I'm currently working on the following problem and I'm stuck:
Let $K \trianglelefteq G$ and let $\bar H \leq G/K$. Let $\pi: G \rightarrow G/K$ denote the quotient map $g \mapsto gK$. Show that $$H=\pi^{-1}(\bar H)=\{g \in G:gK \in \bar H\}$$ is a subgroup of $G$, containing $K$ as a normal subgroup, with $H/K=\bar H$. Show further that if $\bar H \trianglelefteq G/K$ then $H \trianglelefteq G$.
I managed the first parts (please tell me if I made a mistake):
Claim: $H$ is a subgroup.
Proof: $H$ is clearly a subset of $G$, for a subgroup we also need:
- identity: $e \in G$ and $eK=K\in \bar H$ thus $e \in H$.
- inverses: take $g \in H$. Then $gK \in \bar H$ and $(gK)^{-1}=K^{-1}g^{-1}=Kg^{-1}=g^{-1}K \in \bar H$ thus $g^{-1} \in H$.
- closure: take $g_1,g_2 \in H$. Then $g_1K, g_2K \in \bar H$ and $g_1Kg_2K=g_1g_2K \in \bar H$ thus $g_1g_2 \in H$.
Claim: $K$ is contained in $H$.
Proof: Since $e \in H$, we have $k \in K: kK=K=eK \in \bar H$. Thus $k \in H$ for all $k$.
Now I'm stuck. Thank you for your help!
Applying the fact that $K\unlhd G$ to the fact that $K\leq H$, we get that $K\unlhd H$ (basically $h^{-1}kh\in K$ for all $h\in H$ as elements of $H$ are also elements of $G$).
That $H/K=\overline{H}$ is pretty immediate from the definition of $\pi$, as $H/K=\pi(H)=\pi(\pi^{-1}(\overline{H}))=\overline{H}$.
To get that $H\unlhd G$: You've already proven its a subgroup, so we just need to prove closure under conjugation. Let $h\in H=\pi^{-1}(\overline{H})$ and $g\in G$. Then by normality of $\overline{H}$ in $G/K$ we have $$(gK)^{-1}hK(gK)=g^{-1}hgK\in\overline{H}.$$ Therefore, $g^{-1}hg\in\pi^{-1}(\overline{H})=H$ as required for normality.