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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 $X/\lambda_K$.

Tim Lee
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  • Just a couple comments: $\mathsf{SP}(\mathcal{K})$ contains every subgroup of every product of $\mathbf{Z}_2$ and $\mathbf{Z}_3$. In particular, it contains $\mathbf{Z}_2,\mathbf{Z}_2^2,\mathbf{Z}_2^3,\ldots$. – Eran Dec 19 '14 at 02:08
  • Also, $\lambda_\mathcal{K}$ cannot be isomorphic to $\mathbf{Z}_5$ since the first is a congruence and the second is an algebra. – Eran Dec 19 '14 at 02:10
  • Thank you for your corrections. Is it possible to find $\lambda_K$, $U/\lambda_K$, and $X/\lambda_K$ explicitly in the above example? – Tim Lee Dec 19 '14 at 06:46
  • Yes, it is. Just let $\mathbf{U}=\mathbf{Z}_{30}$ and use the definitions. – Eran Dec 19 '14 at 19:38
  • Is it correct that $\lambda_K$ has the following form $\lambda_K={([i],[j])| i-j \mbox{ is divisible by } 5}$, where [] indicates an integer modulo 30? So, $U/\lambda_K=Z_6$? – Tim Lee Dec 20 '14 at 12:04
  • How many congruence classes does your $\lambda_K$ have? What's the definition of $\mathbf{U}/\lambda_K$? – Eran Dec 20 '14 at 21:08
  • I think $\lambda_K$ has five congruence classes to make $U/\lambda_K$ isomorphic to $Z_6$. Can you show me what $\lambda_K$ looks like? – Tim Lee Dec 21 '14 at 11:59

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