Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

1 Answer 1

Since you referred to the center of an algebra (rather than the center of a group), I believe you mean Definition 5.1 on page 35 of the commutator book. But by that definition it is obvious that the center is an equivalence relation.

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.

See also this answer.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.