I recently asked the question at here, but I lost the account as I didn't register it, so I am continuing my question ( isomorphism theorem corollary ).
Given a normal subgroup $N$ of a group $G$, and given any other subgroup $H$ of $G$, let
$q : G → G/N$ be the quotient map. Then
$H · N = \{ hn : h \in H, n \in N \} = q^{−1}(q(H))$
is a subgroup of $G$. If $G$ is finite, the order of this group is
$$|H · N| =
\frac{|H| · |N|}{|H ∩ N|}$$
Further,
$q(H) ≈ H/(H ∩ N)$.Proof: By definition the inverse image $q^{−1}(q(H))$ is
$$\begin{align*}
\{ g \in G : q(g) \in q(H) \} &= \{ g \in G : gN = hN\text{ for some }h \in H \}\\
&= \{ g \in G : g ∈ hN\text{ for some }h \in H \}\\
&= \{ g \in G : g \in H · N \}\\
&= H · N
\end{align*}$$The previous corollary already showed that the inverse image of a subgroup is a subgroup. And if $hN = h'N$,
then $N = h^{−1}h'N$, and $h^{−1}h' \in N$. Yet certainly $h^{−1}h' ∈ H$, so $h^{−1}h' \in H \cap N$. And, on the other hand,
if $h^{−1}h' \in H \cap N$ then $hN = h'N$. Since $q(h) = hN$, this proves the isomorphism. From above, the inverse
image $H · N = q^{−1}(q(H))$ has cardinality
$\operatorname{card} H · N = |\ker \ q| · |q(H)| = |N| · |H/(H \cap N)| =
\frac{|N| · |H|}{|H \cap N|}$Question: I am not sure how we can proceed from $|N| · |H/(H ∩ N)|$ to $\frac{|N| · |H|}{|H ∩ N|}$.
Response: $H\cap N$ is normal in $H$. Each coset of $H\cap N$ in $H$ has cardinality $|H\cap N|$, and the cosets of $H\cap N$ partition $H$, so there are $\dfrac{|H|}{|H\cap N|}$ of these cosets. But these cosets are precisely the elements of $H/(H\cap N)$, so $|H/(H\cap N)|=\dfrac{|H|}{|H\cap N|}$.
But then,
Let $f : G → H$ be a surjective homomorphism of finite groups. Let $Y$ be a subgroup
of $H$. Let $X = f^{−1}(Y) = \{ x ∈ G : f(x) ∈ Y \}$
be the inverse image of $Y$ in $G$. Then
$|X| = |\text{ker} \ f| · |Y|$ Proof: By the isomorphism theorem, without loss of generality $Y = G/N$ where $N = \text{ker} \ f$ is a normal
subgroup in $G$. The quotient group is the set of cosets $gN$. Thus,
$f^{−1}(Y) = \{ xN : f(x) ∈ Y \}$ That is, the inverse image is a disjoint union of cosets of $N$, and the number of cosets in the inverse image is $|Y|$.
I am not getting it. So, in the latter case, you multiply $|Y|$ and $|\text{ker} \ f|$ to get $|X|$, and why is this different from the first case?
Also, how can we then say that $\text{card} H \cdot N = |\text{ker} \ q| \cdot |q(H)|$?
Best Answer
It’s exactly the same reasoning as in my answer to the earlier question. In fact, I was simply proving the special case of this proposition that’s needed for the proof of the theorem. To make the connection between this proposition and its application to the theorem, note that the $G,H,f$, and $Y$ of the proposition correspond to the $G,G/H,q$, and $q(H)$ of the theorem.