I have
$$\sigma = ( 1, 2, 3, 4, 5 )( 6, 10 )( 7, 11 )( 8, 12 )( 9, 13 ),$$
$$\tau = ( 2, 5 )( 3, 4 )( 6, 7, 8, 9, 10, 11, 12, 13 )$$
as my two permutations that generate $G$.
How would I find the generator of the intersection of $$H = \langle \sigma \rangle$$ and $$K = \langle \tau \rangle$$
Would I have to find all the elements of $H$ and $K$ and see which elements coincide or is there a more efficient way (since the permutations are not that small). I wasn't quite understanding some of the other explanations here on the site. Thank you for all of the help!
Since $\vert \sigma \vert = 10$ and $\vert \tau \vert = 8$, the only possibilities for $\vert H \cap K \vert$ are $1$ and $2$. That's because $H \cap K$ is a subgroup of $H$ and of $K$, so its order must divide both $10= \vert H \vert$ and $8= \vert K \vert$, so $\vert H \cap K \vert \mid \gcd(10, 8)=2$.
That means the intersection can only be non-trivial if $\sigma^5 \in K$ and $\tau^4 \in H$. It's pretty easy to see (by calculation) whether that happens. Since it does, that element, along with the identity element, has to be the entire intersection.