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Let $A$ be a set and let $\circ:A\times A\rightarrow B,$ $A\subseteq B$ be a binary operation ($A$ is not necessarily closed under $\circ$). If there exists some unique $e\in A$ such that $e\circ a=a\circ e=a$ for all $a\in A$ and for all $a\in A,$ there exists some unique $b\in A$ such that $a\circ b=b\circ a=e.$ Is there a name for $(A,\circ)?$ $$$$ Essentially, A is a group without the requirement that A be closed under the operation.

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  • $\begingroup$ Related $\endgroup$
    – Shaun
    Commented Jul 25, 2023 at 19:43
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    $\begingroup$ Do you also want to add some form of associativity? (That's an important part of group-ness.) $\endgroup$
    – Noah Schweber
    Commented Jul 25, 2023 at 19:45
  • $\begingroup$ @NoahSchweber Yes $\endgroup$
    – Miles Gould
    Commented Jul 25, 2023 at 20:04

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That table in Wikipedia is helpful: group-like structures.

So your object could be called a groupoid, if we consider the binary operation as a partial operation $B \times B \rightarrow B$. (Although this is larger than your definition, because in most cases the set of pairs for which the operation is defined cannot be expressed as $A \times A$).

Unfortunately, there are more than one definition for groupoid, as explained in various places, e.g. on Wolfram MathWorld.

Note that the combination of non-closedness with associativity is tricky: you have to state whether $a \circ (b \circ c) = (a \circ b) \circ c$ means "when both sides are defined, they are equal", or "when at least one side is defined, the other side is also defined and equal".

A well-known example of groupoid is the fundamental groupoid, in algebraic topology.

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