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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 $\operatorname{card} H · N = |\ker \ q| · |q(H)| = |N| · |H/(H \cap N)| = \frac{|N| · |H|}{|H \cap N|}$

I am not sure how we can proceed from $|N| · |H/(H ∩ N)|$ to $\frac{|N| · |H|}{|H ∩ N|}$.

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1 Answer 1

$H\cap N$ 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 $|H/(H\cap N)|=\dfrac{|H|}{|H\cap N|}$.

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