# Thread: order of an element of a group

1. ## 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.

2. 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.

3. 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 and $0\leq r_2 . 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 $mn$.

4. 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 and $0\leq r_2 . 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.