3
$\begingroup$

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.

$\endgroup$
2
  • 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
1
$\begingroup$

$o(\sigma) = 2$, $o(\tau) = 5$ 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 $\{1,\tau,\tau^2,\tau^3,\tau^4,\sigma,\sigma\tau,\sigma\tau^2,\sigma\tau^3,\sigma\tau^4\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.