Let $p$ and $q$ be primes, and the order of a group $G$ be $pq$. If there is $a,b$ in the group $G$ such that
$$ab^p = b^qa,$$
then show that $b=e.$
My attempt:
Group $G$ is cyclic because $p$ and $q$ are both primes and therefore the group is abelian.
$a^{-1}ab^p= a^{-1}b^qa$ and I get $b^p=b^q$ because $a^{-1}b^qa = b^q$ do to that the group is abelian. Does this imply that $b=e$?
I don't understand why
I am assuming $p$ and $q$ are distinct. If not then $p=q$ and a cyclic group $G$ with $pq=p^2$ elements gives a counterexample; indeed let $b$ be any generator of $G$.
Let $r$ be the smallest positive integer such that $b^r =e$. Then $r\mid pq$. [Indeed, the set $\{b,b^2,\ldots, b^{r-1}, e\}$ of elements generated by $b$ has cardinality precisely $r$ and forms a subgroup of $G$.] So $r$ must be either $1,p,q,$ or $pq$. Now if $r$ is 1, then we are done: then $b^r=b^1=b=e$. So we show that $r$ cannot be $p,q$, nor $pq$.
So suppose that $r = p$. Then $b^r=b^p=e$ so plugging this into $ab^p=b^qa$, we'd get $ae =b^qa$ which would imply $a=b^qa$ which would imply $b^q=e$. So then both $b^p=e$ and $b^q=e$ would hold for prime distinct $p,q$. This and Claim 1 [below] however, together imply that then there would be a positive integer $c<\min\{p,q\}$ so that $b^c=e$. So $r$ as defined cannot be $p$ after all.
Likewise $r$ cannot be $q$ after all either.
So now suppose that $r=pq$. But then this would imply $G$ is a cyclic group generated by $b$ and so $G$ is abelian. But then this gives $ab^p = ab^q$ which in turn gives [assuming $q> p$ WLOG] $a=ab^{q-b}$ which in turn gives $e=b^{q-p}$ which implies $r \le q-p < pq$ a contradiction. So $r$ cannot be $pq$ either.
So the only possibility left now is $r$ must be 1 and so $b^1=b^r=b=e$ it must be.
Below is Claim 1. In fact, Claims 2 and 3 are also true and are included for completeness, but they aren't really needed here.
Claim 1: Let $b$ be a group element, and let $p$ and $q$ be relatively prime positive integers satisfying $b^p=b^q=e$. There is a positive integer $c<\min\{p,q\}$ such that $b^c=e$.
Claim 2: Let $b,p$, and $q$ be as in Claim 1. Then in fact $b^1=b=e$.
Proofs of Claims 1 and 2: Let us assume WLOG that $p<q$. Divide $q$ by $p$ to get $q=mp+c$ where $m$ is an integer, and where $c$ is the remainder $c=p \mod q$.
We now show Claim 1. That $p$ and $q$ are relatively prime implies that when $q$ is divided by $p$ there has to be a nonzero remainder and so $c$ is positive and less than $\min\{p,q\}$. So we now claim that $b^c=e$. Indeed, $b^q=e$ and $q=mp+c$ together give $b^{mp+c}=b^{mp}b^c=e$. This and that $b^p=e$ gives $e=b^q=b^{mp}b^c =(b^{p})^mb^c = e^mb^c = eb^c$. So this string of equations imply $e=eb^c$, which gives $b^c =e$, and so Claim 1 follows.
We now show Claim 2. We use induction on $p+q$. Repeat the reasoning as in the proof of Claim 1 to get $b^c=e$. In fact, because $p$ and $q$ are relatively prime, $c$ as above also satisfies the following: $c$ and $p$ are relatively prime as well. Thus $b^c=e$ and $b^p=e$ for two relatively prime integers $p$ and $c$ satisfying $p+c < p+q$. So by induction it follows that $b^1=b=e$ and so Claim 2 follows as well. $\surd$
In fact the following is also true:
Claim 3: Let $b$ be a group element, and let $p$ and $q$ be [not necessarily relatively prime] positive integers satisfying $b^p=b^q=e$ then $b^{r'}=e$ for $r' =\gcd(p,q)$.
