# Trying to prove a formula for divider of the order of a group element

Let G be a group, $a, b \in G, n \in \mathbb{N}, n \geq 2$. Show that $ba = a^nb \Rightarrow ord (a)$ divides $n^{ord(b)} - 1$, where $ord(x)$ is the order of $x \in G$ with $ord(x) = min\left\{n \in \mathbb{N} : x^n = e\right\}$.

I started with a distinction of cases:

• If $a = e$, $ord(a) = 1$ simply divides every number
• If $b = e \Rightarrow b = b^n \Rightarrow ord(b) = n - 1 \Rightarrow n - 1 \ \vert \ n^1 - 1$

But what can I do if $a, b \neq e$? Can you help me to start with this? Can you help me to go on?

• I think your second case needs revising. It will be clearer if you write the statements separately instead of running them together, but it's not true that $ord(b) = n - 1$ if $b = e$. – NoName Nov 3 '14 at 17:59

You are given that $bab^{-1}=a^n$. Thus $a$ and $a^n$ are conjugates. This means their orders have to be the same. Let $|a|=t$. Then $$t=|a^n|=\frac{t}{\gcd(n,t)}$$
This gives $\gcd(n,t)=1$.
Let $|b|=k$. Now consider $(ba)^k=bababa\ldots=$
• Thanks for your answer and sorry for the late response! What do you mean by conjugate, I could only find a definition for a mapping of complex numbers. Why does this conclude that the orders of $a^n$ and $a$ are the same? I tried to reproduce your steps. Is $|a^n|$ your notation of $ord(a^n)$? – muffel Nov 5 '14 at 20:26
• Two elements $a$ and $b$ are conjugates if there exists an element $g$ in the group $G$ such that $a=gbg^{-1}$ (it forms an equivalence relation on the group and it is very important in group theory). For the second question, yes $|a^n|$ means order of the element $a^n$. – Anurag A Nov 5 '14 at 21:24