by Lozansky
Last Updated August 01, 2020 14:20 PM

**Sorry about the slight mess of a title.**

Let $G$ be a finite group and $N$ a normal subgroup of $G$. If $H$ is a subgroup of $G/N$, prove that $\phi^{-1}(H)$ is a subgroup in $G$ of order $|H| \cdot |N|$ , where $\phi : G \to G/N$ is the canonical homomorphism.

*Attempted solution:*

First of all, $\phi^{-1}(H) = \{ g \in G : \phi(g) \in H \}$. To show that it is a subgroup in $G$, it is sufficient to prove that the set is non-empty and that if $g, h \in \phi^{-1}(H)$, then $gh^{-1} \in \phi^{-1}(H)$. Clearly it is nonempty since $H < G/N$ which implies $H \subset N$. Let $g,h \in \phi^{-1}(H)$. Then $\phi(gh^{-1}) = \phi(g) \phi(h^{-1}) = gNh^{-1}N = gh^{-1}N$, since $N$ is normal in $G$. Note that if $\phi(h) \in H$, then so must $\phi(h^{-1}) \in H$, since $H$ is a subgroup. This proves that $\psi^{-1}(H) < G$.

To prove that the order is $|H| \cdot |N|$, I think it is enough to refer to the fact that $G/N$ contains disjoint subsets of $G$ each of order $N$ (since $G$ is finite) and it is "obvious" that we have $|H|$ such subsets so the order of $\phi^{-1}(H)$ is just the product $|H| \cdot |N|$. However, I'm not sure this is so obvious.

Is this proof actually correct?

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