TITLE: Is Lagrange's Theorem equivalent to AC?
QUESTION [34 upvotes]: Lagrange's Theorem is most often stated for finite groups, but it has a natural formation for infinite groups too: if $G$ is a group and $H$ a subgroup of $G$, then $|G| = |G:H| \times |H|$.
If we assume AC, the we get a fairly straightforward proof (pick a representative of each coset, and then let your map $(G : H) \times H \to G$ be $(gH, h) \mapsto gh$).
However, somebody told me someone told them that the converse is true too, i.e. Lagrange's Theorem gives us AC. (Indeed, the wikipedia page for LT offhandedly mentions they're equivalent without a reference). Does anybody have a proof (or alternatively a proof AC is stronger than LT)?

REPLY [8 votes]: Lagrange's Theorem does not follow from ZF. In fact, the following weaker statement also does not follow:
$\sf LT^{-}$: Let $H$ be a subgroup of $G$. Then $|H|$ divides $|G|$, i.e. there exists a set $A$ s.t. $|H| \times |A| = |G|$.
(The name LT$^{-}$ is not standard, but in the vein of the name LT$^{+}$.)
I stumbled upon the paper The Construction of Groups in models of set theory that fail the Axiom of Choice (Hickman, 1976) where he constructs a model of ZF with an amorphous group (a group whose carrier set is amorphous, i.e. an infinite set that isn't the disjoint union of two infinite sets).
He then goes on to prove several properties about amorphous groups, including the following:

Suppose $G$ is an amorphous group, and $H \leq G$ a finite non-trivial subgroup. (Such a subgroup always exists: every element in $G$ has finite order as $G$ has no $\aleph_0$ subset.)
Suppose that there was a set $A$ and a bijection $f \colon H \times A \to G$. As $H$ is finite, $A$ must be infinite. But then for any $h_1 \neq h_2 \in H$, $f(\{h_1\} \times A)$ and $f(\{h_2\} \times A)$ are infinite disjoint subsets of $G$, contradicting $G$ being amorphous.

So we need some 'choice' (i.e. a statement along the lines of 'no amorphous sets exist' or 'every infinite set is Dedekind-infinite') at the very least to have LT. How much is still unclear.