Group Theory - order of element

**artvandalay11** · December 1st 2009, 06:05 PM

Determine the order of $a^m$ where $a$ is an element of order $n$ 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 $m$ and $n$ are, but maybe I'm wrong.

$a^n=1$ (the identity)

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

**Drexel28** · December 1st 2009, 06:18 PM

Originally Posted by artvandalay11

Determine the order of $a^m$ where $a$ is an element of order $n$ 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 $m$ and $n$ are, but maybe I'm wrong.

$a^n=1$ (the identity)

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

We kind of need this lemma

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

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

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

**aman_cc** · December 1st 2009, 09:39 PM

Originally Posted by Drexel28

We kind of need this lemma

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

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

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

Hi - Is your final result correct?

$\left|a^m\right|=\text{lcm}(m,n)$

I trust $\left|a^m\right|=n/(m,n)$ where (m,n) is gcd(m,n)

**lepton** · December 1st 2009, 10:56 PM

$\text{lcm}(m,n)=(mn)/\gcd(m,n)$

**aman_cc** · December 1st 2009, 11:28 PM

Originally Posted by lepton

$\text{lcm}(m,n)=(mn)/\gcd(m,n)$

Right. But what I'm saying is that order is $n/\gcd(m,n)$

**Drexel28** · December 2nd 2009, 04:50 AM

Originally Posted by aman_cc

Right. But what I'm saying is that order is $n/\gcd(m,n)$

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

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