# Group Theory Simple Proof

I am new at group theory, and I came across a question I would like help with

Suppose we have a set $S$ with the only elements $p,q,r$. Let $a$ and $b$ be two elements of $S$. Consider the following properties of $S$:

1) $aa=a$

2) $ab=ba$

3) $(ab)c=a(bc)$

4)$pa=a$ for every element $a$

Prove that there exists some element in $b \in S$ such that $bp=b, bq=b, br=b$.

Thank you! I am new in Group Theory so i was just looking for some help

-
I assume that you were also told that the "product" of any two elements of $S$ is in $S$. Note then that we know almost all the "multiplication table" for $S$, the only thing missing is what is $qr$. Could $qr$ be $p$? Could it be $q$? $r$? Hint: Use 4). – André Nicolas Jan 28 '12 at 0:34
@André: I don’t think that you can determine which of the non-identity elements $qr$ is. – Brian M. Scott Jan 28 '12 at 0:53
@Brian M. Scott: One cannot, but there is symmetry, so we don't care. – André Nicolas Jan 28 '12 at 1:00
@André: Agreed. It just seemed to me that a natural interpretation of the hint for someone having trouble with the original problem would be that you can determine that product. – Brian M. Scott Jan 28 '12 at 1:06
I changed the tags from [group-theory] and [finite-groups] to [semigroups], because no group can satisfy the given conditions. – Arturo Magidin Jan 28 '12 at 1:25

The wording is a little unclear, but I’m assuming that (1)-(3) hold for every choice of $a,b$, and $c$ in $S$.

You know, using (2) and (4), that $bp=pb=b$ for any choice of $b$, so it’s $bq$ and $br$ that you have to worry about. If $b=q$, you’ll get $bq=qq=q$, but it’s clear whether $br=qr$ will be equal to $q$. A similar problem arises if $b=r$: we don’t know whether $rq=r$ or not. Of course by (2) $rq=qr$, so the real question is what $qr$ is. All we know is that it must be one of $q$ and $r$ if such a $b$ is going to exist. So why not just try letting $b=qr$, whatever it is?

Then $(qr)p=p(qr)=qr$ by (2) and (4), $(qr)q=(rq)q=r(qq)=rq=qr$ by (2), (3), (1), and (2) again, and $(qr)r=q(rr)=qr$ by (3) and (1).

However, there’s a potential trap here: we want $qr$ to be whichever one of $q$ and $r$ makes this work, but how do we know that $qr$ isn’t actually $p$ instead? The answer is in Henry’s comment below: if $qr$ were equal to $p$, we’d have $(qr)r=pr=r$ by (4), but also $(qr)r=q(rr)=qr=p$ by (3) and (1), and $r$ would have to be equal to $p$, which we know is not true. We can be assured, therefore, that $qr\ne p$.

-
+1 for using $b=qr$, but I think you may have to add that $qr \not = p$, since otherwise $(qr)r=r$ while $q(rr)=p$ in contradiction to (3) – Henry Jan 28 '12 at 1:08
@Henry: I was thinking of that as part of finishing off the $(qr)r$ case, but that probably is a bit much to ask. On further consideration, I think that I’ll turn it into a full answer. – Brian M. Scott Jan 28 '12 at 1:17

We complete the multiplication table for $S$, under the assumption that the product of two things in $S$ is also in $S$. (That fact was not explicitly mentioned.)

Could we have $qr=p$? Then $(qr)r=pr=r$. But $(qr)r=q(rr)=qr=p$. That would make $(qr)r$ simultaneously equal to $p$ and $r$, which is impossible.

That leaves the possibilities $qr=q$ and $qr=r$. Suppose first that $qr=q$. Then let $b=q$. We have by the rules $qp=q$, $qq=q$, and $qr=q$, and we are finished.

By symmetry, if $qr=r$ then we can take $b=r$. Or else we can more or less repeat the previous argument, interchanging the roles of $q$ and $r$.

Remark: But we really ought to check that we have not been lied to, that if for example we complete the multiplication by $qr=q$, all the given rules will be satisfied. After all, we could have been given a collection of "rules" that turn out to be inconsistent.

The only thing that really needs checking is $4$). That takes some work. In principle we need to check that $a(bc)=(ab)c$ for all $27$ triples $(a,b,c)$. With some thinking we can cut it down to a lot less checking than that. If there is at least one $p$ among $a$, $b$, and $c$, verification is easy. so we only need to worry when each of $a$, $b$, $c$ is $q$ or $r$. So we are down to $8$ cases, easily checked.

-