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This question is partly inspired by the recent question on measurability and the axiom of choice.

Suppose I come up with a way to define a subgroup of a group, in a way that involves "no arbitrary choices" in some sense. Then chances are, the subgroup will be normal. For example, the commutator subgroup of a group is normal, as is the commutator subgroup of the commutator subgroup.

I seem to recall that there are general theorems that say something like, any subgroup that is first-order definable without parameters is normal. Is this true? If so, are there even stronger theorems of this type?

The motivation is that such a meta-theorem could allow me, when I'm introducing a new kind of subgroup, to sidestep an explicit verification that the subgroup is normal.

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Timothy Chow
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    It is rather a triviality than a theorem that any subset of a group that is 1st order definable without parameters is invariant under all group automorphisms (so in particular, if it's a subgroup, then it's characteristic, hence normal). – YCor Jul 14 '15 at 22:14
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    @YCor That argument seems to apply to second-order and higher-order definitions as well! You could allow definitions of iterated transfinite order, which essentially allows one to build a set-theoretic universe atop the group. – Joel David Hamkins Jul 14 '15 at 22:23
  • Note that this only works for definitions in terms of the abstract structure of the group, not for subgroups of a group you find in mathematics that you can define mathematically. – Will Sawin Jul 15 '15 at 01:50
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    Well, I don't see the point: groups defined mathematically are defined in terms of the abstract structure of the group as well. The main point is the use of parameters (e.g., if we have a group action, when we pick a point stabilizer). On the other hand, even if allowed to quantify on integers or subsets of the group, if we use no parameters (hidden in any kind of way, as is often the case since most usual groups are defined either with a prescribed action, or with distinguished generators/elements) we should only be able to define automorphism-invariant subsets. – YCor Jul 15 '15 at 07:03
  • Yes, I wasn't thinking too hard when I wrote "definable without parameters"; I guess that condition is too strong to be interesting. Is there any interesting way to weaken it? – Timothy Chow Jul 15 '15 at 20:18

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