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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2Try [here](http://math.stackexchange.com/a/136810/23211) – 2012-05-13
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0What does the question have to do with rings and fields? – 2012-05-13
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0@lhf, maybe OP is enrolled in a "groups, rings, and fields" course, and figures any question from the course is automatically a "groups, rings, and fields" question. – 2012-05-13
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0@ymar: I'm not sure it's appropriate to add the semigroups tag - even though obviously this is really a question about semigroups - because the tag being present might imply that that the OP has some familiarity with semigroup theory, which I doubt is the case. – 2012-05-13
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0@TaraB I don't know, but I have noticed that people do add tags this way here. Perhaps a meta question would be a good idea? – 2012-05-13
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0@ymar: I do it myself, but usually when I think that adding the tag will help draw the attention of the right people to answer the question, for example. – 2012-05-13
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0@TaraB I see. I just think the main purpose of tags is to classify questions by their content to make the site easier to browse. And this is actually an important question which other people may want to find. I've been thinking about changing the title too, but I would like at least to make sure Victoria understands what the new title means. – 2012-05-13
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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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1This approach will not work. We're only assuming that $G$ is a monoid, and it's possible to have $y^2 = y$ but $y\neq 1$ in a monoid. – 2012-05-13
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3for $y \in G$ $ \exists x \in G $ s.t $yx=1$ So for $ y \in G $ multiply from right both sides of $ yy=y $ by $x$ then we get $yyx = yx = 1$ which suggests that $y=1$ so it works – 2012-05-13
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0@ÜstünYıldırım: What do you mean by 'which suggests that'? That doesn't sound like a proof to me. – 2012-05-13
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1If $y$ has a right inverse ($yx=1$ for some $x$), then you can cancel any $y$ on the right by right-multiplying by $x$. Doing this to $y^2=y$ gives $y(yx)=yx$ and then $y=1$. – 2012-05-13
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0@anon: Thanks! I was indeed much too hasty. We are _not_ only assuming $G$ is a monoid and so my objection doesn't apply. – 2012-05-13
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0It's funny 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