# Proof that preimage of a subgroup to quotient group is a subgroup

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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