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A group is a monoid for which also:

For each element $a\in G$ there exists an element $a^{-1}$ such that $a^{-1}a = aa^{-1} = e$.

Now instead consider a slightly looser requirement:

For each element $a\in G$ there exists an element $/a$ such that $a/a=e$ and an element $a^{-1}$ such that $a^{-1}a=e$

Let's call such a structure a paragroup.

The question is if there's paragroups that are not groups? I'd guess so, because it would mean that one should have chosen the looser formulation of the invertibility axiom. I guess there's an example of it.

If we loosen the requirement to only reqire one sided inverses, would that really loosen the requirement for paragroups? Or can one prove the existence of RH-inverse from the existence of LH-inverse or vice versa?

skyking
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  • The statement *For each element $a\in G$ there exists an element $/a$ such that $a/a=e$ and an element $a^{-1}$ such that $a^{-1}a$* appears incomplete. – Andreas Caranti Mar 10 '17 at 11:43
  • @AndreasCaranti I don't think so. Why do you think it's incomplete? – skyking Mar 10 '17 at 11:52
  • It's the ending that appears to be missing something, "such that $a^{-1} a$". – Andreas Caranti Mar 10 '17 at 11:53
  • If I may add another comment, writing $a^{-1}$ from the outset appears to *postulate* that the inverse is unique, as you are writing it as a function of $a$. However, it is enough to require that "For each $a \in G$ there exists an element $b$ such that $b a = a b = e$" to be able to *show* that such a $b$ is uniquely determined by $a$, and thus can be safely written as a function of $a$. – Andreas Caranti Mar 10 '17 at 11:57
  • @AndreasCaranti Thanks, sometimes one gets blind to ones own miswrites, I've corrected the question. – skyking Mar 10 '17 at 11:58
  • You're welcome. And it is not that it happens sometimes, rather it happens all the time (including of course to YT). That's why it is essential to have someone proofread what we write. – Andreas Caranti Mar 10 '17 at 11:59
  • @AndreasCaranti That may appear that way, but that need not be implied. We may (at least with AC) chose one of the inverses to be denoted $a^{-1}$. – skyking Mar 10 '17 at 12:02
  • OK, but usually a writing like $f(a)$ or $a^{2}$ assumes that we are talking of a one-valued function of $a$. – Andreas Caranti Mar 10 '17 at 12:05

2 Answers2

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Consider the element $a^{-1}a/a$. If we still have associativity, then: $$a^{-1}a/a=(a^{-1}a)/a=/a$$ and $$a^{-1}a/a=a^{-1}(a/a)=a^{-1}$$ Hence $a^{-1}=/a$.
The relation between left and right inverses depends on how you define the identity. Suppose we have a set $G$ with an associative operation on it. If $G$ contains an element $e$ with the following property:

For all $g \in G$, $eg=ge=g$

Then the statement

For all $x \in G$ there exists an element $x^{-1}$ such that $x^{-1}x=e$

implies the statement

For all $x \in G$ there exists an element $/x$ such that $x/x=e$

Indeed, they are the same element: let $x \in G$; then $$xx^{-1}xx^{-1}=x(x^{-1}x)x^{-1}=xx^{-1}$$ Now apply the (left!) inverse of $xx^{-1}$ to see that $xx^{-1}=e$.

A natural question to ask, then, is:

If $G$ is a set with an associative binary operation, and a left/right identity; what is the relationship between left/right inverses of $G$?

Let's decompose this a little further. Suppose $G$ has two elements, $e$ and $f$, with the following properties:

For all $g \in G$, $eg=g$ and $gf=g$

Then $ef=f$ (left identity) and $ef=e$ (right identity), so $e=f$. So if $G$ has a left identity and a right identity, they are the same (this does not even require associativity!).

Suppose $G$ only has a one-sided identity, i.e. it contains an $e$ such that

For all $g \in G$, $eg=g$.

Must it be the case that $G$ has a right identity? Not in general; consider any set $X$, with the multiplication $xy=y$. This is associative (check!) and every $x$ is a left identity. But there certainly isn't a right identity! On the other hand, suppose that $G$ has this property:

For every $g \in G$, there exists $g^{-1}$ such that $g^{-1}g=gg^{-1}=e$

i.e. we have a left identity and normal inverses (inverse in the sense of the left identity). Then

$$geg^{-1}=g(eg^{-1})=gg^{-1}=e$$ The applying $g$ on the right shows $ge=g$. So $e$ is a right identity.

What about a left identity with only left inverses? Well, writing $g^{-1}$ for the left identity of $g$,

$$gg^{-1}gg^{-1}=g(g^{-1}g)g^{-1}=gg^{-1}$$ Applying the left inverse of $gg^{-1}$ shows $gg^{-1}=e$.
Taking $g=e$ shows that $e^{-1}=e$. Moreover, $$ge=g(g^{-1}g)=(gg^{-1})g=eg=g$$ So $e$ is a right identity, and $G$ is a group.

Finally, for the case of left identity and right inverses, consider our counterexample from before. Every $x$ is a left identity and a right inverse for every other element. But it certainly isn't a group.

preferred_anon
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  • Good answer, but is there something that can be said about the existence of LH-inverse from RH-inverse or vice versa? Or is there a counter example where only LH-inverse or RH-inverse exists? – skyking Mar 10 '17 at 10:27
  • I don't think that one implies the other, but I'm thinking about it now. – preferred_anon Mar 10 '17 at 10:27
  • Ok, I think that answers your question: I suspect that if you allow your identity to only be a left or right identity, then you will be able to find $G$ with left but not right inverses (or vice versa) but I can't think of an example just now. – preferred_anon Mar 10 '17 at 10:41
  • But if you have left and right identities: $e_lx=x$ and $xe_r=x$ you would have $e_l = e_le_r = e_r$ right? – skyking Mar 10 '17 at 10:50
  • Absolutely! I've added a bit more detail. – preferred_anon Mar 10 '17 at 11:59
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I am not sure whether this answer your question, but it can be shown that if

  1. $G$ is a non-empty set,
  2. $\cdot$ is an associative operation on $G$,

and the following conditions hold

  1. there is $e \in G$ such that $a \cdot e = a$ for all $a \in G$, and
  2. for each $a \in G$ there is $b \in G$ such that $a \cdot b = e$,

then $G$ is a group.

See this MSE post, where left identity and left inverses are used.

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Andreas Caranti
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