Help me prove equivalently of regular semigroup and group. Let $S$ be a semigroup. Prove that the following are equivalent:


*

*$\forall a \in S \exists! x \in S$ such that $ax \in E(S)$ where $E(S)$ is the set of all idempotent.

*$\forall a \in S \exists! x \in S$ such that $a=axa$.

*$S$ is regular semigroup containing exactly one idempotent.

*$S$ is a group.
 A: This doesn't have a complete answer, but maybe it'll help others.

A regular semigroup is a semigroup $S$ such that $I_a = \{ x : axa = a \}$ is non-empty for every $a \in S$. If we set $V(a) = \{ x : axa = a \text{ and } xax= x \}$ then $V(a) \subset I_a$ and $V(a)$ is non-empty iff $I_a$ is non-empty.
An idempotent is an element $a \in S$ such that $aa=a$, and the collection of idempotents of $S$ is denoted $E(S)$.
An inverse semigroup is a regular semigroup in which idempotents commute, or equivalently, such that $V(a)$ consists of a single element for all $a \in S$.
If $axa=a$, then $(axa)x = (a)x$ means that $(ax)(ax) = (ax)$ so that $ax \in E(S)$. Similarly, $x(axa) = x(a)$ means that $xa \in E(S)$.
If $S$ is a group, then $I_a = \{ a^{-1} \}$ and $E(S) = \{1\}$, so $4 \implies 3$.
If $I_a$ is non-empty and $E(S) = \{ e \}$ then there is some $x$ such that $ax = xa = e$. Then $ea =(ax)a = a$ and $ae = a(xa) = a$, so $e$ is a two-sided identity and $x=a^{-1}$ is a two-sided inverse, and $S$ is a group. Hence $3 \implies 4$.
Clearly $4 \implies 1,2$ with $x=a^{-1}$ the unique solution each time.
Assume 1. Then for each $x$ there is a unique $x$ such that $axax=ax$. Note that $a(xax) \in E(S)$ so $x=xax$. Hence $(xa)(xa) = (xax)a = xa$ so $xa \in E(S)$, but also $x(axa) \in E(S)$, so $axa =a$. Hence $x \in I_a$. Now suppose $y \in I_a$. Then $aya=a$, so $(ay)^2 = ayay = ay$ and $ay \in E(S)$. Hence $y=x$, and $1 \implies 2$.
Assume 1. If $a \in S$, let $a^{-1}$ denote the unique solution to $aa^{-1} \in E(S)$. As before, if $aa^{-1}aa^{-1} = aa^{-1}$, so $a^{-1} a a^{-1} = a^{-1}$ and $a^{-1} a a^{-1} a = a^{-1} a$, so $(a^{-1})^{-1} = a$. Hence $a a^{-1} a = a$ as well, and $S$ is a regular semigroup. If $axa=a$, then $axax=ax$ so $x=a^{-1}$, so $S$ also satisfies 2 and the similar $xax=x$, so that $S$ is an inverse semigroup, and by a theorem idempotents commute. That means $(ef)^2 = e^2 f^2 = ef \in E(S)$, but $e^{-1}$ is the unique solution to $ex \in E(S)$. Hence $f=e^{-1}=e$, and the idempotents are unique. Hence $1 \implies 3$.
So we have a single implication remaining: $2 \stackrel{?}{\implies} 1$.

Now something very close to 2 does not imply 4. 2’ Suppose for each $a \in S$ there is exactly one $x$ such that both $axa=a$ and $xax=x$.
Semigroups satisfying $2'$ are called inverse semigroups, and need not be groups. For plain old 2 though, I haven't found a non-group inverse semigroup that works.
A: Adding to what Jack Schmidt has written, let me prove that 2 implies 4. 
If for all $a$ there exists a unique $x$ such that $axa = a$, then the same $x$ satisfies $xax = x$ since $a(xax)a = (axa)xa = axa = a$. So we have Jack's 2', i.e., the semigroup is an inverse semigroup. For given $a$, let $a^\ast$ denote that $x$. 
Idempotents in inverse semigroups commute. This seems to be well-known, but a proof can be found here: Do the idempotents in an inverse semigroup commute? 
We now deduce the following: 


*

*$(ab)^\ast = b^\ast a^\ast$ (because $abb^\ast a^\ast ab = a a^\ast a b b^\ast b = ab$, using the fact that $b b^\ast$ and $a^\ast a$ are idempotent and therefore commute). 

*If $e$ is idempotent, then $e^\ast = e$ (since $eee = e$). 

*For any $x$, if $axa = a$ and $bxb = b$, then $a = b$ (since $axa = a$ implies $xax = x$ implies $a = x^\ast$, and similarly $b = x^\ast$). 
Now let $e$ be any idempotent and let $a$ be any element. We will show that $ae = a$. Indeed, we have $ae a^\ast ae = aee a^\ast ae = ae e^\ast a^\ast ae = ae (ae)^\ast ae = ae$ using, in turn, idempotence of $e$, 2., 1., and definition of $(ae)^\ast$. We also have $a a^\ast a = a$; therefore $ae = a$ by an application of 3. 
By similar reasoning, $ea = a$. Thus any idempotent $e$ is an identity element, or rather let us say the identity element $1$. Since $a a^\ast$ and $a^\ast a$ are idempotent, we have $a a^\ast = 1 = a^\ast a$ and we are in a group. 
I'll stop here since I see someone else has now posted an answer. 
A: I prefer to solve this problem in this way : $1=>2=>3=>4=>1$
$1=>2$ :
$ax \in E(S)$ means $\forall k \in S$ ; $k(ax)=(ax)k=k$ . because $x$ is unique and $a\in S$ so let $k = a$ then $a(ax)=(ax)a=a$ so $axa=a$
$2=>3$ :
First of all it's better to use $a^*$ insist of $x$.
so $2$ will be like this : $\forall a \in S \exists! a^* \in S$ such that $a=aa^*a$.
We claim that $aa^*$ is the unique idempotent.
Prove : that is enough to show that $\forall b\in S ; (aa^*)b=b(aa^*)=b$
I) $aa^*a=a => aa^*aa^*=aa^*$

II) $aa^*aa^*=aa^* => aa^* (aa^*) aa^* = aa^* aa^* =>$because of $(I) $$ aa^* (aa^*) aa^* = aa^* => (aa^*)^* = aa^*$ 

III) Suppose that $\exists X , X\in S$ and $a^*X=a^*$ 
So we have $a^*X=a^* => aa^*X=aa^* => aa^* (X) aa^* =aa^*aa^*=>$ because of $(I)$ $aa^* (X) aa^* =aa^*$ that means $X = (aa^*)^* = aa^*$(because of $(II)$) so we can say $a^*aa^*=a^*$

IV) We will show that there is $X\in S$ that $\forall A,B \in S;$  $ BX=A$
$BX=A => A^*BX=A^*A => A^*B(X)A^*B =(A^*AA^*)B =>$  because of $(III)$ $A^*B(X)A^*B =A^*B => X=(A^*B)^* \in S$ and also $B(A^*B)^* = A$

V) For all $a,b,x\in S$
Let $A=aa^*$ and $B=ax$
So we have : $(IV)$ $B(A^*B)^* = A => ax((aa^*)^*ax)^*=aa^* => $ because of $(II)$ $ax((aa^*)ax)^*=aa^* => ax((aa^*a)x)^*=aa^* => ax(ax)^*=aa^*$
Let $x = (b^*a)^* => ax = b $(because of IV)
Now we put that on main equation : $ax(ax)^*=aa^* => bb^*=aa^*$

VI) We will show that there is $X\in S$ that $\forall A,B \in S;$  $ XB=A$
$XB=A => XBA^*=AA^* => BA^*(X)BA^*=B(A^*AA^*)=>$ because of $(III)$ $BA^*(X)BA^*=BA^* => X=(BA^*)^* \in S$ and also $(BA^*)^*B=A$

VII) For all $a,b,x\in S$
Let $A=aa^*$ and $B=xa^*$
So we have : $(VI)$ $(BA^*)^*B=A => (xa^*(aa^*)^*)^*xa^*=aa^* => $because of $(II)$ $(xa^*aa^*)^*xa^*=aa^*=>$ because of $(III)$ $(xa^*)^*xa^*=aa^*$
Let $x=(a^*b^*)^* => xa = (a^*b^*)^* a = b$ (because of $VI$)
Now we put that on main equation : $(xa^*)^*xa^*=aa^* => b^*b=aa^*$

VIII) $(V)$ and $(VII) => \forall b\in S ; bb^*=b^*b$

Now we claim that $aa^*$ is the unique idempotent.
$\forall b\in S $ 
$b=bb^*b=b(bb^*)=$(because of $(V)$)$baa^*$
$b=bb^*b=(bb^*)b=$(because of$(VIII)$)$(b^*b)b=(aa^*)b$
So $b(aa^*)=(aa^*)b=b$
$3=>4$:
To prove this that's enough to show that $\forall a\in S : \exists! x \in S ; ax=xa=e$
$3=>\forall a\in S : \exists! x \in S ; axa=a $:
$axa=a => a(xa)=a => xa = e$
$axa=a => (ax)a=a => ax = e$
So : $ax=xa=e=>$ $x$ is inverse for $a$ and $x$ is unique $=> S$ is a group.
$4=>1$:
$S$ is a group so $\forall a\in S ; \exists!$ $ a^{-1};aa^{-1}=e \in E(S)$ so in $1$ let $x=a^{-1}$
