To finish this problem, finally I have to show that where and
is a quotient group. Can you help me?
Hello,
I have this problem to solve and I have no idea how to do it...
Show that the class of Abelian groups has the CEP. Does the class of lattices have the CEP?
An algebra A has the congruence extension property (CEP) if for every and there is a such that . A class K of algebras has the CEP if every algebra in the class has the CEP.
How can I solve this?
Thanks for all your hints.
Clearly, a congruence on an abelian group B, is just a quotient group B/H, for some subgroup H, that is to say that [b] = [b]' iff b - b' is in H. It's not hard to see that H = [0], since the identity 0 must lie in any subgroup of B (I am writing [b] instead of b/θ, as the congruence class of an element b in B).
To define a congruence φ on A such that φ∩(BxB) = θ, simply consider φ defined by [a] = [a'] iff a - a' in H, that is: (a,a') in φ if a-a is in 0/θ (that is, we are considering the quotient group A/H: H is a subgroup of B, and thus a subgroup of A. In abelian groups, every subgroup is normal).
So now we have two subsets of BxB: θ, and φ∩(BxB).
Suppose (b,b') is in θ. Then b - b' is in H (which is the congruence class containing 0), and as elements of A, we STILL have b - b' in H, as the operation of A restricted to B IS the operation of B. Hence (b,b') is in φ, and thus trivially in φ∩(BxB).
On the other hand, suppose (a,a') is in φ∩(BxB). This means that (a,a) is in φ, AND (a,a') is in BxB. Hence a,a' are in B, and since the defining relation of the congruence in A is the SAME subgroup as in B, we have a - a' in H, that is (a,a') is in θ.
The error in your reasoning seems to be the assumption that 0/θ is a quotient of B (or A), it is not: it is a subgroup of these.
We can prove this:
Let 0/θ be the subset of B congruent to 0. Since congruences are equivalence relations, we have 0 in 0/θ by reflexiveness.
Suppose b,b' are in 0/θ. Since we have a congruence, we have [b+b'] = [b] + [b'] = [0] + [0] = [0+0] = [0], so b+b' is likewise congruent to 0, showing closure.
Finally, let b be any element of 0/θ. Then [0] = [b + -b] = [b] + [-b] = [0] + [-b] = [0 + -b] = [-b], so -b is likewise in 0/θ.
Thus 0/θ is a subgroup of B.