Showing that a quotient group $G/N$ is isomorphic to $\mathbb{Z}_3$

Question

I have permutations $\sigma=(135)(27)$, and $\tau = (27)(468)$. $G =\langle \sigma,\tau \rangle$ and $N$ is the smallest subgroup of $G$ that contains $\tau$, so $N = \langle \tau \rangle$. $|\sigma| = |\tau| = 6$. I have to show that $G/N$ is isomorphic to $\mathbb{Z}_3$, but I don't know how to do this in an 'easy' practical way (not theoretical approach).

Normally I start by looking at the quotient groups order, and go by that but I don't see any obvious approach to determine $G$'s order. To start off with, how do I find $G$'s order in a quick way?

Answer 1

Hint: $G = \langle (135),(27),(468)\rangle$ and $N = \langle (27),(468)\rangle$.

Answer 2

Your group is abelian becaus $\sigma$ and $\tau$ commute, the 2-Sylow is the same to the 2-sylow of $\langle\sigma \rangle$ and $\langle\tau\rangle $ because $\sigma^3=\tau^3$, also your groupe have 3-sylow isomorphic to $Z_3\times Z_3$ and then $G/\langle\tau\rangle$ is $G/\langle\sigma\rangle$ isom $Z_3$