Given that $G$ is a subgroup of $S_{13}$ (symmetric group) which can be generated with the following two permutations, how can I prove that every element in $G$ can be written as a product of $\sigma$s and $\tau$s?
$$\sigma = ( 1, 2, 3, 4, 5 )( 6, 10 )( 7, 11 )( 8, 12 )( 9, 13 ),$$
$$\tau = ( 2, 5 )( 3, 4 )( 6, 7, 8, 9, 10, 11, 12, 13 )$$
My attempt:
Find orders of $\sigma$ and $\tau$ : $10$ and $8$ respectively (using lcm).
Then also, $G$ is not abelian since $\tau^{-1} \sigma \tau \neq \sigma$.
$\tau^{-1} \sigma \tau = \sigma^{-1}$ implies that $\tau \sigma = \sigma^{-1} \tau = \sigma^{9} \tau$
letting $0 \leq i \leq 9$ and $0 \leq j \leq 7$, we get the product of the form $\sigma^i \tau^j$.
I am not sure if I have actually proved what I needed to. Would really appreciate some help on making sure the proof is correct. Thank you!
You have there an anti-commutativity relation: $ \sigma\tau=\tau\sigma^9 $ (note that I corrected it). Since $\sigma $ and $\tau $ generate, every element of the group can be written as a word in $\sigma $ and $\tau $.
Then the anti-commutativity relation allows you to rewrite those words into the form $\sigma^i\tau^j $ (thus the order is at most $80$).