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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.

Arturo Magidin
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Victoria
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    Try here –  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? –  May 13 '12 at 12:39
  • @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. – Tara B May 13 '12 at 12:44
  • @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. –  May 13 '12 at 13:05
  • @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. – Tara B May 13 '12 at 13:18

2 Answers2

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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.

Dustan Levenstein
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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

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 –  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
  • It'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. – Tara B May 13 '12 at 13:40