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When a group action $G \times X \rightarrow X$ is defined with a group $G$ and a set $X$, why is there not a special name for the set $X$? I know that this is referred to as a $G$-set, but the set $X$ itself does not seem to have a special name without the "$G$" specification.

To illustrate the point: when we have the appropriate action of a ring $R$ on a set $M$ such that $R \times M \rightarrow M$, then the set $M$ is a "module" without specification of the "$R$", and then we can refer to the $R$-module if we want to be more specific. But we cannot refer to the "set" in the same way, it has to be "$G$-set". So I am wondering is there such a name that was used in this case?

user50229
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    You could refer to these generically as "sets with a group action", but there's nothing you can say about the general case anyway. (Whereas by comparison, every module is an abelian group.) – Zhen Lin Aug 13 '13 at 10:57
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    Not really, I don't think. Mathworld calls $X$ a "group set" but I have not seen that elsewhere and it doesn't cite any references. Perhaps the closest convention is calling $X$ a "space" after it has already defined to be a $G$-set. – anon Aug 13 '13 at 10:57
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    But we should only call $M$ simply a module when the ring $R$ is understood. And being a module at least implies some structure apart from the action of $R$. Being a $G$-set does not provide any structure apart from the action of $G$, so there is nothing special about a set on which some unspecified group acts. – Tobias Kildetoft Aug 13 '13 at 10:57
  • I am not entirely sure what your point is. Groups act on so many things. For example, they act on other groups, and there is the theory of groups acting on trees, on graphs, on $\operatorname{CAT}(0)$ spaces... You name it, a group acts on it. On the other hand, a module is necessarily an abelian group. – user1729 Aug 13 '13 at 10:57
  • @anon Interesting! I've never seen that but it would seem to be an answer to the question. – user50229 Aug 13 '13 at 11:00

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If $G$ is a semigroup then one use names "polygon" [Kilp M., Knauer U., Mikhalev A.V., Monoids, acts and categories] or "operand" [A. H. Clifford, G. B. Preston, The algebraic theory of semigroups] for $G$-sets (i.e. for the sets on which $G$ acts). So you can use them also, if you would like.

Boris Novikov
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The language of monoidal categories offers a unification of various notions spread out over mathematics. $G$-sets and $R$-modules are really the same notions. In every monoidal category, there is the notion of a monoid object, namely an object $G$ equipped with morphisms $1 \to G$ and $G \otimes G \to G$, satisfying unitality and associativity. Besides, there is the notion of a module object over that monoid object, namely an object $X$ equipped with a morphism $G \otimes X \to X$ satisfying unitality and associativity in a certain sense. For the cartesian monoidal category of sets, the monoid objects are precisely the monoids $G$ in the usual sense, and $G$-modules are usually called $G$-sets. But really, you could just stick to the general formalism and call them $G$-modules$^{(1)}$. For the monoidal category of abelian groups with the usual tensor product, the monoid objects are the rings in the usual sense, and $R$-modules in the general sense above are precisely the $R$-modules in the usual sense.

Back to the general case: For a $G$-module $(X,G \otimes X \xrightarrow{a} X)$ I would call $X$ the underlying object. It is not good to call it a module, because the action $a$ has been forgotten.

$^{(1)}$ You might object that this contradicts with the common meaning of $G$-modules as $R[G]$-modules, where $R$ is a ring which is understood. But in my opinion this only shows that this usual usage is not good. Why does one say $G$, the monoid (or often group) object in the category of sets, where one actually means the induced monoid object $R[G]$ in the category of $R$-algebras? It is important and also quite useful to distinguish objects of different categories. When $G$ is a monoid, it is understood to be a monoid object in the category of sets, and nothing else, so that every construction, in particular the definition of a module, refers to the category of sets. Therefore "$G$-modules" is not ambiguous.

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Martin Brandenburg
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  • Thanks for your informative answer. So from this am I right to say that with the usual $R$-module, you would call $M$ the underlying object, which is the Abelian group in this case? I notice in most introductory texts (with no category theory) that they do refer to the underlying object as module, just as we call a group by its underlying set $G$, do you then think this is a bad convention? – user50229 Aug 13 '13 at 21:15
  • Yes it is bad to ignore forgetful functors. For an $R$-module $(M, R \otimes M \to M)$ I would call $M$ the underlying abelian group. It is also quite common to denote it as the underlying additive group. The underlying set of a group should, of course, not be confused with the group itsself. – Martin Brandenburg Aug 14 '13 at 08:19