If $F$ is a finite group, then any finitely generated element of ${\rm Var}(F)$ is finite This is Robinson "A Course in the Theory of Groups" Ex. 2.3.10
Let $F$ be a finite group.
Notation: $(F)$ denotes the class of all groups either trivial or isomorphic to $F$.
${\rm Var}(F)$ is the intersection of all varieties containing $(F).$
Show, if $G$ is a finitely generated member of ${\rm Var}(F),$ then $G$ is itself finite.

I believe it's true that $G$ must be an image of a subgroup of a Cartesian product of $(F)$-groups.  So w.l.o.g., $G$ is an image of a subgroup of a Cartesian product of copies of $F$ itself.  Is this correct?
First, as a simpler case, suppose $G'$ is a f.g. cartesian product, ${\rm Cp}_{\lambda\in\Lambda}F_\lambda$ of copies of (finite) $F$.  Suppose $X = \{X_1...X_n\}$ is finite and $G' = \langle X\rangle$.  It's clear that if two components $\lambda$ and $\mu$ are such that $X_{i,\lambda} = X_{i,\mu}$ for all $i\in\{1..n\}$ then those components are essentially identical, and could be collapsed into one.  But $X$ and $F$ are finite, therefore only a finite number of components can be different, and can't be collapsed.  So w.l.o.g. we can assume $\Lambda$ is finite, and therefore $G'$ is finite.
However, I don't know how to extend this argument (if it can be extended) to subgroups, let alone images of subgroups.  Any help appreciated.
 A: Your first question: yes, for any class of groups $\mathcal{X}$, Birkhoff's HSP Theorem says that $\mathrm{Var}(\mathcal{X})=\mathbf{HSP}(\mathcal{X})$, that is, it consists of groups that are homomorphic images of subgroups of cartesian products of groups in $\mathcal{X}$. When $\mathcal{X}=\{F\}$ a single group, then you get homomorphic images of subgroups of cartesian powers of $F$.
You are essentially done. You've proven that a finitely generated subgroup of a direct (cartesian) power of $F$ is finite. All that remains is to show that a finitely generated subgroup of a quotient of such a subgroup is also finite.
Let $G\leq \mathop{\mathrm{Cp}}_{\lambda\in\Lambda}F$ be a subgroup of a cartesian power, let $N\triangleleft G$, and let $K=G/N$. Let $x_1N,\ldots,x_kN\in K$. We want to show that $A=\langle x_1N,\ldots,x_kN\rangle$ is finite.
Let $\mathfrak{K}=\langle x_1,\ldots,x_k\rangle$ for any particular choice of coset representatives of the generators of $K$. Note that $\mathfrak{K}N$ is the subgroup of $G$ corresponding to $A$, so that $A\cong \mathfrak{K}N/N$. By the Second Isomorphism Theorem (or Third, depending how you number them), $\mathfrak{K}N/N\cong \mathfrak{K}/\mathfrak{K}\cap N$. Because $\mathfrak{K}$ is a finitely generated subgroup of $G$, it is a finitely generated subgroup of the cartesian power of $F$, and hence is finite by what you've already proven. Hence, $\mathfrak{K}/\mathfrak{K}\cap N$ is a quotient of a finite group, hence is itself finite. Thus, $A$ is finite, which is what we wanted to prove. $\Box$
