# 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

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How come $\alpha\in S_4$ and 9 appears in the product of cycles of $\alpha$ ? – Amr Feb 10 '13 at 23:50
@Amr you are right, my mistake, fixed it – bobdylan Feb 10 '13 at 23:53

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 $\alpha^2\in<\beta>$

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Thats excellent, im checking if $\alpha^2 \in \langle B \rangle$ but i have an intuition that this can never be since $\beta$ can be expressed as 6, 2 cycles and $\alpha$ can be expressed as 5, 2 cyles so it has to be the trivial case, Is my logic correct? – bobdylan Feb 11 '13 at 2:23

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.

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