order of an element of a group

**manik** · April 29th 2010, 02:17 PM

I need the Proof:

a, b are two elements of a group G such that ab=ba.
If o(a)=m & o(b)=n and m,n are relatively prime to each other,then
o(ab)=mn.

**GnomeSain** · April 29th 2010, 03:07 PM

Originally Posted by manik

I need the Proof:

a, b are two elements of a group G such that ab=ba.
If o(a)=m & o(b)=n and m,n are relatively prime to each other,then
o(ab)=mn.

It's obvious that $(ab)^{mn} = e$ .

To show that mn is the smallest such number, note that the order of ab must be a multiple of both a and b. Depending on how strictly your teacher grades, you may want to explicitly justify this statement. Since m and n are relatively prime the smallest such number is their LCM, which is mn.

**Bruno J.** · April 29th 2010, 06:54 PM

Originally Posted by GnomeSain

It's obvious that $(ab)^{mn} = e$ .

To show that mn is the smallest such number, note that the order of ab must be a multiple of both the orders of a and b. Depending on how strictly your teacher grades, you may want to explicitly justify this statement. Since m and n are relatively prime the smallest such number is their LCM, which is mn.

Yes, and that's what the problem essentially is about!
Suppose $(ab)^k=1=a^kb^k$ (where the last equality is justified because $a,b$ commute). Use the division algorithm to write $k=q_1m+r_1=q_2n+r_2$ , with $0\leq r_1 <m$ and $0\leq r_2 <n$ . Then $a^{r_1}b^{r_2}=1$ . Now raise this expression to the power $m$ , eliminating $a$ from the left side; you obtain $b^{r_2m}=1$ . Now we know $b^{m}$ has the same order as $b$ (prove it, using $(m,n)=1$ ), and therefore we must have $n\mid r_2$ , which implies $r_2=0$ . Similarily we must have $r_1=0$ . Therefore $k$ is divisible both by $m,n$ and therefore divisible by $mn$ .

**GnomeSain** · April 30th 2010, 12:55 PM

Originally Posted by Bruno J.

Yes, and that's what the problem essentially is about!
Suppose $(ab)^k=1=a^kb^k$ (where the last equality is justified because $a,b$ commute). Use the division algorithm to write $k=q_1m+r_1=q_2n+r_2$ , with $0\leq r_1 <m$ and $0\leq r_2 <n$ . Then $a^{r_1}b^{r_2}=1$ . Now raise this expression to the power $m$ , eliminating $a$ from the left side; you obtain $b^{r_2m}=1$ . Now we know $b^{m}$ has the same order as $b$ (prove it, using $(m,n)=1$ ), and therefore we must have $n\mid r_2$ , which implies $r_2=0$ . Similarily we must have $r_1=0$ . Therefore $k$ is divisible both by $m,n$ and therefore divisible by $mn$ .

Thanks Bruno. I was very careless.

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