Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\frac{|N| · |H|}{|H ∩ N|}$.

share|cite|improve this question

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.