Let $G$ be a set with associative binary operation and a unit. Assume that for every $g\in G$ there exists $x \in G$ with $gx = 1$. Prove that $xg = 1$ is a consequence.
If every element of a monoid is right invertible, then every element is invertible
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0@ymar: Fair enough. Yes, it's difficult about the title. The things I would naturally think of are not likely to be meaningful to a student in a 'groups, rings and fields' course. – 2012-05-13
2 Answers
The statement is that every $g \in G$ has a right inverse $x$, ie, $gx = 1$. Now the same statement holds in turn for $x$: let $g'$ (suggestively named) be a right inverse for $x$, so that $xg' = 1$. Then on the one hand, using associativity $gxg' = (gx)g'= 1\cdot g' = g'$, but on the other hand, $gxg' = g(xg') = g\cdot 1 = g$. So $g = g'$, and $x g = x g' = 1$.
An equivalent conceptual way of thinking of this is as follows: The statement that $x$ is a right inverse of $g$ is identical to the statement that $g$ is a left inverse of $x$. So $x$ has a left inverse, and is assumed as always to have a right inverse. Therefore it must simply have a two-sided inverse.
1.- Try to prove that $y\in G\,,\,yy=y\Longrightarrow y=1$
2.- Now prove, using your notation, that $xg\cdot xg=xg$
DonAntonio
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0It's fu$n$ny how I would have accepted Üstün's proof quite happily if it had said 'which implies that' instead of 'which suggests that', because I would have then read it properly. Probably a good lesson for me. – 2012-05-13