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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)?

Jacob
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    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. – daruma Jun 11 '21 at 20:25
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    @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?" – user1729 Jun 11 '21 at 20:27
  • @user1729 Exactly. – Jacob Jun 11 '21 at 20:34

1 Answers1

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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 $G$.

Shaun
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Sasha
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