# Group structure on $G/H$ when $H$ is not normal

Let $$G$$ be a group, $$H$$ a subgroup, and $$G/H$$ the set of left cosets of $$H$$ in $$G$$. We can give this set an operation $$*$$, defined by $$(g_1H) * (g_2H) = (g_1g_2)H$$, which is well-defined if and only if $$H$$ is normal in $$G$$. In this case $$(G/H, *)$$ is a group.

However, even if $$H$$ is not normal, $$G/H$$ can still be given a group structure since every non-empty set admits a group structure (assuming the Axiom of Choice). I assume that there is no "natural" way to assign a group structure to $$G/H$$ that ends up being useful. In what sense, if any at all, can this last sentence be made precise? Can it be made more general (e.g. categorical)?

• I don't really see the point of why you would want to invoke axiom of choice. Sure, you can make $G/H$ a group but this group structure has absolutely nothing to do with the group structure of $G$. It's completely unnatural, in my opinion. Jun 11 at 20:25
• @daruma isn't the point of the question: "We can do it via AoC, and this is unnatural. Is there sometimes a more natural way, which does have something to do with the group?" Jun 11 at 20:27
• @user1729 Exactly. Jun 11 at 20:34

You can say this maybe as follows. There is no group structure on $$G/H$$ for which the projection map $$G \to G/H$$ is a group homomorphism (unless $$H$$ is normal in $$G$$).
You can take the categorical thing, the group via which all morphisms from $$G$$ which are trivial on $$H$$ factor. This will give you $$G/H^{\prime}$$ where $$H^{\prime}$$ is the "normal closure" of $$H$$ in $$G$$.