To find a subgroup, I calculated the entire Cayley Table for $D_5$ and looked for wherever there was an Identity. Then, I calculated my subgroup of $S_5$ accordingly.
I know that $D_5$ =
{$I, R_1, R_2, R_3, R_4, M_0, M_1, M_2, M_3, M_4$}
with the corresponding subgroup:
{$I, (123), (124), (142), (132), (12), (13), (14), (23), (24)$}
Now, I know this isn't correct because when I take (23)(12) it should give (142) [because $M_3M_0 = R_3$], but this is clearly incorrect. I know the last 5 elements must be transpositions because in $D_5$ the final five elements are inverses of themselves.
I was wondering if there was a succinct method of finding a subgroup of $S_5$ that is isomorphic to $D_5$ without going through a bunch of unnecessary calculations.
Thank you.
$D_{10} = \langle (12345), (25)(34) \rangle \leq S_5$ (My $D_{10}$ is your $D_5$ - I prefer this notation where the subscript is the number of elements).
The full subgroup structure of $S_5$ can be found here.
In order to construct this subgroup, the abstract group $D_{10}$ has presentation $\langle a , b \mid a^5 = b^2 = 1, bab^{-1} = a^{-1} \rangle$, so start by trying to find two elements satisfying these relations. For example, the first element must have order 5, and so must be a 5 cycle.