Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer

1.- Try to prove that $y\in G\,,\,yy=y\Longrightarrow y=1$

2.- Now prove, using your notation, that $xg\cdot xg=xg$


share|cite|improve this answer
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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.