# Subgroup generated by two cycles in $S_5$.

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.

• The important observation here is that $\sigma\tau\sigma^{-1} = \tau^{-1}$. Aug 17 '19 at 20:22
• Possibly useful: each of these is a symmetry of the regular penatabon with vertices sequentially labeled. 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 $$\{1,\tau,\tau^2,\tau^3,\tau^4,\sigma,\sigma\tau,\sigma\tau^2,\sigma\tau^3,\sigma\tau^4\}$$.