Let $\sigma=(13579)(24)(68)$ and $\tau=(15937)(26)(48)$ be elements in $S_9$. Prove that $$\langle \sigma,\tau\rangle\cong \mathbb{Z}/5\mathbb{Z}\times V_4.$$
In a previous question, I showed that $\sigma^2=\tau^6$. I know that, in order to find an isomorphism, I need to either find a bijective homomorphism or to make use of some theorem in my book which says that if $H_1,H_2$ are subgroups of $G$ that commute, have intersection $\{e\}$ and where every $g\in G$ can be written as $g=h_1 h_2$ with $h_1\in H_1$, $h_2\in H_2$, then $G\cong H_1\times H_2$. However the subgroups $\langle\sigma\rangle$ as well as $\langle\tau\rangle$ have order 10, and $\mathbb{Z}/5\mathbb{Z}$ has order 5 and $V_4$ order 4.
Could someone provide a hint, I know this should be easy.
If a permutation is a product of $5$-cycles and $2$-cycles, then there are powers you can take it to to extract out the $5$-cycles and $2$-cycles. For instance,
$$ \sigma^5=(24)(68). $$
Use this to show that $\langle\sigma,\tau\rangle=\langle (24)(68),(26)(48),(13579),(15937)\rangle$. Note the first two permutations generate a subgroup of $\mathrm{Perm}(\{2,4,6,8\})$ and the other permutations each generate a subgroup of $\mathrm{Perm}(\{1,3,5,7,9\})$; the result will be an internal direct product.