congruence-relationsuniversal-algebra
Let $2=\{0,1\}$, $+$ denote addition modulo $2$, $g\left(x\right)=x+1$. Consider the following Algebra $A=\langle 2,+,g\rangle$. I want to calculate congruence relation on $A\times A$. Can someone explain how to do it!
Congruence relation $\theta$ is an equivalence relation on algebra together with property that if $x_i\theta y_i$ for $i=1,2,3..,n$ then $f\left(x_1,x_2,…,x_n\right)\theta f\left(y_1,y_2,…,y_n\right)$ for all $f$ $n$-ary operations on algebra.
I think you're right and you've found a confusion in the book.
The usual definition of a congruence on a relational structure would not have the condition (RP*) that you quoted but rather (RP#): If $a_1\equiv b_1,\dots,a_n\equiv b_n$ and $R_i(a_1,\dots,a_n)$ then $R_1(b_1,\dots,b_n)$. (Here $n$ is the number of argument places of $R_i$.) This (RP#) would indeed have the claimed, undesirable consequence if equality were one of the relations $R_i$. Indeed, using (RP#) with equality as $R_i$, we'd be able to infer from $x\equiv y$ (taking $a_1,b_1,a_2$ all to be $x$ and taking $b_2$ to be $y$) that $x=y$. So the equivalence relation $\equiv$ could only be equality.
The book's unusual requirement (RP*) seems to be designed specifically for the situation where each $R_i$ (now having $n+1$ argument places) is intended to represent an $n$-place function. As far as I can see, it does not cause any problem when the equality relation is among the relations $R_i$.
Also, "intended to represent an $n$-place function" might explain the use of the word "algebra". It looks to me as if the authors were sometimes thinking of algebras and sometimes of relational structures, and the two topics got mixed together confusingly.
That is not correct.
Consider a chain with $3$ (or more) elements, say
$$0 \prec a \prec 1.$$
There, you have the congruence $\theta = \Theta(a,1) = \{(0,0),(a,a),(a,1),(1,a),(1,1)\}$, and $0$ is not related either with $a$ or $1$, but $0 \leq a$ and $a \leq 1$, so your conjecture would yield $(0,a) \in \theta$.
Regarding your main question, you seem to seek a concept of congruence on a poset that would restrict to a lattice congruence if that poset happens to be a lattice.
Again, that doesn't exist.
Take for example, as a lattice $\mathbf L$ the one whose Hasse diagram is depicted next:
Its congruence lattice is isomorphic to the lattice itself (easy to check).
Now, drop one of the operations (either $\wedge$ or $\vee$, it doesn't matter) and you get a semi-lattice with the same Hasse diagram (so, the same poset), but its congruence lattice is the following one:
Hence the congruence lattice really depends on the operations.
More generally, if $\mathbf L$ is any lattice, then the congruence lattice of $\mathbf L$ is distributive.
But if $\mathbf L'$ is the semi-lattice that arises from dropping one of the lattice operations, then the congruence lattice of the semi-lattice $\mathbf L'$ is distributive iff $\mathbf L$ is a chain.
(This follows from a well-known result which states that a semi-lattice has a distributive congruence lattice iff that semi-lattice doesn't contain the four-element lattice above as a sub-semi-lattice.)
Best Answer
Your operation $g$ is irrelevant to compute the congruences of $A \times A$ because it's one of the projections of $+$.
So the congruence lattice of $A \times A$ is the same as the one of the group $\mathbb Z_2 \times \mathbb Z_2$, which is the lattice $M_3$.
Here, having the linked diagram as reference, \begin{align} 0 &= \{((a,b),(a,b)):a,b\in A\},\\ 1 &= \{((a,b),(c,d)): a, b, c,d \in A\},\\ x &= \{((a,b),(a,b')): a,b,b' \in A\},\\ y &= \{((a,b),(a',b)): a,a',b \in A\},\\ z &= \{((a,a),(b,b)): a,b \in A\}. \end{align}
The $0$ and $1$ congruences hold for all algebras: in the $0$, each element is related with itself and no other; in the $1$, every element is related with every other one.
In all algebras of the type $A \times A$, where $A$ is an algebra, we always get the congruences which are kernels of the projections, that is, we relate elements which have a common coordinate.
In this case, these are $x$ (where the first coordinate is fixed) and $y$ (where the second coordinate is fixed).
The last case, $z$, is still a congruence relation here, but it's not for all algebras $A$.
To see that these are precisely the congruence relations of $A \times A$, given that it is equivalent to a group, it is enough to see that that lattice is isomorphic to the subgroup lattice of $\mathbb Z_2 \times \mathbb Z_2$ (notice that, since $\mathbb Z_2$ is Abelian, so is $\mathbb Z_2 \times \mathbb Z_2$, whence all of its subgroups are normal).
Indeed, $x$ corresponds to the subgroup generated by $(0,1)$; $y$ to the subgroup generated by $(1,0)$ and $z$ to the subgroup generated by $(1,1)$.