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When reading Counting Rules of Nambu-Goldstone Modes (arXiv:1904.00569), I find the statement

The set of unbroken symmetries forms a subgroup $H$ of $G$. Other elements of $G$, $G \setminus H$ as a set, is said to be broken.

That is, given a subgroup $H \subset G$, there is a set $G\setminus H = \{g \in G | g \notin H \}$. Is there another (better) name for this set? It will not be a group, since it will not be closed under whatever group action it inherits from $G$, since the identity will be contained within $H$ and not $G \setminus H$.

I believe this is not the same notation being discussed in Meaning of $\setminus$ notation in Group Theory, where $\setminus$ is the right coset. Furthermore, this is distinct from the complement of a subgroup since that is another group.

A similar idea occurs in Lie algebras, where if one has a subalgebra $\mathfrak{h} \subset \mathfrak{g}$, there is a set $\mathfrak{g}\setminus \mathfrak{h} = \{g \in \mathfrak{g} | g \notin \mathfrak{h} \}$. (In relativistic quantum field theory, we have Goldstone's theorem which states that we get a massless boson associated with each "broken generator" of a spontaneously broken symmetry, i.e., a massless boson for each element of $\mathfrak{g} \setminus \mathfrak{h}$).

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    $\begingroup$ I have not seen such a set referred to with a special name. I would call it the complement of $H$ (in $G$). By the way, it is never a subgroup because it does not contain the identity element. $\endgroup$
    – Torsten Schoeneberg
    Commented Oct 21, 2022 at 4:24
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    $\begingroup$ Perhaps it could be noted that although $G \setminus H$ is the complement in the set-theoretic sense, in group theory a complement of a subgroup $H$ of a group $G$ is a subgroup $K$ such that $H K = G$ and $H \cap K$ is trivial; a complement in the latter sense need not exist in general. $\endgroup$
    – Andreas Caranti
    Commented Oct 21, 2022 at 13:06
  • $\begingroup$ It occurs to me that there might be a natural isomorphism between $G \setminus H$ in the set-theoretic sense to the right cosets $G \setminus H$. If that's true, I can see why there wouldn't be a name (or commonly-made distinction) for the former idea. $\endgroup$
    – kc9jud
    Commented Oct 22, 2022 at 21:31
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    $\begingroup$ @kc9jud There is no such isomorphism. (In general, if $G$ is finite with $n$ elements and $H$ is a subgroup consisting of $k$ elements, then each coset $gH$ has $k$ elements as well, there are $n/k$ such cosets in the set $G\setminus H$ (group theory version -- note that this denotes the set of all right cosets, not an individual coset) and there are $n - k$ elements in $G \setminus H$ (set difference). Usually, these numbers will be different, so there can't be an isomorphism.) $\endgroup$
    – Eike Schulte
    Commented Oct 22, 2022 at 21:36
  • $\begingroup$ @EikeSchulte ah, that makes sense -- I'm so used to thinking about Lie groups so those kinds of counting arguments aren't what I think of first... $\endgroup$
    – kc9jud
    Commented Oct 22, 2022 at 21:51

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