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One of the group axioms is the existence of an inverse, which is usually stated as: For each $a\in G$ there exists an element $b$ in $G$ such that $ab=ba=e$ where $e$ is the identity element of the group.

When we only deal with matrix multiplication, we know that it suffices to show $ab=e$ in order to deduce that both matrices are invertible and that $b$ is the inverse of $a$.

My question is:

Is it possible to write the group axiom of inverses with only one equality? I.e., to say that for each $a\in G$ there exists an element $b$ in $G$ such that $ab=e$ where $e$ is the identity element of the group. Will this equality always imply that also $ba=e$ as in matrices? Alternatively, is it possible to construct a set $G$ with an operation $\star$ that will satisfy associativity, existence of identity and this new definition of existence of inverses but will fulfill the original statement $ab=ba=e$ (only $ab=e$)?

Thank you

Dr. John
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  • See https://math.stackexchange.com/questions/65239/right-identity-and-right-inverse-in-a-semigroup-imply-it-is-a-group?noredirect=1&lq=1 and https://math.stackexchange.com/questions/433546/is-a-semigroup-g-with-left-identity-and-right-inverses-a-group for examples of how little we can and cannot include in the axioms – Milten Jun 11 '24 at 07:55
  • I think this is essentially a duplicate of this post. But note that assuming the axioms right identity together wioth left inverses do not imply that we have a group. – Derek Holt Jun 11 '24 at 07:56
  • Note also: In a general ring, it is not true that $ab=1$ implies $ba=1$, so (finite-dimensional) matrices are a special kind of ring in this regard – Milten Jun 11 '24 at 07:58

2 Answers2

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There is an alternative definition for group $G$ with operation $\circ$ :

$1:$ $G$ is closed under $\circ$

$2$: $G$ is associative.

$3:$ $G$ has left(right) identity.

$4$: $\forall a\in G,$ $a$ has left(right) inverse.

$\textbf{The case for left:}$

$\forall a\in G$, $ea=a$.

Suppose that the left inverse of $a$ is $a'$ and the left inverse of $a'$ is $a''$.

$e=a''a'=a''(ea')=a''(a'aa')=(ea)a'=aa'\implies a=ea=(aa')a=a(a'a)=ae$

Bowei Tang
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Yes. Suppose you have a monoid $M$ with right inverses.

Then let $a\in M$ and choose $b\in M$ such that $ab=e$. Now choose $c\in M$ such that $bc=e$. Than $abc=c$ and $abc=a$ so $c=a$ and $ab=ba=e$.

Iq-n-dI
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  • Interesting! So the second equality is indeed redundant. Thank you! – Dr. John Jun 11 '24 at 07:52
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    This derivation assumes that $e$ is a left and right identity. Following Bowei Tang's answer this can be avoided. – Kurt G. Jun 11 '24 at 08:23
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    Exact, this is nice. – Iq-n-dI Jun 11 '24 at 08:28
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    @Dr.John Careful. It's not the equality $ab=e$. You can have an associative operation with identity $e$ such that for some $a$ and $b$ it happens $ab=e$ but $ba\ne e$. You need to talk not about "the identity" $ab=e$ (which in general doesn't not imply $ba=e$), but about the whole statement: $\color{red}{\forall a \exists b : } ab =e$ – jjagmath Jun 11 '24 at 08:44
  • @jjagmath : thank you for this remark. Can you give me an example of what might go wrong if we just take a $b$ such that $ab=e$? I am not sure I understand – Dr. John Jun 11 '24 at 15:36
  • @Dr.John Take the set $G = {f : \Bbb N \to \Bbb N}$ with the composition $\circ$ as operation. $G$ is closed under $\circ$, $\circ$ is associative, with identity $e = id_{\Bbb N}$, the identity function on $\Bbb N$. Consider the functions: $f(n) = n+1$ and $g$ defined as $g(0) =0$, $g(n)=n-1$ for $n>0$. Then $g\circ f =e$ but $f\circ g \ne e$. So the equality $g\circ f= e$ does not imply the equality $f\circ g =e$. – jjagmath Jun 11 '24 at 17:59
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    @Dr.John The important point is: if an element has a left inverse, it can happen that it doesn't have a right inverse. But if all elements have a left inverse, then each of those turn out to be also the right inverse. – jjagmath Jun 11 '24 at 18:12
  • @jjagmath: Thank you! – Dr. John Jun 12 '24 at 12:54