# Why $Z(A)$ is an equivalence relation on $A$?

For every algebra $A$, the center $Z(A)$ is a congruence on $A$. Why is $Z(A)$ an equivalence relation on $A$?

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I don't understand. If "congruence" means what I think it means then I don't know what you mean when you say that a subset of $A$ (as opposed to a subset of $A \times A$) is a congruence. –  Qiaochu Yuan Dec 18 '12 at 8:32

So perhaps your confusion arises from talking about the center of a group rather than a general or "universal" algebra. In that case, for a group $G$, the center $Z(G)$ is not really a congruence relation of $G$. However, as a normal subgroup, $Z(G)$ is a single class of a congruence relation of $G$ -- namely, the class containing the identity element of $G$.
In general, each normal subgroup $N$ of a group $G$ can be identified with a congruence relation, $\theta_N$, of $G$. The congruence classes of $\theta_N$ are the cosets of $N$. Thus, the lattice of normal subgroups of $G$ is isomorphic to the lattice of congruence relations of $G$. It is because of this identification that a normal subgroup is sometimes loosely referred to as a "congruence" of the group.