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?



Join prime pseudovarieties
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.






