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Let $E$ be a set , and let $f_a$ and $g_a$ be two functions such that for all $a \in E$ :

$$f_a : E \rightarrow E$$
defined by $ f_a(x)=ax$, and

$$g_a : E \rightarrow E$$ defined by $ g_a(x)=xa$.

The Question is : Suppose that $f_a$ and $g_a$ are surjective for all $a \in E$, and suppose $e$ is a neutral element in $E$, show that all elements of $E$ are invertible.

I tried starting like this :

Suppose that : $x(x^{-1}) = e$

So : $(x^{-1})x(x^{-1}) = (x^{-1})e$

$e(x^{-1}) = (x^{-1})e$

So that means $f(x^{-1}) = g(x^{-1})$ are both in $E$.

I think that my answer is not logical because I don't think I showed the proper proof.

Can I get some help on how to start this answer or maybe how I can go about solving it please? Thanks in advance.

Paul
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MathLearner
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  • Could you write exactly what the problem states? What you've doesn't really make much sense. If f(x) = ax for all a in E then for a, b in E f(x) = ax = bx => a = axx^-1 = bxx^-1 = b which is impossible. – fleablood Nov 16 '15 at 20:28
  • Or at least f(x) is not a function in that f(x) can equal both ax and bx which are not equal. If so I don't know what an "application" is and I don't know what "inversable" means. – fleablood Nov 16 '15 at 20:31
  • In the questions it has f with a little a at the bottom of it. So it's like f_a(x) if you know what I mean, I didn't know how to write it in Latex sorry. – MathLearner Nov 16 '15 at 20:35
  • Ok, I re-edited the questions with the proper writing. – MathLearner Nov 16 '15 at 20:37
  • Ah, that makes a significant difference! It's not a single f for all elements but different f's for different elements. – fleablood Nov 16 '15 at 20:38
  • I'm still not sure what the question is asking. If E is a group then by definition all elements are invertable. – fleablood Nov 16 '15 at 20:45

1 Answers1

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We have to show that every element in $E$ is invertible, So let $a \in E$, required to show $a$ is invertible in $E$.

As $e \in E$, and $f_a$is surjective, then there exists $x \in E$, such that $f_a(x)=e$, so that $ax=e$, hence $a$ is left invertible.

On the other hand, as $e \in E$, and $g_a $ is surjective, then there exists $x' \in E$, such that $g_a(x')=e$, so that $x'a=e$, hence $a$ is right invertible.

Hence $a$ is invertible .

Nizar
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    So in other words, E was not defined to be a group? – fleablood Nov 16 '15 at 20:49
  • @fleablood I editeed the quetion in a suitable form , however the OP didn't submit by edition yet ! – Nizar Nov 16 '15 at 20:51
  • Also, $x = x'$ in that case, since $(x'a)x = x'(x'a)$ (under the assumption of associativity, but if it's a group...) – Sake Nov 16 '15 at 20:59
  • @PaulPicard just let me check my knowledge, should $x$ and $x'$ be equal to say that $a$ is invertible ?? and sorry I dont get what did you wrote above. – Nizar Nov 16 '15 at 21:01
  • @PaulPicard I don't think $E$ is a group, otherwise every element is invertible . – Nizar Nov 16 '15 at 21:02
  • @Nizar No, $a$ is invertible in $E$ if it has a left and a right inverse. So what you wrote is right. ($x$ and $x'$ can be different) – Sake Nov 16 '15 at 21:02
  • @PaulPicard so in your comment did you mean $(x'a)x=x'(xa)$, or I misunderstood you ? – Nizar Nov 16 '15 at 21:04
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    @Nizar Probably misunderstood since $ax = x'a = e$ means that $x'ax = x'x'a$. – Sake Nov 16 '15 at 21:06