The following theorem is found in the book "Universal Algebra, Fundamentals and Selected Topics" by Clifford Bergman (pp.98).
Theorem 4.28. Let $U$ be free for $K$ over $X$. Then, $U/\lambda_k$ is free in $SP(K)$ over $X/\lambda_K$.
The associated definition is as follows. (Superscripts are omitted in Theorem 4.28.)
Definition 4.26 Let $K$ be a class of algebras and A an algebra of the same type. We define
$\Lambda_K^A=\{\theta \in Con A: A/\theta \in S(K)\}$,
$\lambda_K^A=\wedge\Lambda_K^A$.
Now, my question is described as follows:
Let $Z_n$ denote the cyclic group of order $n$. The group $Z_{30}$ is free for $K=\{Z_2,Z_3\}$ over $\{1\}$. Then, the group $Z_{30}$ is free for $A_6=HSP(K)$, which happens to be the variety of Abelian groups satisfying the identity $6x \approx 0$.
To construct an algebra free in $A_6$, I need to use Theorem 4.28, where $U=Z_{30}$, $K=\{Z_2,Z_3\}$, and $X=\{1\}$.
The textbook says that $Z_6$ is free in $A_6$. I think $SP(K)=\{\{e\}, Z_2, Z_3, Z_6\}$, but I am having hard time finding $U/\lambda_K$ and $X/\lambda_K$.
It seems like $\lambda_K$ is isomorphic to $Z_5$, since $Z_{30}/Z_{5}$ is isomorphic to $Z_6$.
I will appreciate if someone helps me find $U/\lambda_K$ and $X/\lambda_K$.