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Let $\mathcal{C}$ be a collection of (isomorphism classes of) finite groups with the following properties:

  1. If $G\in\mathcal{C}$ and $H$ is a homomorphic image of $G$, then $H\in\mathcal{C}$
  2. If $G\in\mathcal{C}$ and $H\subseteq G$, then $H\in\mathcal{C}$
  3. If $G,H\in\mathcal{C}$, then $G\times H\in\mathcal{C}$

$\mathcal{C}=\text{collection of all finite solvable groups}$ is such a collection. Let's call these collections closed.

Let $\mathcal{B}\subseteq\mathcal{C}$ be called a basis if:

  1. Every $G\in\mathcal{C}$, $G\cong\prod_i E_i$ for some finite subset $\{E_i\}\subseteq\mathcal{B}$ (where $\prod$ refers to the external direct product).

  2. Every $E$ in $\mathcal{B}$ is indecomposable; it cannot be written as a direct product of proper subgroups.

Example: The closed collection formed by the finite abelian groups have a basis, namely the cyclic groups of prime-power order.

Question: Which closed collections have a basis?

Edit: @BenjaminSteinberg pointed out in the comments that one could simply take the indecomposable groups in $\mathcal{C}$ as a basis. He also noted that what I called "closed collections" are better known as "pseudovarieties".

Question: What are some concrete examples of pseudovarieties whose bases have an "interesting" or "nice" description? I know I'm being vague here, but I think this question is interesting anyway.

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  • $\begingroup$ These are called pseudovarieties of groups $\endgroup$
    – Benjamin Steinberg
    Commented Oct 17, 2023 at 0:58
  • $\begingroup$ If you have a pseudovariety of nilpotent groups you can take all the indecomposable p-groups (for some prime p) in the pseudovariety as a base. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 17, 2023 at 1:00
  • $\begingroup$ I guess by Remak’s theorem you can always take the indecomposable groups in your pseudovariety as a base (see en.wikipedia.org/wiki/Krull–Schmidt_theorem) $\endgroup$
    – Benjamin Steinberg
    Commented Oct 17, 2023 at 1:03
  • $\begingroup$ Oh yes, my question was too naive; it does make sense to take all the indecomposable groups as a base. I was hoping to see some more examples of pseudovarieties whose bases have a "nice" description (like in the case of finite nilpotent groups and finite abelian groups) $\endgroup$
    – semisimpleton
    Commented Oct 17, 2023 at 1:06
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    $\begingroup$ @JoshuaZ, the indecomposable groups in the pseudovariety form the unique minimal basis. They must be included because they cannot be decomposed as a direct product and the Remak theorem shows they suffice $\endgroup$
    – Benjamin Steinberg
    Commented Oct 17, 2023 at 1:28

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