Group such that factors in any product-decomposition are reducible

Question

Motivation. Let us call a group $G = (G,\cdot)$ (product-)reducible if there are groups $H_1, H_2$, each having more than $1$ element, with $G \cong H_1\times H_2$. Otherwise, $G$ is said to be irreducible. If $G$ is finite, then $G$ is isomorphic to a product of irreducible groups. This is because in the finite case, proper subgroups of a group have strictly smaller cardinality, so you can't have infinitely descending chains of proper subgroups. - In the infinite case, things may look different.

Infinite case. Is there an infinite group $G$ such that whenever $\kappa \geq 0$ is a cardinal and $G_\alpha$ are non-trivial groups for $\alpha\in\kappa$ such that $G \cong \prod_{\alpha\in\kappa}G_\alpha$, then for all $\alpha\in\kappa$, the group $G_\alpha$ is reducible?

Answer 1

The question is: is there an infinite group whose nontrivial direct factors are all directly decomposable? The answer is 'Yes'. First I will explain why the answer is 'Yes' to the analogous question for Boolean algebras (BAs).

Claim 1. The following are equivalent for a nontrivial BA $B$:

1. $B$ has an indecomposable direct factor.
2. $B$ has a direct factor of size $2$.
3. $B$ has an atom.

Reasoning. For the equivalence of (1) and (2), a nontrivial BA is indecomposable if and only if it has size $2$. (For any BA $C$ of more then $2$ elements, if $a\in C-\{0,1\}$, then the complementary ideals $(a)$ and $(\neg a)$ are factor ideals for a nontrivial direct decomposition.)

For (2)$\Rightarrow$(3), if $B$ has a direct factor of size $2$, then we may assume that $B=C\times D$ where $|C|=2$, and when this happens $a=(1,0)\in C\times D = B$ is an atom. For (3)$\Rightarrow$(2), if $a\in B$ is an atom, then the complementary ideals $(a)$ and $(\neg a)$ are factor ideals for a direct decomposition, where $B/(\neg a)$ has size $2$. \\\

This is enough to show that any atomless BA has the property that no nontrivial direct factor is directly indecomposable. One can repackage this so that it is a statement about groups.

Claim 2. If $A$ is a finite, simple, nonabelian group and $B$ is an atomless BA, then the group that is the Boolean power $A[B]^*$ has the property that no nontrivial direct factor is directly indecomposable.

[The Boolean power $A[B]^*$ is the subalgebra of the direct power $A^{B^*}$ (where $B^*$ is the Stone space of $B$) whose elements $f\in A^{B^*}$ are the continuous functions $f\colon B^*\to A$ where $A$ is equipped with the discrete topology.]

Reasoning. The Maurer-Rhodes Theorem yields that if $A$ is a finite, simple, nonabelian group and $A_A$ is the expansion of $A$ by all constants, then $A_A$ is a primal algebra. Saying that $A_A$ is primal means that it is finite and every finitary operation $g\colon (A_A)^n\to A_A$ is a term operation of $A_A$. The work of T. K. Hu shows that any variety generated by a primal algebra $P$ is categorically equivalent to the variety of Boolean algebras. The equivalence is the composition of two dualities: Stone duality, $B\mapsto B^*$, composed with the Boolean power functor $B^*\mapsto P[B]^*$. In particular, this equivalence guarantees that if $B$ is any BA and $P$ is any primal algebra, then $B$ and $P[B]^*$ have isomorphic congruence lattices, and under this isomorphism the factor congruences will correspond. Thus, if $B$ is atomless and $P$ is the primal algebra $A_A$, then $B$ and $(A_A)[B]^*$ have isomorphic congruence lattices and corresponding direct decompositions.

Now pass from $A_A$ to $A$ by dropping the constants that were added. This has some effect: the reduct $A$ of $A_A$ is no longer primal, the algebra $A[B]^*$ has richer subalgebra structure than $(A_A)[B]^*$, etc. But the congruences of $(A_A)[B]^*$ and $A[B]^*$ are the same, and the factor congruences are the same. This is enough to establish the claim. \\\

Answer 2

$\DeclareMathOperator\supp{supp}$Let $X$ be a Cantor space, and let $S$ be a nonabelian simple group (or more generally a nontrivial group in which every nontrivial conjugacy class has a trivial centralizer, e.g., the symmetric group $S_n$ for $n\ge 5$). Let $G=C(X,S)$ be the group of continuous functions $C\to S$ ($S$ being viewed as discrete group).

(Note that if $S$ is countable, so is $G$.)

I claim that every nontrivial direct factor of $G$ is isomorphic to $G$ (and hence decomposable). To see this, I check that every direct product decomposition $G=A\times B$ is induced by a finite clopen partition of the Cantor set, i.e., for some clopen partition $X=Y\sqcup Z$, one has $A=C(Y,S)$ and $B=C(Z,S)$. (Thus, $G$ is nontrivial, decomposable, and every nontrivial direct factor of $G$ is isomorphic to $G$ and hence decomposable as well.)

Lemma (immediate — this is where the assumption on $S$ is used): for $g\in G_S$, the centralizer of the conjugacy class $c_g$ of $g$ is the set of $h$ with support disjoint from $g$.

Consider the supports of elements of $A\cup B$. They form a clopen covering of $K$. By compactness, one can extract a finite covering. Noting that supports of elements of $A$ and $B$ are disjoint (by the lemma), we obtain a finite covering of $X$ of the form $X=\bigcup_{i\in I}\supp(a_i)\sqcup \bigcup_{j\in J}\supp(b_j)$, with $a_i\in A$, $b_j\in B$, and $I$, $J$ finite. Write $Y=\bigcup_{i\in I}\supp(a_i)$, $Z=\bigcup_{j\in J}\supp(b_j)$. If $b\in B$, then for every $i$, $b$ centralizes $c_{a_i}$, and hence, by the lemma, $\supp(b)\cap\supp(a_i)$ is empty. Since this holds for all $i$, $\supp(b)\subseteq Z$. Similarly $\supp(a)\subseteq Y$ for all $a\in A$. Hence $A\subseteq C(Y,S)$ and $B\subseteq C(Z,S)$. Since $G=A\times B$, these have to be equalities: $A=C(Y,S)$ and $B= C(Z,S)$.

Answer 3

There are also abelian examples (these are called superdecomposable abelian groups).

For example, by Theorem 5.1 of

Corner, A. L. S., Every countable reduced torsion-free ring is an endomorphism ring, Proc. Lond. Math. Soc., III. Ser. 13, 687-710 (1963). ZBL0116.02403.

there is a countable torsion-free example.