Suppose $T$ and $U$ are subgroups of a group $G$, $G = T \ltimes U$ is a semidirect product, and $H$ is a subgroup of $G$ containing $U$. Prove that $H = (H \cap T) \ltimes U$.
First, in this notation is $U$ meant to be the normal subgroup in the construction? In Dummit & Foote, the sign is flipped and the normal group is on the left. My understanding is that wherever the semidirect product sign crosses, the group on that side is the normal one. In that case, I would begin by noting that since $T \cap U = \{1\}$, and $U$ is normal in $G$, then $U$ is also normal in $H$, and $H \cap T \cap U$ is also trivial. But I have to check that my multiplication is properly defined too. This is where I really need to know for sure which of the groups is supposed to be normal.
It's not much you need to show, it suffices to show that we can write any $h \in H$ uniquely as:
$h = tu$ with $t \in H \cap T$ and $u \in U$.
But since $G = TU$ (because it is a(n internal) semi-direct product of $T$ and $U$), and $H$ is a subgroup of $G$, we know there is such a $T,U$ decomposition of $G$; that is, we know we can write:
$h = tu$ uniquely for $t \in T$, and $u \in U$.
So if we can show that $t$ is also in $H$, we're done.
But $t = hu^{-1}$, and both $h,u \in H$.