# Group Theory - order of element

• Dec 1st 2009, 06:05 PM
artvandalay11
Group Theory - order of element
Determine the order of $\displaystyle a^m$ where $\displaystyle a$ is an element of order $\displaystyle n$ in a group G

I don't really know what they book is looking for here. It seems to me this will break down into a bunch of cases depending on what $\displaystyle m$ and $\displaystyle n$ are, but maybe I'm wrong.

$\displaystyle a^n=1$ (the identity)

and yeah that's pretty much all I got so far
• Dec 1st 2009, 06:18 PM
Drexel28
Quote:

Originally Posted by artvandalay11
Determine the order of $\displaystyle a^m$ where $\displaystyle a$ is an element of order $\displaystyle n$ in a group G

I don't really know what they book is looking for here. It seems to me this will break down into a bunch of cases depending on what $\displaystyle m$ and $\displaystyle n$ are, but maybe I'm wrong.

$\displaystyle a^n=1$ (the identity)

and yeah that's pretty much all I got so far

We kind of need this lemma

Lemma: Let $\displaystyle g\in G$ with $\displaystyle |g|=n$. If $\displaystyle g^{m}=e$ then $\displaystyle n|m$.

Proof: By the division algorithim $\displaystyle m=qn+r\quad q\in\mathbb{Z},0\le r<n$. So then $\displaystyle g^m=g^{qn+r}=\left(g^n\right)^qg^r=g^r=e$ and since $\displaystyle 0\le r<n$ and $\displaystyle n$ is the least positive integer such that $\displaystyle g^k=e$ we must have that $\displaystyle r=0$. Therefore $\displaystyle m=qn\quad q\in\mathbb{Z}$ and the conclusion follows. $\displaystyle \blacksquare$

So basically we need the least number $\displaystyle k$ such that $\displaystyle n|mk$, i.e. we need $\displaystyle \left|a^m\right|=\text{lcm}(m,n)$
• Dec 1st 2009, 09:39 PM
aman_cc
Quote:

Originally Posted by Drexel28
We kind of need this lemma

Lemma: Let $\displaystyle g\in G$ with $\displaystyle |g|=n$. If $\displaystyle g^{m}=e$ then $\displaystyle n|m$.

Proof: By the division algorithim $\displaystyle m=qn+r\quad q\in\mathbb{Z},0\le r<n$. So then $\displaystyle g^m=g^{qn+r}=\left(g^n\right)^qg^r=g^r=e$ and since $\displaystyle 0\le r<n$ and $\displaystyle n$ is the least positive integer such that $\displaystyle g^k=e$ we must have that $\displaystyle r=0$. Therefore $\displaystyle m=qn\quad q\in\mathbb{Z}$ and the conclusion follows. $\displaystyle \blacksquare$

So basically we need the least number $\displaystyle k$ such that $\displaystyle n|mk$, i.e. we need $\displaystyle \left|a^m\right|=\text{lcm}(m,n)$

Hi - Is your final result correct?
Quote:

$\displaystyle \left|a^m\right|=\text{lcm}(m,n)$
I trust $\displaystyle \left|a^m\right|=n/(m,n)$ where (m,n) is gcd(m,n)
• Dec 1st 2009, 10:56 PM
lepton
$\displaystyle \text{lcm}(m,n)=(mn)/\gcd(m,n)$
• Dec 1st 2009, 11:28 PM
aman_cc
Quote:

Originally Posted by lepton
$\displaystyle \text{lcm}(m,n)=(mn)/\gcd(m,n)$

Right. But what I'm saying is that order is $\displaystyle n/\gcd(m,n)$
• Dec 2nd 2009, 04:50 AM
Drexel28
Quote:

Originally Posted by aman_cc
Right. But what I'm saying is that order is $\displaystyle n/\gcd(m,n)$

Thanks, typo. It was supposed to be $\displaystyle \frac{\text{lcm}(m,n)}{m}=\frac{n}{(m,n)}$