# order of an element of a group

• Apr 29th 2010, 02:17 PM
manik
order of an element of a group
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.
• Apr 29th 2010, 03:07 PM
GnomeSain
Quote:

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 $\displaystyle (ab)^{mn} = e$.

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.
• Apr 29th 2010, 06:54 PM
Bruno J.
Quote:

Originally Posted by GnomeSain
It's obvious that $\displaystyle (ab)^{mn} = e$.

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 $\displaystyle (ab)^k=1=a^kb^k$ (where the last equality is justified because $\displaystyle a,b$ commute). Use the division algorithm to write $\displaystyle k=q_1m+r_1=q_2n+r_2$, with $\displaystyle 0\leq r_1 <m$ and $\displaystyle 0\leq r_2 <n$. Then $\displaystyle a^{r_1}b^{r_2}=1$. Now raise this expression to the power $\displaystyle m$, eliminating $\displaystyle a$ from the left side; you obtain $\displaystyle b^{r_2m}=1$. Now we know $\displaystyle b^{m}$ has the same order as $\displaystyle b$ (prove it, using $\displaystyle (m,n)=1$), and therefore we must have $\displaystyle n\mid r_2$, which implies $\displaystyle r_2=0$. Similarily we must have $\displaystyle r_1=0$. Therefore $\displaystyle k$ is divisible both by $\displaystyle m,n$ and therefore divisible by $\displaystyle mn$.
• Apr 30th 2010, 12:55 PM
GnomeSain
Quote:

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

Thanks Bruno. I was very careless.