Abelian definable groups.

Question

In a complete theory, is it true that every infinite intersection of definable groups that is abelian is contained in an abelian definable group?

Answer 1

In fact, more is true; if $(G,\cdot)$ is a definable group, then any "type-definable" abelian subgroup of $G$ is contained in a definable abelian subgroup of $G$. To see this, let $U$ be a saturated model. Suppose that $(\phi_i(x))_{i\in I}$ is a (small) collection of formulas such that $\bigcap_{i\in I}\phi_i(U)$ is an abelian subgroup of $G(U)$. Then, by saturation of $U$, the partial type $\{\phi_i(x),\phi_i(y):i\in I\}\cup\{xy\neq yx\}$ is inconsistent, as a realization would yield two non-commuting elements of $\bigcap_{i\in I}\phi_i(U)$. Thus, by compactness, there is $\theta(x)$ a finite conjunction of some $\phi_i$ such that $U\models \forall x\forall y(\theta(x)\wedge\theta(y)\to xy=yx)$. If each $\phi_i$ defines a subgroup of $G$, as is possible in your question hypothesis, then we are already done, as $\theta$ now defines an abelian subgroup of $G$ containing our type-definable group.

More generally, there is a standard trick we can use. Let $H=C_G(C_G(\theta))$ be the `double centralizer' of $\theta$; explicitly, $H$ is defined by the formula $$\chi(x)\equiv\forall y\big[[\forall z(\theta(z)\to yz=zy)]\to xy=yx]\big].$$ Now check that $H$ is an abelian subgroup of $G$ containing $\theta$, from which the result follows.

Exercise: If $G$ is a group, and $X\subseteq G$ is a set with the property that any two elements of $X$ commute, the $C_G(C_G(X))$ is an abelian subgroup of $G$ containing $X$.