Abstract Algebra: Orders of Groups

**Jhevon** · March 4th 2008, 10:30 PM

Hello All,

I'm stuck on part (b)

Problem:

(a) Prove that if $a$ and $b$ are elements of an Abelian group $G$ , with $\circ (a) = m$ and $\circ (b) = n$ , then $(ab)^{mn} = e$ ( $e$ denotes the identity element in my text). Indicate where you use the condition that $G$ is Abelian.

(b) With $G,~a$ and $b$ as in part (a), prove that $\circ (ab)$ divides $\circ (a) \circ (b)$ .

(c) Give an example of an Abelian group $G$ and elements $a$ and $b$ in $G$ such that $\circ (ab) \ne \circ (a) \circ (b)$ . Compare part (b)

Things that might come in Handy:

Let $G$ be a group and $a \in G$ .

We define the order of $a$ , denoted $\circ (a)$ , to be the smallest, positive integer $n$ such that $a^n = e$ . If no such integer exists, we say $a$ has infinite order.

$a^n$ of course means $a$ operating on itself $n$ times. that is, $a^n = \underbrace{aaaa \cdots a}_{n \mbox{ times}} = a*a*a*a* \cdots *a$

Tell me if you guys need anymore information.

What I have tried:

(a) Proof: Let $a,~b,~G$ be as in part (a). Since $G$ is Abelian, $(ab)^{mn} = a^{mn}b^{mn}$

Now, $a^{mn}b^{mn} = a^{mn}b^{nm} = (a^m)^n(b^n)^m = e^ne^n = ee = e$

QED

(b) This is where I'm stuck. Here's what I did so far.

Let $a,~b,~G$ be as in part (a). And let $\circ (ab) = p$ .

Then $m,~n, \mbox{ and }p$ are all the smallest, positive integers such that: $a^m = e$ , $b^n = e$ and $(ab)^p = a^pb^p = e$ . We want to show that $p|mn$ , that is, $mn = kp$ for some $k \in \mathbb{Z}$ .

Now, $a^mb^n = ee = e = a^pb^p$

...

that's it, I'm not sure where to go from there. I've tried several things, but they all end up doing nothing.

(c) My example was $(\mathbb{Z}_3, \oplus)$ with $a = [1]$ and $b = [2]$

Thanks guys and gals

**ThePerfectHacker** · March 5th 2008, 06:07 AM

Originally Posted by Jhevon

(b) With $G,~a$ and $b$ as in part (a), prove that $\circ (ab)$ divides $\circ (a) \circ (b)$ .

Just use a theorem. Suppose $x$ is a group (any group not just abelian) $G$ and $x$ has finite order $k$ . If $n$ is any positive integer so that $x^n = e$ then $k|n$ . Now apply this theorem here.

(c) Give an example of an Abelian group $G$ and elements $a$ and $b$ in $G$ such that $\circ (ab) \ne \circ (a) \circ (b)$ . Compare part (b)

Consider $G=\mathbb{Z}_5$ and $a=[2],b=[3]$ .

**JaneBennet** · March 5th 2008, 07:48 AM

(b)
If $\mathrm{o}(x)=p$ and $x^k=e$ , write $k=qp+r$ where $0\leq r<p$ . Hence $e=x^k=(x^p)^qx^r=x^r$ and it follows that $r=0$ since p is the smallest positive integer such that x to that power is the identity.

(c)
Take a = b = any element of order 2. Then $\mathrm{o}(a\circ b)=1$ but $\mathrm{o}(a)\cdot\mathrm{o}(b)=4$ .

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