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Suppose that $(M, \circ)$ is a set $M$ over which there is defined a binary operation $\circ$ so that we have:

1) For every $(a,b) \in M \times M$ we have $a \circ b \in M$

2) For every $a \in M$ there exist exactly $k$ different elements $a_1^{-1},...,a_k^{-1} \in M$ so that we have $a \circ a_i^{-1}=a_i^{-1} \circ a = 1$, for every $i=1,...,k$

3) We have $1 \circ a = a \circ 1 =a$ for every $a \in M$

This would be a generalization of a loop concept because, as is easily seen, loops are obtained when $k=1$

Do these generalisations exist for every $k \in \mathbb N$? How to construct them if they do?

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  • $\begingroup$ Could you give example of such system for any k=/=1 please? $\endgroup$
    – kakaz
    Commented Dec 26, 2017 at 21:53
  • $\begingroup$ @kakaz I do not have an example, also, I do not know what exactly to think about even the existence of such structures. $\endgroup$
    – user114642
    Commented Dec 26, 2017 at 22:14
  • $\begingroup$ Is there any motivation for this axiomatization? $\endgroup$ Commented Dec 26, 2017 at 22:15
  • $\begingroup$ @JosephVanName Other than the motivation for the generalisation and for "playing around", no. $\endgroup$
    – user114642
    Commented Dec 26, 2017 at 22:16
  • $\begingroup$ Since the (abstract-algebra) tag is deprecated - see the tag-info - it would be nice to think about some more suitable tags. (Among other things, the question currently does not have any top-level tag.) $\endgroup$ Commented Dec 27, 2017 at 9:31

1 Answer 1

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No such $M$ exists for any $k>1$. Here is why: Obviously, by your axioms, the element $1\in M$ must be unique. Then $1\circ 1^{-1}=1$, hence $1=1^{-1}$, and $1^{-1}$ is unique as well.

But if you want, say, exactly two inverses to $a\in M$ for all $a\neq 1$ only, then it is possible by defining a three-elemet set $M=\{a_1,a_2, 1\}$ with $a_i\circ a_j=1$ and $a_i\circ 1=1\circ a_i=a_i$. Similarly, you can construct such a $(k+1)$-element set of loops $\{M,\circ \}$ for every $k$.

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  • $\begingroup$ Are there some structures with more than one identity? $\endgroup$
    – user114642
    Commented Dec 26, 2017 at 23:18
  • $\begingroup$ @AntoinePalAdeen: No, not with axiom (3). $\endgroup$ Commented Dec 26, 2017 at 23:20
  • $\begingroup$ Of course, but without it? $\endgroup$
    – user114642
    Commented Dec 26, 2017 at 23:24
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    $\begingroup$ Without axiom (3) you will have to modify axiom (2) to make sense of it. $\endgroup$ Commented Dec 26, 2017 at 23:42
  • $\begingroup$ mathoverflow.net/questions/289539/… $\endgroup$
    – user114642
    Commented Dec 30, 2017 at 0:55

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