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Let $\sigma$, $\tau$ be two permutations of $S_n$.

We know that $\langle \sigma \rangle \cap \langle \tau \rangle$ is a cyclic subgroup and (from Lagrange's Theorem) that its order divides $$\gcd(\left|\langle \sigma \rangle\right|, \left|\langle \tau \rangle\right|).$$

Do we have other results that offer us a way to find $k$, $h$ such that $$\langle \alpha\rangle = \langle \sigma \rangle \cap \langle \tau \rangle = \langle \sigma^k \rangle = \langle \tau^h \rangle,$$ that is, to find the order and a generator of the intersection?


For example: how would we proceed if we consider in $S_{15}$ $$\sigma = (1,2,3,4)(5,6,7,8,9)(10,11,12)(13,14)$$ $$\tau = (1,3) (2,4)(5,6,7,8,9)(10,12,11)(13,14,15) $$ and want to find $\langle \sigma \rangle \cap \langle \tau \rangle$? From the Lagrange's theorem we only know that its order divides $30$.

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When proving that the intersection of two cyclic subgroups is again cyclic, one can do it constructively by showing that (I'm just using your notation above)

$$\sigma^k, k :=\min\{m \in \mathbb{N}_{\geq 1} \mid \sigma^m \in \langle \alpha \rangle\}$$

is a generator of $\langle \alpha \rangle$.

So regarding your example, we are looking for a minimal $m$, such that $\sigma^m = \tau^n$ for an $n \in \mathbb{Z}$. Since we are dealing with cycles, we can assume that $n \in \mathbb{N}$. Now $\sigma^m(15) = 15$ $\forall m$, so looking at the last cycle of $\tau$, which is $(13, 14, 15)$, $n$ must be a multiple of 3. Then $\tau^n(j) = j$ $\forall j \in \{10,...,15\}$. In order to achieve this for $\sigma^m$ as well, $m$ must be a multiple of 2 and 3, so $6\mid m$. Assume $m = 6$. This leaves the 5-cycle $(5,6,7,8,9)$ unaltered, so $n \equiv 1 \bmod 5$.

In fact

$$\sigma^{6} = (1,3)(2,4)(5,6,7,8,9) = \tau^{21}.$$

Now $\langle \alpha \rangle = \langle \sigma^6 \rangle$ with $\mathrm{ord}(\sigma^6) = 10$.

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  • $\begingroup$ Hi, I'm trying to solve a similar problem and I'm trying to apply the reasoning you explained. The only thing that I haven't clear is the point where you say $\sigma^m(15) = 15$. Why do you affirm this? Because $\sigma$ fixes 15 or because of the two permutations belonging to $S_{15}$? Thanks. $\endgroup$ – PCNF Jan 26 '19 at 15:54
  • $\begingroup$ @PCNF A bit old, sorry, but still: Yes, because $\sigma$ fixes $15$. $\endgroup$ – johnnycrab Feb 20 '19 at 8:20

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