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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.

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Try here –  user23211 May 13 '12 at 10:13
    
What does the question have to do with rings and fields? –  lhf May 13 '12 at 12:27
    
@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. –  Gerry Myerson May 13 '12 at 12:32
    
@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. –  Tara B May 13 '12 at 12:33
    
@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? –  user23211 May 13 '12 at 12:39

2 Answers 2

up vote 4 down vote accepted

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.

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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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This 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. –  Tara B May 13 '12 at 12:35
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for $y \in G$ $ \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 –  Ustun May 13 '12 at 12:58
    
@ÜstünYıldırım: What do you mean by 'which suggests that'? That doesn't sound like a proof to me. –  Tara B May 13 '12 at 13:15
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If $y$ 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 $y=1$. –  anon May 13 '12 at 13:22
    
@anon: Thanks! I was indeed much too hasty. We are not only assuming $G$ is a monoid and so my objection doesn't apply. –  Tara B May 13 '12 at 13:25

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