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

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 $$H$$ contains at least the identity. 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 $$\phi^{-1}(H) < G$$.

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?

Your first part is correct, but after writing $$\phi(gh^{-1})=\phi(g)\phi(h^{-1})=gNh^{-1}N=gh^{-1}N$$ you should say that this implies that $$gh^{-1}\in \phi^{-1}(H)$$.
As for the second thing, to be more precise, you can choose $$\lvert H\rvert=k$$ many disjoint coset representatives: $$x_1N,\ldots, x_kN$$ and then note that since these are pairwise distinct representatives: $$x_iN\cap x_jN=\varnothing$$ unless $$x_i=x_j$$. Hence, $$\phi^{-1}(H)=\bigsqcup_{i=1}^kx_i N$$ has cardinality $$\lvert H\rvert\cdot\lvert N\rvert.$$ However, your idea is correct.