Existence of a canonical bijection between $G/H \times H/K$ and $G/K$

Question

Suppose $G$ is a group and $K<H<G$. All of the constructions of a bijection $G/H \times H/K \to G/K$ that I saw go more or less as follows. Choose a set of representatives ${g_i}$ for $G/H$, then the map $(g_i H,hK) \mapsto g_i h K$ is bijective.

This map depends on the choice of the representatives: suppose for example that the chosen representative for the coset $H$ is $1$, then the image of $(H,K)$ clearly changes if we pick another representative $g \in H - K$.

My question is whether there exists a way to contrsuct a bijection which is independent of the representatives.

To give some motivation for the question, I am interested in the cases where the group has additional structure (e.g. a topological/algebraic group), and I was wondering whether we can construct a morphism between those cosets.

Answer 1

Consider the following example: $G$ is the non-cyclic group of order $4$ (the Klein 4-group), $H$ is one of its three subgroups of order $2$, and $K$ is the trivial subgroup. Both of the quotients $G/H$ and $H/K$ are (canonically, in fact uniquely) isomorphic to $\mathbb Z/2$, so your question comes down to whether $G$ is canonically isomorphic $(\mathbb Z/2)\times(\mathbb Z/2)$. There are two isomorphisms between $G$ and $(\mathbb Z/2)\times(\mathbb Z/2)$, but neither is canonical. Indeed, these two isomorphisms correspond to each other under the (unique) non-trivial automorphism of $G$ that fixes $H$, whereas a canonical isomorphism would be invariant under automorphisms of $G$.

EDIT: I should have written bijection rather than isomorphism in a few places. Fortunately, the argument still works, because for Klein 4-groups, any identity-preserving bijection is an isomorphism. And a putative canonical bijection could be made identity-preserving by interchanging a pair of value, which preserves canonicity.

Answer 2

Since you've mentioned the axiom of choice, and Andreas Blass gave a nice concrete and very finite example, allow me to provide a much greater failure.

It is consistent, in the absence of choice, that $\Bbb{R/Q}$ has cardinality which is strictly greater than $\Bbb R$ itself. On the other hand, it is provable without using any choice that $\Bbb{R/Z}$ is just the unit circle and has the same cardinality as $\Bbb R$, and that $\Bbb{Q/Z}$ is just a quotient of a countable group, so it is countable.

So, if $\Bbb{R/Q\times Q/Z}$ has any bijection with $\Bbb{R/Z}$, it must mean that $\Bbb{R/Q}$ has the same size as $\Bbb R$. As this consistently fail, that means that it is consistent for there to be no bijection whatsoever.

Now, if your definition of "canonical" is somehow one that implies one can prove its existence even without the axiom of choice, then there can be no canonical bijection in this case.