Subdirectly irreducible algebra of language F with cardinality $\geq 2^\kappa$

Question

If language $F$ has cardinality $\kappa$ ($\kappa$ is some infinite cardinal) and arity of every operation in F is 1, then doesn't exist subdirectly irreducible algebra of language F with cardinality $>2^\kappa,$ but exist subdirectly irreducible algebra of language F with cardinality $2^\kappa.$ How to prove it?

Answer 1

Let $\mathbf G = \langle G, +, - , 0 \rangle$ be a group with $|G| = \kappa$.
Define $\mathbf A = \langle A, (\alpha_g)_{g \in G}\rangle$, where $A = \{0,1\}^G$, and each $\alpha_g$ is a unary map on $A$ given by: for each $a \in G$ and each $f \in A$, $$\alpha_0(f)(a) = f(0),$$ and, for $x \neq 0$, $$\alpha_x(f)(a)=f(a+x).$$ Let $f_0,f_1:G\to\{0,1\}$, with $f_0(g)=0$ and $f_1(g)=1$, for all $g \in G$ (the constant maps).
Notice that $\alpha_0(f) = f_{f(0)}$, and so $\alpha_0(f)$ is either $f_0$ or $f_1$.

We will see that $\mathbf A$ is subdirectly irreducible with monolith $\mu=\Theta(f_0,f_1)$, the principal congruence generated by the pair $(f_0,f_1)$.
This is equivalent to show that if $\theta$ is a non-trivial congruence of $\mathbf A$ then $\mu \subseteq \theta$, that is $(f_0,f_1)\in\theta$.

So let $\theta$ be any non-trivial congruence of $\mathbf A$.
Thus, there exist $f,g \in A$ with $f \neq g$ and $(f,g) \in \theta$.
Since $f \neq g$, there exist $a \in G$ such that $f(a) \neq g(a)$;
suppose, without loss of generality, that $f(a)=0$ and $g(a)=1$.

Now, where $\equiv$ denotes congruence modulo $\theta$, we have $$f \equiv g \Longrightarrow \alpha_a(f) \equiv \alpha_a(g) \Longrightarrow \alpha_0(\alpha_a(f)) \equiv \alpha_0(\alpha_a(g)).$$ From $\alpha_a(f)(0)=f(a+0)=f(a)=0$, it follows that $\alpha_0(\alpha_a(f))=f_0$;
from $\alpha_a(g)(0)=g(a+0)=g(a)=1$, it follows that $\alpha_0(\alpha_a(g))=f_1$.

Hence $(f_0,f_1) \in \theta$, and therefore $\mu \subseteq \theta$.
Since $\theta$ was any non-trivial congruence, it follows that $\mu$ is the monolith of $\mathbf A$, and $\mathbf A$ is subdirectly irreducible.

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Now we prove that for $\mathbf S$ subdirectly irreducible, $|S| \leq 2^{\kappa}$.
If $\mathbf S$ is subdirectly irreducible, with monolith $\mu = \Theta(a,b)$, with $a,b \in S$ such that $a \neq b$, then for any pair of elements $c,d \in S$, if $c \neq d$ then there exists a term $t$ in the language of $F$ such that
\begin{equation}\label{equation} t(c)=a \quad\text{ and }\quad t(d)=b.\tag{1} \end{equation}

Let $T$ be the set of term operations in the language of $F$.
Since each such operation is built from the composition of finitely many basic operations and projections, and $F$ is infinite, it follows that $$|T| \leq \sum_{n\geq0}|F|^n = \sum_{n\geq0}|F|=|F|.$$ Now from \eqref{equation} the map $\Phi:S\to\mathcal P(T)$ given by $$x \mapsto \{t \in T: t(x)=a\}$$ is injective, whence $$|S| \leq 2^{|T|} = 2^{|F|} = 2^{\kappa}.$$