Group Q: Suppose |a|=24, find a generator for <a^m> in

[daon]

Group Q: Suppose |a|=24, find a generator for <a^m> intersection <a^n>.

The books also asks us to try this example: <a^21> intersect <a^10>, which I found to be <a^6>.

I wrote a c++ program to do this, so I found several examples including the one above. However, I cannot find a pattern.

Here are some results.. I made a function f: (m x n) -> (<a^m> int <a^n>) where |a|=24, if that makes sense.

f(21,10) = e, 6, 12, 18 = <a^6>
f(8,16) = e, 8, 16 = <a^8>
f(12,14) = e, 12 = <a^12>
f(4,18) = e, 12 = <a^12>
f(3,9)=e, 3, 6, 9, 12, 15, 18, 21 = <a^3>

I thought it would have something to do with GCD or LCM.

-Daon

 

[daon]

I think it is the LCM, as expirimentation shows that <a^210>= <a^10> n <a^21> = <a^6>. And this is true since gcd(210,24) = gcd(6,24). Is this right?

edit: I don't have to prove this, but how would one attempt this proof? I guess the proposition would be something like:

<a^n> = <a^x> n <a^y> if and only if n is a multiple of the lcm(x,y).

Would that be correct?