# Thread: Problem with order of element

1. ## Problem with order of element

Let a be in a group G with |a| = m. If n is relatively prime to m, show that a = b^n for some b in G.

My Proof so far:

Now the order of a is m, so I have a^m = e, the identity. n is relatively prime to m, so I have ng + r = m for some integers g and r, with m > r > n.

umm... what now?

Let a be in a group G with |a| = m. If n is relatively prime to m, show that a = b^n for some b in G.

My Proof so far:

Now the order of a is m, so I have a^m = e, the identity. n is relatively prime to m, so I have ng + r = m for some integers g and r, with m > r > n.

umm... what now?
I should think that this would be easier to do if you start from the idea that G contains the subset $\{ e, a, a^2, ~ ... ~, a^{m -1} \}$. Then show that one of these elements must be your "b" for a given n.

Just a thought.

-Dan

3. n is relatively prime to m, so I have
Well, everything goes well to that point Now by Euclidean division, there are integers g and r, such that gn+rm=1. So $\alpha^{gn+rm}=\alpha\Rightarrow \alpha^{gn}\alpha^{rm}=\alpha\Rightarrow (\alpha^{g})^n(\alpha^{m})^{r}=\alpha\Rightarrow (\alpha^{g})^{n}=\alpha$, that is $\beta=\alpha^{g}$.

Ps. Topsq:
Then show that one of these elements must be your "b" for a given n.
Suppose $\alpha=\beta^n=\alpha^i$ for some i<n, and prove you chose a bad start

4. We haven't learn the Euclidean division yet, so I don't think I'm allow to use that. And how to you have the (a^m)^r?

thanks

5. We haven't learn the Euclidean division yet
Really? Then what is this:

so I have ng + r = m
Use that rule and amaze everyone by showing gn+rm=1 for some g,r!

And how to you have the (a^m)^r?
a^m=e and e^r=e.

Let the order the order of $a$ be $m$. And consider the cyclic subgroup $\left< a\right> = \{a,a^2,...,a^m\}$. Now the order of this cyclic group is $m$. Thus, the generators of this group are $a^k$ where $k$ is relatively prime to $m$. In particular, $k=n$ by hypothesis. That means $\left< a^n \right> = \left< a \right>$. Now $a^n =\{ a^n,a^{2n},...,a^{mn} \}$ is a premutation of $\{ a,a^2, ... ,a^m\}$ thus $a$ is found among $\{a^n,...,a^{nm}\}$ and hence $a=a^{nj} = \left( a^j \right) ^n$. Let $b=a^j$ for some $j$. And the proof is complete.