4
$\begingroup$

Consider the following sequence of assertions, each of which implies the next. Nothing below has any topology, we just have sets and discrete groups.

  1. If $F \hookrightarrow E \twoheadrightarrow B$ is fibre bundle of sets (this just means every fibre has the same cardinality), then $|E| = |F \times B|$
  2. If $H \hookrightarrow E \twoheadrightarrow B$ is a principal bundle, then $|E| = |H \times B|$
  3. If $H$ is a subgroup of $G$, then $|G| = |H \times G/H|$
  4. If $H$ is a normal subgroup of $G$, then $|G| = |H \times G/H$|

Question: Do all these assertions fail without the axiom of choice? Or, which of them hold? I actually only care about (3), but for some reason I thought adding these extra statements might somehow clarify things.

$\endgroup$
  • 1
    $\begingroup$ Seems doubtful. $(\mathbb Q,+)$ is a normal subgroup of $(\mathbb R,+),$ and $|\mathbb R/\mathbb Q|\le|\mathbb Q\times\mathbb R/\mathbb Q|,$ but I don't see how to prove $|\mathbb R/\mathbb Q|\le|\mathbb R|$ without the axiom of choice. Of course the fact that I don't know how to prove it doesn't prove anything. $\endgroup$ – bof Nov 6 '16 at 3:15
  • 1
    $\begingroup$ @Alephnull Are you saying that $|\mathbb R/\mathbb Q|$ is obviously less than or equal to $|\mathbb R|$? Without the axiom of choice? How do you prove that? $\endgroup$ – bof Nov 6 '16 at 3:34
  • 2
    $\begingroup$ @Alephnull Everybody knows that there is a surjective map from $\mathbb R$ to $\mathbb R/\mathbb Q.\ $ The statement $|\mathbb R/\mathbb Q|\le|\mathbb R|$ means that there is an injective map from $\mathbb R/\mathbb Q$ to $\mathbb R.$ In general, the existence of a surjection from $A$ to $B$ does not imply the existence of an injection from $B$ to $A$ in set theory without the axiom of choice. I thought you might know some clever way to prove it in this particular case. $\endgroup$ – bof Nov 6 '16 at 3:43
  • 2
    $\begingroup$ @Alephnull I'm sure there's no way to choose a representative of each coset without AC, because that would mean you could construct a nonmeasurable set without AC. Conceivable, there is some way to construct an injection $f:\mathbb R/\mathbb Q\to\mathbb R$ without picking a representative from each coset. But that seems unlikely. $\endgroup$ – bof Nov 6 '16 at 3:51
  • 1
    $\begingroup$ @Alephnull I don't chat. Anyway, I thought we were finished. Your first comment was based on a misconception which has now been cleared up, right? $\endgroup$ – bof Nov 6 '16 at 4:13
9
$\begingroup$

Yes, the axiom of choice is needed, to some extent.

As bof mentions in the comments, it is always the case that $\Bbb{ Q\hookrightarrow R\twoheadrightarrow R/Q}$. However, as explained by Andrés E. Caicedo on MathOverflow, it is consistent that $\Bbb{R/Q}$ cannot be linearly ordered, as a set. In that case, it is impossible that $\Bbb{|R|=|Q\times R/Q|}$, as that would imply that you can linearly order $\Bbb{R/Q}$.

This should be a counterexample for all the cases.

If you are in the situation where all the equivalence classes in $G/H$ have the same cardinality as $H$ (e.g. like in the $\Bbb{R/Q}$ case), then as explained in the second part of this answer on MathOverflow, you can get the equality from the assumption that "cardinal multiplication is just 'repeated summation'", which is equivalent to the Partition Principle. However, it is still open if the Partition Principle is equivalent to the axiom of choice.

So it might be the case that having $|G|=|H\times G/H|$ implies the axiom of choice, even just from this equality in the Abelian case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for settling this, and also for the tour through some interesting MathOverflow threads! $\endgroup$ – Mike F Nov 6 '16 at 6:32
  • $\begingroup$ You're welcome. Let me also add a remark to that last sentence, that it is often the case that $\bf Ab$ sort of mimics $\bf Set$. You can look at Andreas Blass' paper "Injectivity, projectivity, and the axiom of choice" (also MR0542870 if you can read German) for some interesting examples. $\endgroup$ – Asaf Karagila Nov 6 '16 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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