$a,b$ are elements in a group $G$. Let $o(a)=m$ which means that $a^m=e$, $\gcd(m,n)=1$ and $(a^n)*b=b*(a^n)$. Prove that $a*b=b*a$.
Hint: try to solve for $m=5,n=3$.
I am stuck in this question and can't find an answer to it, can anyone give me some hints?
Thanks.
On
Saying that the order of $a$ is $m$ is much more than saying that $a^m=e$. On the other hand, if $a^m=1$ (for $m>0$) then the order of $a$ is a divisor of $m$, so $\gcd(m,n)=1$ also implies that $\gcd(o(a),n)=1$.
You have $a^nb=ba^n$, so also $ba^{2n}=ba^na^n=a^nba^n=a^na^nb=a^{2n}b$. This hints that $$ ba^{nr}=a^{nr}b $$ for every integer $r$. Prove it (induction for positive $r$ and a simple trick for negative $r$). Now finish without looking at the spoiler (use Bézout’s identity).
If $mx+ny=1$, for integers $x$ and $y$, then $a=a^{mx+ny}=a^{mx}a^{ny}=a^{ny}$ and therefore$$ba=ba^{ny}=a^{ny}b=ab$$
$a^n b = ba^n \implies a^n = b^{-1} a^n b = (b^{-1} ab)^n = c^n$. There are $x,y \in \mathbb Z$ such that $nx + my = 1$. We have $a^{nx} = c^{nx} = c^{1-my}$, so $a^{nx} c^{my} = c$. But $c^{my} = (c^m)^y = ((b^{-1} a b)^m)^y= (b^{-1} a^m b)^y = (b^{-1} b)^y = e$. Also $a^{nx} = a^{1-my} = a ((a^{-1})^m)^y = a e = a$ (the order of $a$ is the same as that of $a^{-1}$). Thus $a = c = b^{-1}a b$ and so $ba = ab$.