A class of groups $\mathcal{V}$ is called a variety if there exists a set of words $W$ such that $G$ belongs to $\mathcal{V}$ if and only if $w(g_1,g_2,\ldots)=1$ for all $w\in W$ and $g_1,g_2,\ldots\in G$. In Robinson's book ("A course in the theory of groups") the following theorem of Birkhoff (2.3.4 and 2.3.5) is stated and proved:
A class $\mathcal{V}$ of groups is a variety if and only if $\mathcal{V}$ is closed under taking quotients and subdirect products.
Let $\mathcal{V}$ be the smallest variety containing the symmetric group $S_3$. By Birkhoff's theorem, the members of $\mathcal{V}$ can be constructed iteratively from $S_3$ by taking quotients and subdirect products. However, I cannot imagine how the subgroup $C_3\le S_3$ arises in this way (it must be in $\mathcal{V}$ by the definition of varieties). Any ideas?
Let $D$ be an infinite countable set. Start from the unrestricted direct power $S_3^D=C_3^D\rtimes C_2^D$. Let $C_2^{(D)}$ be the restricted direct power (=direct sum). Let $G$ be the semidirect product $C_3^D\rtimes C_2^{(D)}$; this is a subdirect product. Write $C_3^{)D(}$ for the near direct power $C_3^D/C_3^{(D)}$. Then $G$ admits the quotient $H=C_3^{)D(}\rtimes C_2^{(D)}$. But one readily sees that the action of $C_2^{(D)}$ on $C_3^{)D(}$ is trivial. So $H$ is the direct product $H=C_3^{)D(}\times C_2^{(D)}$, which admits $C_3$ as quotient.
However, one can't obtain $C_3$ from $S_3$ when considering only quotients and finite subdirect products. Indeed, for a finite group $K$ in the variety generated by $S_3$, one has the equivalence between (a) the abelianization of $K$ is a 2-group and (b) the center of $K$ is a 2-group, or (b') the centralizer of every element of order 3 is a proper subgroup. Then (a) passes to quotients, and (b') passes to subdirect products.