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Let $S$ be a semigroup. Show that $S$ is a group if and only if $aS=S=Sa$ for all $a\in S$.

Since it is if and only statement, we have to show that if $S$ is a group then $ aS=S=Sa$, which I already know how to do.

The other part: if you have $aS=S=Sa$, then prove that $S$ is a group.

(NB: See the comments for the original thoughts on the problem.)

Shaun
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ADAM
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  • What are your thoughts? –  Mar 16 '16 at 04:48
  • A group (G,·) is a nonempty set G together with a binary operation · on G such that the following conditions hold: (i) Closure: For all a,b in G the element a · b is a uniquely defined element of G. (ii) Associativity: For all a,b,c in G, we have a · (b · c) = (a · b) · c. (iii) Identity: There exists an identity element e G such that e · a = a and a · e = a for all a G. (iv) Inverses: For each a G there exists an inverse element $a^{-1}$ G such that a · $a^{-1}$= e and $a^{-1}$ · a = e. – ADAM Mar 16 '16 at 04:50
  • You're supposed to prove an "if and only if" statement. That means you need to prove both that $S$ is a group $\implies aS=S=Sa$ and that $aS=S=Sa\implies S$ is a group. Try tackling these one at a time: Can you show that if $S$ fulfills the conditions you posted, that it follows that $aS=S=Sa$? – Mark Schultz-Wu Mar 16 '16 at 04:53
  • I understand that I have to do it in two directions. Would you apply the first one for me and I will try to work in the others. And would you give me an idea about the other direction. Thank you for helping – ADAM Mar 16 '16 at 05:05
  • @Mark Since $S$ is a group it's closed to any $\forall a,s \in S \ \ as \in S$ , but how can we says that $aS=Sa$ That'd be showing the cosets are equal right? So we'd be showing it's normal correct? Am I even on the right path.... – Fmonkey2001 Mar 18 '16 at 01:17

3 Answers3

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Here is an outline, see if you can fill in the details.

Pick a specific $a\in S$. Then $a\in aS$, so $a=ae$ for some $e$.

(Aim: this means that $S$ has no possible identity element except $e$, so we have to prove that $e$ actually is an identity.)

For any $x\in S$ we have $x\in Sa$, say $x=ya$. Therefore $$xe=yae=ya=x\ .$$ This proves that $e$ is a right identity, that is, $xe=x$ for all $x\in S$.

See if you can explain for yourself why every element has a right inverse: that is, for all $x$ there exists $y$ such that $xy=e$.

To show that the right inverse is also a left inverse, let $xy=e$ and $yz=e$, then simplify $$yxyz$$ in two ways to show that $yx=e$.

Finally, show that $e$ is also a left identity.

David
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  • The right identity makes sense and the left identity was simple too, especially since you gave the way to do the right identity. I had a question on the right inverse equaling the left inverse. You let y be the right inverse of x and the left inverse of z. Why though did you write yxyz? After we show y is a left inverse of z and right inverse of y would we have to show that x=z? Don't know if my question makes sense, but i'm not sure on which part of this I'm not understanding. I'll post back if something clicks. – Fmonkey2001 Mar 18 '16 at 01:11
  • Considering $yxyz$ is a "bright idea" which I saw somewhere and I don't think I can give any prior justification. But if you simplify it in two ways you will find that it gives $yx=e$. – David Mar 18 '16 at 01:15
  • I'll trust you on that. I'll also try to find where it came from for fun. I'll work on the simplification now. Thanks for the help.

    Are we setting yxyz equal to e?

     – Fmonkey2001 Mar 18 '16 at 01:18
  • @david would you please help me out with left identity. – ADAM Mar 22 '16 at 00:08
  • Assuming you have shown that every $x$ has a (left and right) inverse $y$, we have $ex=xyx=\cdots,$. See if you can finish this. – David Mar 22 '16 at 00:25
  • for the left identity would I do. for any $a \in S$ we have $a \in aS$ , say $x=xa$ then $ex=eax=ax=x$. Thus. $ex=x$ which prove that the left and right identities are equal i.e $xe=ex$ – ADAM Mar 22 '16 at 00:54
  • @David I really do appreciate it if you would help me out with that. – ADAM Mar 22 '16 at 01:34
  • I think you have used $ea=a$, which is only true if you already know $e$ is a left identity. Try simplifying $xyx$ as I suggested before. – David Mar 22 '16 at 01:35
  • I did not get the point yet. would you please say it any different way. – ADAM Mar 22 '16 at 01:42
  • If $xy=e$ can you simplify $yx$? Can you simplify $xe$? – David Mar 22 '16 at 02:06
  • $yx=e$ since we have associative and $xe=x$ – ADAM Mar 22 '16 at 02:56
  • So I would have $ex=xyx=xe=x$ – ADAM Mar 22 '16 at 03:24
  • If so now I have the left and right identities are equal $ex=xe$ ? – ADAM Mar 22 '16 at 03:25
  • Got it ;) $,!$ – David Mar 22 '16 at 05:58
  • Thanks a lot. Would you explain the right inverse a little bit. – ADAM Mar 22 '16 at 12:53
  • @David would you explain why every element has a right inverse in the form that you have given ? – ADAM Mar 22 '16 at 14:37
  • any one could help me out with that why every element has a right inverse in the form $xy=e$ – ADAM Mar 22 '16 at 17:43
  • Let us continue this discussion in chat. – ADAM Mar 22 '16 at 19:07
  • Fix $x$. By assumption $xS=S.$ We have $e\in S=xS$, so there exists $y$ such that $xy=e$. – Shaun Apr 13 '17 at 14:02
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Let $a \in S$. Since $aaS = S = Saa$, there exist $b, c \in S$ such that $(aa)b = a$ and $c(aa) = a$.

Setting $e = ca$, one gets $e = ca = c(aa)b$ and $c(aa)b = ab$. Therefore $e = ca = ab$ and $ee = (ca)(ab) = c(aa)b = e$. In particular $e$ is idempotent.

Let now $x \in S$. Since $eS = S$, there exist $y, z \in S$ such that $ey = x$ and $ze = x$. Then $ex = eey = ey = x$ and $xe = zee = ze = x$. Thus $S$ is a monoid with identity $e$. Finally, as $xS = S$ and $Sx = S$, there exist $u, v \in S$ such that $xu = e$ and $vx = e$. Thus $S$ is a group.

J.-E. Pin
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This is a summary proof partially inspired by @David.

Proof.

($S$ has an id) Let $a$ be any item in $S$. Because $aS=S$, exists $e\in S$ such that $ae=a$ ($e$ is the right-id of $a$). Need to show that $e$ is actually the right-id of all items in $S$ (not just $a$).

To do this, let $x$ be any item in $S$. Since $S=Sa$, exists $y\in S$ such that $x=ya$. Therefore, $xe=yae=ya=x$. That is to say, $e$ is the right-id of $x$ (or any item in $S$). Samely, we can show that $S$ has a left-id $e'$.

Since $e'$ is the left-id, $e'e=e$. Since $e$ is the right-id, $e'e=e'$. That is to say, $e=e'e=e'$, and hence $S$ has a unique id $e$.

(Any item in $S$ has an inverse) Let $a\in S$. Becuase $aS=S$, exists $a^{-1}\in S$ such that $aa^{-1}=e$ ($a$ has a right-inverse). Samely, since $Sa=S$, $a$ has a left-inverse $A^{-1}$ such that $A^{-1}a=e$.

Since $aa^{-1}=e$, $A^{-1}aa^{-1}=A^{-1}e$, $ea^{-1}=A^{-1}$, $a^{-1}=A^{-1}$. That is to say, $a$ has a unique inverse.

($S$ is a group) Follows immediately from above. $\blacksquare$

apen
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