after M. V. Lomonosov

Received
9 X 1962

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