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
On
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.
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 $qr\ne p$.