Congruence relation of an 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.
 A: 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 $(1,1)$.
