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Let $\mathbf A$ be an (universal) algebra and let us denote $\mathrm{Clo}(\mathbf A)$ the set of term operations of $\mathbf A$. Let $\alpha, \beta$ be congruences of $\mathbf A$. We say that $\alpha$ has $\beta,\beta$-term condition iff $$ \forall t\in \mathrm{Clo}(\mathbf A)\ \ \forall (a,b)\in \beta\ \ \forall (\mathbf c,\mathbf d)\in \beta:\ \ (t(a,\mathbf c),t(a,\mathbf d))\in \alpha \Longrightarrow (t(b,\mathbf c),t(b,\mathbf d))\in\alpha $$ We denote by $[\beta,\beta]$ the least congruence $\gamma$ such that $\gamma$ has $\beta,\beta$-term condition. It is easy to see that $[\beta,\beta]\leq \beta$.

Laet $\alpha\leq\beta$ be congruences on $\mathbf A$. We say that $\beta$ is Abelian over $\alpha$ iff $\alpha$ has $\beta,\beta$-term condition. This should be equivalent to $[\beta,\beta]\leq\alpha$. Indeed, left to right implication holds by definition of $[\beta,\beta]$. Why does the implication right to left hold? I seem to be overlooking something simple.

I use the terminology used in the Bergman's book "Universal Algebra". The fact in my question is noted on page 263 there.

Ondrej Draganov
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  • The something simple you are overlooking: $(t(a,\vec{c}),t(a,\vec{d}))\in[\beta,\beta]\leq\alpha\implies(t(b,\vec{c}),t(b,\vec{d}))\in[\beta,\beta]\leq\alpha$. – Eran Jun 06 '18 at 13:54
  • @Eran This, for one, I am aware of :o) However, I need to have: $$ (t(a,\mathbf c),t(a,\mathbf d)) \in \alpha \Longrightarrow (t(b,\mathbf c),t(b,\mathbf d)) \in \alpha. $$ So I assume $(t(a,\mathbf c),t(a,\mathbf d)) \in \alpha$, from which I do NOT have $(t(a,\mathbf c),t(a,\mathbf d)) \in [\beta,\beta]$. – Ondrej Draganov Jun 06 '18 at 14:03
  • you're right. my bad. – Eran Jun 06 '18 at 14:38

1 Answers1

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You are not overlooking anything. There exist situations where $[\beta,\beta]\leq\alpha$, but $\beta$ is NOT abelian over $\alpha$.

Let me discuss one special case, where $\beta = 1_{\bf A}$ is the universal congruence on ${\bf A}$ and $[\beta,\beta]=0$. This means: ${\bf A}$ is an abelian algebra. If you could deduce that $\beta$ is abelian over $\alpha$ for an arbitrary congruence $\alpha$ on ${\bf A}$, you could go one more step and deduce that $\beta/\alpha = 1_{{\bf A}/\alpha}$ is an abelian congruence on ${\bf A}/\alpha$; i.e., that ${\bf A}/\alpha$ is an abelian algebra. To summarize, if the statement you are trying to prove was true, then it would imply that quotients of abelian algebras are abelian.

Here is a counterexample. Let ${\bf A} = \langle \{0,1,2,3\}; \ast\rangle$ be defined so that

    $3\ast 3 = 2$,
    $3\ast x = x\ast 3 = 1$ if $x\neq 3$, and
    $x\ast y = 0$ otherwise.

You can check that ${\bf A}$ is an abelian algebra that has a nonabelian quotient ${\bf A}/\alpha$, where $\alpha$ is the congruence on ${\bf A}$ generated by $(0,1)$.

In fact, it is easy to see that nonabelian quotients of abelian algebras must exist, even without constructing a specific one: Every algebra is a quotient of an absolutely free algebra, and every absolutely free algebra is abelian. Thus nonabelian quotients of abelian algebras (like nonabelian groups) are easy to produce.

Having said this, what we are discussing belongs to the pathology of commutator theory. If

    ${\bf A}$ belongs to a congruence modular variety, or
    ${\bf A}$ is finite and $\alpha$ is covered by $\beta$,

then $[\beta,\beta]\leq\alpha$ DOES imply that $\beta$ is abelian over $\alpha$. (The proofs of these statements are not trivial.)

Keith Kearnes
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