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A completely simple semigroup has by definition an idempotent $e$. If $x\in S$, then $eex=ex$ and so $ex=x$ and similarly, we see that $e$ is a right identity. Thus $S$ is a monoid and the identity is the unique idempotent of $S$. Let us write $1$ for the identity. Let $x\in S$. Then $uxy=1$ for some $u,y$ by simplicity. Then $xyuxyu=xyu$ and so $xyu=1$ as $1$ is the unique idempotent. Similarly, $yuxyux=yux$ and hence $yux=1$ because each idempotent is the identity. Thus $x$ is invertible with inverse $yu$. So $S$ is a group.

A completely simple semigroup has by definition a (primitive) idempotent $e$ and $SxS=S$ for all $x\in S$. If $x\in S$, then $eex=ex$ and so $ex=x$ and similarly, we see that $e$ is a right identity. Thus $S$ is a monoid and the identity is the unique idempotent of $S$. Let us write $1$ for the identity. Let $x\in S$. Then $uxy=1$ for some $u,y$ by simplicity. Then $xyuxyu=xyu$ and so $xyu=1$ as $1$ is the unique idempotent. Similarly, $yuxyux=yux$ and hence $yux=1$ because each idempotent is the identity. Thus $x$ is invertible with inverse $yu$. So $S$ is a group.

A completely simple semigroup has by definition a primitive idempotent $e$. If $x\in S$, then $eex=ex$ and so $ex=x$ and similarly, we see that $e$ is a right identity. Thus $S$ is a monoid and the identity is the unique idempotent of $S$. Let us write $1$ for the identity. Let $x\in S$. Then $uxy=1$ for some $u,y$ by simplicity. If $xyu=1$, then $yuxyu=yu$ and so $yux=1$ by cancellation. Thus $x$ has inverse $yu$. If $xyu\neq 1$, then $u,v=xy$ generate a bicyclic submonoid, which has infinitely many idempotents (the elements $v^nu^n$). This contradicts that each idempotent is the identity. So this second case doesn't happen.

  So $S$ is a group.

A completely simple semigroup has by definition an idempotent $e$. If $x\in S$, then $eex=ex$ and so $ex=x$ and similarly, we see that $e$ is a right identity. Thus $S$ is a monoid and the identity is the unique idempotent of $S$. Let us write $1$ for the identity. Let $x\in S$. Then $uxy=1$ for some $u,y$ by simplicity. Then $xyuxyu=xyu$ and so $xyu=1$ as $1$ is the unique idempotent. Similarly, $yuxyux=yux$ and hence $yux=1$ because each idempotent is the identity. Thus $x$ is invertible with inverse $yu$. So $S$ is a group.

A completely simple semigroup has by definition a primitive idempotent $e$. If $x\in S$, then $eex=ex$ and so $ex=x$ and similarly, we see that $e$ is a right identity. Thus $S$ is a monoid and the identity is the unique idempotent of $S$. Let us write $1$ for the identity. Let $x\in S$. Then $uxy=1$ for some $u,y$ by simplicity. If $xyu=1$, then $yuxyu=yu$ and so $yux=1$ by cancellation. Thus $x$ has inverse $yu$. If $xyu\neq 1$, then $u,v=xy$ generate a bicyclic submonoid, which has infinitely many idempotents. This contradicts that each idempotent is the identity. So this second case doesn't happen.

So $S$ is a group.

A completely simple semigroup has by definition a primitive idempotent $e$. If $x\in S$, then $eex=ex$ and so $ex=x$ and similarly, we see that $e$ is a right identity. Thus $S$ is a monoid and the identity is the unique idempotent of $S$. Let us write $1$ for the identity. Let $x\in S$. Then $uxy=1$ for some $u,y$ by simplicity. If $xyu=1$, then $yuxyu=yu$ and so $yux=1$ by cancellation. Thus $x$ has inverse $yu$. If $xyu\neq 1$, then $u,v=xy$ generate a bicyclic submonoid, which has infinitely many idempotents (the elements $v^nu^n$). This contradicts that each idempotent is the identity. So this second case doesn't happen.

So $S$ is a group.