Find all the elements in the intersection $\langle \alpha\rangle \cap \langle \beta \rangle$ Find all the elements in the intersection $\langle \alpha\rangle \cap \langle \beta \rangle$ of    $\langle \alpha \rangle , \langle \beta \rangle$ where $\alpha, \beta \in S_9$.
$\beta =(521)(96347)(8), \alpha = (15)(37964)(8)(2)$
I know that $|\beta|=15, |\alpha|=10$,
so $\langle \beta \rangle = ( \beta,  \beta^1..., \beta^{15})$ and $\langle \alpha \rangle = ( \alpha,  \alpha^1..., \alpha^{9})$ and so I could just check manually, but is there a more efficient solution that i am not seeing?
Any input would be appreciated,
Thanks
 A: I will assume that you meant $S_9$. By Lagrange's theorem we know that $|<\alpha>\cap<\beta>|$ divides $|<\alpha>|,|<\beta>|$. Thus, $|<\alpha>\cap<\beta>|$ divides $10,15$. Hence $|<\alpha>\cap<\beta>|$ divides 5. Now there are two cases. Either $|<\alpha>\cap<\beta>|$  is trivial (Now things become easy) or $|<\alpha>\cap<\beta>|$  is cyclic of order 5. In the second case it suffices to find only one non-identity element $x\in <\alpha>\cap<\beta>$  to find $<\alpha>\cap<\beta>$, because $<\alpha>\cap<\beta>=\{e,x,x^2,x^3,x^4\}$ (This is because 5 is prime)   
If $<\alpha>\cap<\beta>$  is not trivial:
Let $x\in|<\alpha>\cap<\beta>|$. Thus, $x=\alpha^i$ for some $i\in\{1,2,...,9\}$. Since $(\alpha^i)^5=e$, thus $i\in\{0,2,4,6,8\}$. Hence $<\alpha>\cap<\beta>\leq<\,\alpha^2>$. Since $|<\alpha>\cap<\beta>|=|<\alpha^2>|=5$, thus $<\alpha^2>=<\alpha>\cap<\beta>$. Hence, $\alpha^2\in\alpha>\cap<\beta>$.
Conclusion: $<\alpha>\cap<\beta>$  is not trivial implies $\alpha^2\in<\beta>$. So you just need to check if $\alpha^2\in<\beta>$
A: Your observation on the orders of the two permutations is correct, $|\alpha|=10$ and $|\beta|=15$. By Lagrange's theorem, $|<\alpha>\cap<\beta>|$ must divide the greatest common divisor of these two numbers, which is the prime number 5. To see if there is a common element in $<\alpha>\cap<\beta>$ raise $\alpha$ to a power that will make all but the 5-cycle singletons. Do the same for $\beta$. Now look to see if the two 5-cycles represent powers of the same permutation. The only operations you need to use are compositions of permutations in cycle notation.
