3
$\begingroup$

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)?

$\endgroup$
3
  • 6
    $\begingroup$ 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. $\endgroup$
    – daruma
    Jun 11 at 20:25
  • 6
    $\begingroup$ @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?" $\endgroup$
    – user1729
    Jun 11 at 20:27
  • $\begingroup$ @user1729 Exactly. $\endgroup$
    – Jacob
    Jun 11 at 20:34
9
$\begingroup$

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 $G$.

$\endgroup$
0

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.