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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 $\text{card} H \cdot N = |\text{ker} \ q| \cdot |q(H)|$?

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You can register, then ask someone to flag the moderator to merge your questions together. –  William Aug 22 '12 at 8:25
    
@William thanks. –  user1613156 Aug 22 '12 at 8:27
    
Unfortunately, I don't think that I should merge the questions together: while they're related, this question is on the face of it an extension of the other. In addition, it has been answered. I also don't think I should merge the user accounts together: They have different names, different email addresses, and share the point of access with many other users. –  mixedmath Aug 24 '12 at 4:00

1 Answer 1

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.

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