Proof Verification: Identity is the only idempotent element

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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 $f=e=g$.