Find a subgroup of $S_{17}$ containing $\sigma$ isomorphic to $\Bbb Z / 3 \Bbb Z \times \mathbb{Z} / 4 \mathbb{Z} \times \mathbb{Z} / 5 \mathbb{Z}$ Let $S_{17}$ be the permutation group on $\{1,2,3,...,17\}$ and let $\sigma =(2 \ 5 \ 3 \ 6)(11 \ 3 \ 6 \ 5 \ 7)(7 \ 11 \ 5 \ 8\ 3)(1\ 4) \in S_{17} $.
Find a subgroup of $S_{17}$ that contains $\sigma$ and is isomorphic to $\mathbb{Z} / 3 \mathbb{Z} \times \mathbb{Z} / 4 \mathbb{Z} \times \mathbb{Z} / 5 \mathbb{Z}$.
I am not sure how to tackle this question, earlier I had to write $\sigma$ as a product of disjoint cycles, I obtained $\sigma = (3 \ 11 \ 7 \ 6)(2 \ 5 \ 8)(1 \ 4)$.
 A: If we take three disjoint permutations of orders $3,4$ and $5$ then the subgroup generated by them will be isomorfic to $\mathbb Z_3\times \mathbb Z_4 \times \mathbb Z_5$. We just need to make sure they generate $\sigma$.
For order $3$ we can take $(2,5,8)$
For order $4$ we can take $(12)(3,11,7,6)$
For order $5$ we can take $(12,13,14,15,16)$
This subgroup will contain $\sigma$ as it is the product of the first two generators.
A: Good idea to simplify $\sigma$. $\sigma$ has order 12. Note that 9, 10, 12, 13, 14, 15, 16, 17 are unused.
Note $\sigma^4 = (2\;5\;8)^4 = (2\;5\;8)$ has order 3, let this be the generator of $\mathbb 3 / \mathbb Z$.
Also $\sigma^{3} = (3\;11\;7\;6)^3 (1\;4)^3 = (3\;6\;7\;11) (1\;4)$ has order 4, let this be the generator of $\mathbb 4 / \mathbb Z$.
Lastly you can simply add any 5 cycle $\tau$ of the unused elements to generate $\mathbb Z/5 \mathbb Z$.
Now $G = \langle \sigma^4, \sigma^3, \tau \rangle$ is isomorphic to $\mathbb{Z} / 3 \mathbb{Z} \times \mathbb{Z} / 4 \mathbb{Z} \times \mathbb{Z} / 5 \mathbb{Z}$ and $\sigma$ can be recovered from the generators as $\sigma = (\sigma^3)^3 \sigma^4$.
