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A pseudovariety $\mathbf{V}$ of groups is join prime if for any pseudovarieties $\mathbf{V}_1, \mathbf{V}_2, \ldots,\mathbf{V}_m$, the implication $$\mathbf{V} \subseteq \mathbf{V}_1 \vee \mathbf{V}_2 \vee \cdots \vee \mathbf{V}_m \quad \Longrightarrow \quad \mathbf{V} \subseteq \mathbf{V}_i$$ holds for some $i$. A finite group is join prime if it generates a join prime pseudovariety.

It is known that all groups of order up to 7 are join prime. So it is natural to ask: is the dihedral group $D_4$ of order 8 join prime?

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Yes, the pseudovariety generated by $D_4$ is join prime (and the argument shows that the same is true for the pseudovariety generated by $8$-element quaternion group). The result follows from two observations:

(1) the class ${\mathbf P}$ of finite groups whose Sylow $2$-subgroups are abelian forms a pseudovariety (i.e., this class is closed under finite products, the formation of subgroups and the formation of quotients), and
(2) any pseudovariety not contained in ${\mathbf P}$ contains $D_4$ (and $Q_8$).

Assuming Items (1) and (2), and the obvious fact that $D_4\not\in {\mathbf P}$, we argue as follows: if ${\mathbf V}(D_4)\subseteq {\mathbf V}_1\vee \cdots \vee {\mathbf V}_m$, then by Item (1) there is some $i$ such that ${\mathbf V}_i\not\subseteq {\mathbf P}$. By Item (2), ${\mathbf V}_i$ contains $D_4$, so ${\mathbf V}(D_4)\subseteq {\mathbf V}_i$.

I explain how to prove Item (2). Assume ${\mathbf V}$ is a pseudovariety containing some group $G$ with a nonabelian Sylow $2$-subgroup. We may assume that $G$ is chosen with $|G|$ minimal, and that ${\mathbf V}={\mathbf V}(G)$. Necessarily $G$ is a nonabelian, subdirectly irreducible $2$-group with monolith $M = \langle z\rangle\subseteq Z(G)$ where $z^2=1$, and $G/M$ is abelian. In particular, $G$ is $2$-step nilpotent.

Since $G/M\models [x,y]\approx 1$ and $M\models x^2\approx 1$ we get that $G\models [x,y]^2\approx 1$.

It follows from commutator collection that the set of laws of any finite $2$-step nilpotent group may be axiomatized by: the group laws, the law $[[x,y],z]\approx 1$, an exponent bound $x^m\approx 1$, and an exponent bound on the commutator subgroup $[x,y]^n\approx 1$. Thus, if the exponent of our group $G$ is $2^r$, then since $G$ is nonabelian and satisfies $[x,y]^2\approx 1$ we get that ${\mathbf V}(G)$ is exactly the class of all finite, $2$-step nilpotent $2$-groups satisfying $[x,y]^2\approx 1$ and $x^{2^r}\approx 1$. If $2^r\geq 4$, this class contains $D_4$.

But we must have $2^r\geq 4$, since otherwise $2^r\mid 2$ and $G\models x^2\approx 1$. This can't happen since $G$ is nonabelian.

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