$G = S_7$, $a=(2,5,7)$, $b=(5,7)(4,3,1,6)$, $H = \langle a, b \rangle$, and let $K=\langle b \rangle$. Find a cyclic subgroup of $H$ of order 3 and 6

Question

This is how I tried solving this problem. I wrote out all 12 elements of H (which is definitely not the correct way of going about this problem) and found which elements have order 3 and 6. If we let those elements be generators then will generate a subgroup of $H$?

For example $(e, (2,7,5))$ is an element of $H$ of order 3. Is $<(e, (2,7,5))>$ a cyclic subgroup of $H$ order 3. Likewise, is $<(2,5,7),(4,1)(3,6)>$ a cyclic subgroup of order 6?

Answer 1

$\langle (257)\rangle $ is a cyclic subgroup of $H$ of order $3$.

$\langle (257)(41)(36)\rangle $ will be a cyclic subgroup of order $6$.

And $\langle (257), (41)(36)\rangle =\langle (257)(41)(36)\rangle $.