# Proving the order of a group element

1. Oct 31, 2011

### Chaos2009

1. The problem statement, all variables and given/known data

The problem states that $G$ is a group, $a \in G$ s.t. $o\left(a\right)=dk$, and $x \in G$ s.t. $x^{d}=a$. Prove that $o\left(x\right)=d^{2}k$ where $d,k > 1$.

Note: $o\left(x\right)$ denotes the order of element $x$ in $G$.

2. Relevant equations

3. The attempt at a solution

So, it's rather easy to show that $x^{d^{2}k} = e$ (where $e$ is the identity element of $G$), but I'm having trouble proving that it is the order of $x$.

My thoughts were to try to assume $x^{n}=e$, then divide $n$ by $d$ to get $n=qd+r \Rightarrow x^{qd+r}=x^{qd}x^{r} = {x^{d}}^{q}x^{r} = a^{q}x^{r} = e$. If I could somehow prove $r=0$, I think I could finish the proof, however, I cannot see how to proceed. If someone could point me in the right direction, it would be much appreciated.

2. Oct 31, 2011

### Deveno

instead of dividing n by d, try dividing it by dk.....(the idea is to prove the order of x is a multiple of dk).

then show that if xdkq = e, with 0 < q < d, you get a contradiction (now use what you know about a).