Example of subdirectly irreducible algebra that contains a proper sub-algebra

Question

I read that an algebra $\textbf{A}$ is subdirectly irreducible if whenever $\textbf{A}$ is isomorphic to a sub-algebra of a direct product $\Pi_{i \in I}\textbf{C}_i$, it has to be the case that $\textbf{A}$ is isomorphic to $\textbf{C}_j$ for some $j \in I$.

I would like to ask, is it possible for there to be a subdirectly irreducible algebra that contains proper sub-algebras ? Is there any example in the case of groups ?

Answer 1

Your definition of subdirectly irreducible is not quite right: the correct conclusion is that $\mathbf{A}$ is isomorphic to a subalgebra of $\mathbf{C}_j$ for some $j\in I$, and that the isomorphism is induced by the embedding into the direct product.

In fact, the usual definition is the following (e.g., Jacobson's Basic Algebra II):

Definition. An $\Omega$-algebra $\mathbf{A}$ is subdirectly irreducible if and only if the intersection of all nontrivial congruences of $\mathbf{A}$ is nontrivial.

For instance, in the case of groups this would become:

Definition. A group $G$ is subdirectly irreducible if and only if it contains a minimum nontrivial normal subgroup $N$ (i.e., a subgroup $N$ such that $N\triangleleft G$ and if $K\triangleleft G$, then either $K=\{e\}$ or $N\leq K$).

And for rings,

Definition. A ring $R$ is subdirectly irreducible if and only if it contains a a minimum nontrivial ideal.

And for modules,

Definition. An $R$-module $M$ is subdirectly irreducible if and only if the intersection of all nontrivial submodules of $M$ is nontrivial.

In the case of groups, for instance, we would have:

Theorem. A group $G$ is subdirectly irreducible if and only if for every $f\colon G\to \prod_{i\in I}G_i$, if $f$ is one-to-one then there exists $j\in I$ such that $\pi_j\circ f\colon G\to C_j$ is one-to-one. In particular, $G$ is isomorphic to a subalgebra of $C_j$.

Proof. Assume first that $G$ has a minimal nontrivial normal subgroup $N$, and that $f\colon G\hookrightarrow \prod_{i\in I}G_i$ is an embedding into a direct product. Consider the maps $\pi_i\circ f\colon G\to G_i$. The kernel of $\pi_i\circ f$ is either trivial or contains $N$. Since $f$ is one-to-one, the intersection of all kernels must be trivial. Thus, at least one kernel $\mathrm{ker}(\pi_j\circ f)$ does not contain $N$, which means that it is trivial. Hence, $\pi_j\circ f$ is an embedding and $G$ is isomorphic to a subgroup of $G_j$.

Conversely, the family of maps $\pi_K\colon G\to G/K$ as $K$ ranges over all nontrivial normal subgroups of $G$ induces a morphism $f\colon G\to \prod_{\{e\}\neq K\triangleleft G} G/K$. The kernel of this map is the intersection of all nontrivial normal subgroups of $G$. Since none of the maps $\pi_K\colon G\to G/K$ are embeddings, the morphism $f$ is not an embedding, hence the intersection of all nontrivial normal subgroups of $G$ is nontrivial, proving that $G$ has a minimum nontrivial normal subgroup, as claimed. $\Box$

Examples.

1. The groups $\mathbf{C}_{p^n}$, $p$ a prime, $n\geq 1$, are subdirectly irreducible. When $n\gt 1$, they all have proper nontrivial subgroups.

2. The Prüfer $p$-group $\mathbf{C}_{p^{\infty}}$ is subdirectly irreducible.

3. The groups $S_n$, $n\geq 2$, are subdirectly irreducible. For $n\neq 4$, the minimum normal subgroup is $A_n$; for $n=4$ it is the Klein $4$-subgroup $\{e, (12)(34), (13)(24), (14)(23)\}$.

4. Fields are subdirectly irreducible; any non-prime field contains a proper subfield.