Example of a subgroup of $S_6$, of order $8$.

Question

Give an example of a subgroup of $S_6$, of order $8$.

(if useful, use $\sigma=(1~4)(2~5)(3~6),~\tau=(1~3~5)(4~6~2),$ where $\sigma \tau\sigma=\tau$).

Attempt. Every cyclic subgroup of $S_6$ can not have order $8$, since this would require a permutation whose order is $8$, which is not possible. Of course some combination of permutations $\sigma,\tau$ could provide such subgroup. But I didn't manage it so far.

Thank you for the help in advance.

Answer 1

I tried, but I couldn't really get why they talk about $\tau = (1\ 3\ 5)(4\ 6\ 2)$.

Consider instead $\sigma = (1~2)$ and $\tau = (3~4~5~6)$.

Notice how $\sigma$ has order $2$ and $\tau$ has order $4$. Now notice that $\sigma \tau = \tau \sigma$.

Now take $G = \langle \sigma, \tau \rangle = \{1, \tau, \tau^2, \tau^3, \sigma, \sigma\tau, \sigma\tau^2, \sigma\tau^3 \}$ has order $8$ as desired.

Answer 2

If you take the subgroup generated by three elements (a,b), (c,d) and (e,f), each of order 2 and they commute with each other so it is isomorphic to Elementary abelian 2- group of order 8.