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Thread: order of an element of a group

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by manik View Post
    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.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by GnomeSain View Post
    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 $\displaystyle mn$.
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  4. #4
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    Quote Originally Posted by Bruno J. View Post
    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.
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