An exercise about $X$-invariant in automorphism groups.

Question

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 $U \leq X \leq G$ to the set of $U$-invariant subgroups $Y$ satisfying $U\cap N \leq Y \leq N$.

I have no idea about how to solve it, and any help is sincerely appreciated.

PS: I've always been confused about the concept of "$X$-invariant", so I would like to be told more details. Thanks!

Answer 1

Proof.

For each $X$, $U$ is a subgroup of $X$, hence normalizes $X$; $N$ is a normal subgroup of $G$, so $U$ certainly normalizes $N$. Thus, $U$ normalizes $X\cap N$, i.e., $X\cap N$ is $U$-invariant. Now we can define a map \begin{align*} \varphi~:~\mathfrak{A}=\{ X\mid U\le X\le G \} &\longrightarrow \mathfrak{B}=\{Y\mid Y \mbox{ is a }U\mbox{-invariant subgroup}\}\\X&\longmapsto X\cap N. \end{align*} The map is well defined since $U\leq N_G(X\cap N)$.

For the injectivity: $\forall\, X_1\cap N,X_2\cap N\in \mathfrak{B}$, if $X_1\cap N=X_2\cap N$, then by Dedekind Identity, $X_1=U(X_1\cap N)=U(X_2\cap N)=X_2$.

The map is also surjective. $\forall\, Y\in \mathfrak{B}$, $u\in U$, since $Y$ is $U$-invariant, $Y^u=u^{-1}Yu=Y$, $Yu=uY$. Hence, $YU=UY$, $YU\leq G$. Note that $U\cap N\leq U$, $NU=G$, $(U\cap N)U\subseteq YU\subseteq NU\Longrightarrow U\leq YU\leq G$. So we can apply Dedekind Identity to $YU$, $YU=UY=U(UY\cap N)\Longrightarrow Y=UY\cap N$. Since $UY\in \mathfrak{A}$, each $Y$ has a corresponding $X\in \mathfrak{A}$, therefore $\varphi$ is surjective.

The bijection is hence proved.