I am trying to find subgroups of $D_{10}$. I can easily find the subgroups of order $2$ - $\{e, t\}$, $\{e, st\}, \ldots$ etc. where $s$ is the rotation of each vertices and $t$ is the reflection along the vertical line through 1.
I know that $\langle t, st\rangle$ is also a subgroup. Now $\langle t, st\rangle = \{e, t, st, s^4, s\}$ but $s^4t$ and $s^2t$ are not in the group, therefore $\langle t, st\rangle$ is not a group! Please help me with my logic.
I was also wondering if $a, b \in D_{10}$ and $c, d\in D_{10}$ if $ab= cd$, does this count as closure? even if we were to write out the elements in $D_{10}$ and $a.b$ didn't match any of them?
Thank you
Note that since $s, t\in \langle t, st\rangle$, it must be the case that $\langle t, st\rangle \supseteq \langle s, t\rangle = D_{10}$.
Your computation of the group generated by $t$ and $st$ was incomplete.
For your second question, I'm not entirely clear about what you are asking. Closure means that the product of any two elements of the set (group) is again in the set. More formally: If $a, b\in D_{10}$, then $ab\in D_{10}$.