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Can anyone please verify this proof:

Let there be two idempotent elements $f$ and $g$. Then $ff=f$ and $gg = g$. Then, by definition of the identity element of a group: $f$ is the identity and $g$ is also an identity But the identity in a group is unique, this implies $f = g = e$ Thus, there is only one idempotent element in the group and that is $e$.

Note: I've already proved that the identity is unique previously Also, the proofs for this theorem use the strategy $x*x = x$ and then go on to prove that $e = x$. If my aforementioned proof is incorrect, can you tell me why and why this proof might be better?

Thanks.

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  • $\begingroup$ Recall the definition of the identity element. Can you really conclude from $ff = f$ that $f$ is the identity? $\endgroup$
    – Dune
    Commented Nov 24, 2017 at 12:24
  • $\begingroup$ Anything multiplied by the identity gives the other element it is being multiplied with, so why can't i conclude that f is the identity? $\endgroup$
    – Alea
    Commented Nov 24, 2017 at 12:32
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    $\begingroup$ Two idempotent elements of... what? In a group your reasoning is valid, because we have the cancellation law. In a ring we don't have that always, and the claim is actually false. Mind you, as you showed in a group there is only one idempotent element, so the concept is kinda meaningless. OTOH idempotent elements of rings abound, and they are the key to a lot of structure theory of rings. $\endgroup$ Commented Nov 24, 2017 at 12:45
  • $\begingroup$ @A.Asad: Because you haven't showed that anything multiplied by $f$ gives the other element. $\endgroup$
    – Dune
    Commented Nov 24, 2017 at 12:47

3 Answers 3

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Nowhere you proved that $f$ satisfies $fx=x$ for all $x\in G$, nor you did for $g$. In fact, the same things you said could be written under the weaker assumption that $G$ is just a monoid and, in that case, the thesis would not hold. You really need to use the existence of an inverse.

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$$f^2=f$$ gives $$f^2f^{-1}=ff^{-1}$$ or $$f=e.$$

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More generally, if an element $u$ in a group acts like the identity even for one element, then $u=e$.

Indeed, if $u$ and $v$ are such that $uv=v$, then $u=e$ because $u=ue=uvv^{-1}=vv^{-1}=e$.

So, $ff=f$ and $gg=g$ both imply $f=e=g$.

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