Let $G$ be a finite group. Let $N \trianglelefteq G$ and $U \leq G$ such that $G = NU$. Then there exists a bijection, preserving inclusion, from the set of subgroups $X$ satisfying $U ≤ X ≤ G$ to the set of $U-$invariant subgroups $Y$ satisfying $U \cap N ≤ Y ≤ N$.
This is a problem from Kurzweil-Stellmacher book. I dont know how to do it. Thanks for any help.
Hint: you should use Dedekind's Lemma:
Let $U$ and $V$ be subgroups of a group $G$ and let $U \subseteq X \subseteq G$, with $X$ also a subgroup. Then $U(V \cap X)=UV \cap X$.
Note that $UV$ here is just a subset of $G$, not necessarily a subgroup! Anyway, in your case define a map $\phi$ from the set {$X$: $X$ is a subgroup of $G$ with $U \subseteq X \subseteq G$} to the set {$Y$: $Y$ is a $U$-invariant subgroup of $N$ with $N \cap U \subseteq Y \subseteq N$} by $\phi:X \mapsto X \cap N$. Observe that $\phi$ is well-defined by the normality of $N$. The injectivity of this map can be proved with Dedekind's Lemma. For the surjectivity look at an $X=YU$, of which you have to show it is a subgroup (remember $Y$ was $U$-invariant!) and again apply the lemma.