# Thread: question on order of element

1. ## question on order of element

$G$ be a group. [tex]a,b /in G[\math]
Order of $a = m$
Order of $b = n$

Prove there exists an element in $G$ with order = LCM $(m,n)$. Let this LCM be $l$

My approach:
Consider $ab$
Let order of $ab$ be $o$
Obviously $o|l$ (as $(ab)^l=1$)
Only two cases are possible
Case 1: $m|o$ AND $n|o$ - we are done here
Case 2: $m$ doesn't divide $o$ AND $n$ doesn't divide $o$
In this case we have $a^x$ = $b^y$ where $0 and $0
Stuck here for quite some time.

Is my approach correct? If yes, hint on way fwd please?

2. Originally Posted by aman_cc
$G$ be a group. [tex]a,b /in G[\math]
Order of $a = m$
Order of $b = n$

Prove there exists an element in $G$ with order = LCM $(m,n)$. Let this LCM be $l$

My approach:
Consider $ab$
Let order of $ab$ be $o$
Obviously $o|l$ (as $(ab)^l=1$)
Only two cases are possible
Case 1: $m|o$ AND $n|o$ - we are done here
Case 2: $m$ doesn't divide $o$ AND $n$ doesn't divide $o$
In this case we have $a^x$ = $b^y$ where $0 and $0
Stuck here for quite some time.

Is my approach correct? If yes, hint on way fwd please?
this is true in abelian groups only!

3. Originally Posted by NonCommAlg
this is true in abelian groups only!
Apologies - G is abelian as well.
Any hints? Am I on the right track?

4. Originally Posted by aman_cc
Apologies - G is abelian as well.
Any hints? Am I on the right track?
first prove it for the case $\gcd(m,n)=1.$ (very easy!) can you do that?

5. Originally Posted by NonCommAlg
first prove it for the case $\gcd(m,n)=1.$ (very easy!) can you do that?
Yes I can do that. In that case required element is $ab$
If order is $o$
We have $m|on$ and $n|om$
Under the $\gcd(m,n)=1$ condition I get
$m|o$ and $n|o$ and we are done (as per case 1 in my logic above). Correct?

6. Originally Posted by aman_cc
Yes I can do that. In that case required element is $ab$
If order is $o$
We have $m|on$ and $n|om$
Under the $\gcd(m,n)=1$ condition I get
$m|o$ and $n|o$ and we are done (as per case 1 in my logic above). Correct?
[so $mn \mid o(ab)$ and we also have $(ab)^{mn}=a^{mn}b^{mn}=1$ and so $o(ab) \mid mn.$ thus $o(ab)=mn=\text{lcm}(a,b).$] now we can prove the claim for the general case: clearly we may assume that $m \nmid n.$

let $\text{lcm}(m,n)=\prod_{j=1}^r p_j^{k_j}$ be the prime factorization of $\text{lcm}(m,n).$ let $A=\{j: \ p_j^{k_j} \mid m \} \neq \emptyset,$ because $m \nmid n.$ let $s=\prod_{j \in A} p_j^{k_j}$ and $t=\frac{\text{lcm}(m,n)}{s}.$ obviously $s \mid m, \ t \mid n,$ and $\gcd(s,t)=1.$ therefore,

since $o(a^{m/s})=s$ and $o(b^{n/t})=t,$ we have $o(a^{m/s}b^{n/t})=st=\text{lcm}(m,n).$

7. Originally Posted by NonCommAlg
[so $mn \mid o(ab)$ and we also have $(ab)^{mn}=a^{mn}b^{mn}=1$ and so $o(ab) \mid mn.$ thus $o(ab)=mn=\text{lcm}(a,b).$] now we can prove the claim for the general case: clearly we may assume that $m \nmid n.$

let $\text{lcm}(m,n)=\prod_{j=1}^r p_j^{k_j}$ be the prime factorization of $\text{lcm}(m,n).$ let $A=\{j: \ p_j^{k_j} \mid m \} \neq \emptyset,$ because $m \nmid n.$ let $s=\prod_{j \in A} p_j^{k_j}$ and $t=\frac{\text{lcm}(m,n)}{s}.$ obviously $s \mid m, \ t \mid n,$ and $\gcd(s,t)=1.$ therefore,

since $o(a^{m/s})=s$ and $o(b^{n/t})=t,$ we have $o(a^{m/s}b^{n/t})=st=\text{lcm}(m,n).$
Thanks I get this. In the case where gcd(m,n) <> 1 we found a element with order equal to LCM. Can we comment on o(ab) as well? (In terms of prime factorization you did)

All I could show was

$\frac{mn}{\gcd(m,n)^2}|o(ab)$

Is the above correct? Can we get more specific?

Thanks