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Apart from the title, have below question too.

As left and right inverse coincide for bijective functions, so in groups there should be always a single inverse.

It should be only for proving if the given abstract structure is a group or not, need show left inverse equals the right inverse.

But, if try to take simple example to study left and right inverse, and the equality thereof?

So is there a finite set where the difference can be obvious. I doubt it is impossible in groups.

Or, at least show what is a left inverse / right inverse in some context.

Say, in $\lbrace\mathbb{Z}_n, +\rbrace$; then with $n=7$ if it can be shown, or for non-prime modulo $n=6$.

Shaun
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jiten
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    Since the integers modulo $n$ are a group, the two inverses coincide. If you want to think in terms of functions, you’ll have that a left inverse exists (but maybe not a right) or a right inverse exists (but maybe not a left) but if both exist you obtain left=right. – FShrike Jun 12 '22 at 08:30
  • @FShrike in $\lbrace\mathbb{Z}_7, +\rbrace$, $3= 4^{-1}$ and $4= 3^{-1}$. So, $3+4=0, 4+3=0$. So, is there left inverse and right inverse here, of $4$, hence also of $3$? – jiten Jun 12 '22 at 08:32
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    See https://groupprops.subwiki.org/wiki/Equality_of_left_and_right_inverses_in_monoid#:~:text=In%20a%20monoid%2C%20if%20an%20element%20has%20two%20distinct%20right,its%20two%2Dsided%20inverse). – Sourav Ghosh Jun 12 '22 at 08:33
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    If an element $x$ has a left inverse $y_1$ (i.e. $y_1x=1$) and a right inverse $y_2$ (i.e. $xy_2=1$) then they must coincide because $y_1=y_1\cdot 1=y_1(xy_2)=(y_1x)y_2=1\cdot y_2=y_2$. –  Jun 12 '22 at 08:34
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    On the other hand, this may be a moot point: every monoid is a monoid of functions. Let $(X,\cdot, 1)$ be a monoid. Then, map $x\in X$ to the function $f_x:X\to X$ such that $f_x(y)=x\cdot y$. The mapping $x\mapsto f_x$ is a monomorphism of $X$ to $(X^X,\circ,\text{id}_X)$ and its image (the submonoid of the latter) is isomorphic to the original monoid. So in general algebraic structures you won't see something that you won't already see on the functions! –  Jun 12 '22 at 08:37
  • @StinkingBishop please post an answer. – jitender Jun 12 '22 at 09:00
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    The associative property is crucial in $y_1=y_1\cdot 1=y_1(xy_2)=(y_1x)y_2=1\cdot y_2=y_2$ – lhf Jun 12 '22 at 11:42

1 Answers1

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In every monoid you can prove that, if an element has both the left and the right inverse, then the two must coincide.

Namely, if in a monoid $(X,\cdot,1)$ an element $x\in X$ has a left inverse $y_1$ (i.e. $y_1\cdot x=1$) and a right inverse $y_2$ (i.e. $x\cdot y_2=1$) then $y_1=y_1\cdot 1=y_1\cdot(x\cdot y_2)=(y_1\cdot x)\cdot y_2=1\cdot y_2=y_2$, due to associativity.

On the other hand, this may be a moot point: you wanted to explore whether the thing you noticed on the monoids of functions applies to more generaal monoids. However, every monoid is actually (isomorphic to) a monoid of functions. Let $(X,\cdot 1)$ be a monoid. Then, map $x\in X$ to the function $f_x:X\to X$ such that $f_x(y)=x\cdot y$. The mapping $x\mapsto f_x$ is a monomorphism of $X$ to $(X^X,\circ,\text{id}_X)$ and its image (the submonoid of the latter) is isomorphic to the original monoid. So in general algebraic structures you won't see anything that you won't already see on the monoids of functions.