I'm trying to determine, for $\sigma := (15)(24), \tau := (12345) \in S_5$, what $\langle \sigma, \tau \rangle$ is.

By definition, this would be the subgroup of $S_5$ containing all finite products of the elements $\sigma, \tau, \sigma^{-1}, \tau^{-1}$.

It seems like an extremely tedious task to actually go through all of these, although you could obviously do it in a finite time.

What I have done is compute the commutator $[\sigma, \tau] = (14253)$. Now this contains $\sigma, \tau, \sigma^{-1}, \tau^{-1}$ as a product, so I have the feeling this might provide a shortcut somehow, but I don't yet see how.

  • 2
    $\begingroup$ The important observation here is that $\sigma\tau\sigma^{-1} = \tau^{-1}$. $\endgroup$
    – Derek Holt
    Aug 17 '19 at 20:22
  • $\begingroup$ Possibly useful: each of these is a symmetry of the regular penatabon with vertices sequentially labeled. $\endgroup$ Aug 17 '19 at 20:23

$o(\sigma) = 2$, $o(\tau) = 5$ and $\sigma\tau\sigma = \tau^{-1}$. Thus $\langle\sigma,\tau\rangle = \langle\sigma,\tau\mid\sigma^2 = \tau^5 = 1,\sigma\tau\sigma = \tau^{-1}\rangle\cong D_{10}$, the dihedral group of order $10$. Therefore, the group is $\{1,\tau,\tau^2,\tau^3,\tau^4,\sigma,\sigma\tau,\sigma\tau^2,\sigma\tau^3,\sigma\tau^4\}$.


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