---- > [!theorem] Theorem. ([[third isomorphism theorem]]) > Suppose that $N$ is a [[normal subgroup]] of a [[group]] $G$, and let $\phi:G \to G / N$ denote the [[kernel iff normal subgroup|natural projection homomorphism]]. > 1. There is an [[inclusion map|inclusion]]-preserving [[bijection]] $\{\text{subgroups of $G$ containing N}\} \leftrightarrow \{ \text{subgroups of $G / N$} \}$ > given by $H \mapsto \phi(H)=H / N$ > for $H$ a [[subgroup]] of $G$, with the inverse map given by $\overline{H} \mapsto \phi ^{-1}(\overline{H})$ > for $\overline{H}$ a [[subgroup]] of $G / N$. *Moreover*, in this [[bijection]] [[normal subgroup]]s correspond to [[normal subgroup]]s. > 2. **The third isomorphism theorem.** Suppose $H\trianglelefteq G$ contains $N$. Then $N$ is [[normal subgroup|normal in]] $H$, $H / N$ is a [[normal subgroup]] of $G / N$, and there is a natural [[group isomorphism|isomorphism]] $\frac{G / N}{H /N} \cong G / H.$ > [!basicexample] > - ![[CleanShot 2023-09-18 at 19.15.01.jpg]] > - From Aluffi: ![[CleanShot 2024-06-17 at [email protected]]] > [!proof]- Proof. ([[third isomorphism theorem]]) > # 1 > > ^f1bc64 > > Call $\Phi$ the [[bijection]] outlined in the theorem statement. We need to show $\Phi$ preserves [[subgroup]]s, is injective, and is surjective. > > Preserves [[subgroup]]s: We know $H \leq G \implies \phi(H)=H / N$; we want to show $H / N \leq G / N$. Since $N \trianglelefteq G$, $N \trianglelefteq H$. Thus $H / N$ is indeed a ([[quotient group|quotient]]) [[group]]. Its elements consist of the left [[coset]]s of $N$ which are represented by elements in $H$. Every element of $H$ is an element of $G$, so each such [[coset]] is also represented by an element in $G$ and thus belongs to $G / N$. So $H / N \leq G / N$. > > [[Injection]]: Suppose $H_{1} / N=H_{2} / N$. Then for all $h_{1} \in H_{1}$, there exists $h_{2} \in H_{2}$ such that $h_{1}N = h_{2}N$. Since $\begin{align} > h_{1}N=h_{2}N \iff & h_{2}^{-1}h_{1}N = N \\ > \iff & h_{2}^{-1}h_{1} \in N \\ > \iff & h_{1} \in h_{2} N \\ > \iff & h_{1}= h_{2}h_{2}' > \text{ for some $h_{2}' \in H$}\\ > \iff & h_{1} \in H_{2}, > \end{align}$ > where in the penultimate step we used that $N \subset H$. This shows $H_{1} \subset H_{2}$. A completely symmetric argument shows $H_{2} \subset H_{1}$. Thus $H_{2}=H_{1}$ and so injectivity holds. > > [[Surjection]]: Let $\overline{H} \leq G / N$. Set $K:=\{ g \in G : gN \in \overline{ H} \}$. Note that $K \geq N$. We have $\begin{align} > \Phi(K):= \phi(K) = & K / N \\ > = & \{ kN: k \in K \} \\ > = & \{ kN : kN \in \overline{H} \} \\ > = & \overline{H}, > \end{align}$ > thus since $\Phi(K)=\overline{H}$ and $\overline{H}$ was arbitrary we conclude $\Phi$ is a [[surjection]]. > > > > Normal to normal: We want to show that the [[bijection]] given maps [[normal subgroup]]s to [[normal subgroup]]s. We will show that $aH=Ha \ \ \fa a \in G \implies Q\{ hN : h \in H \}Q^{-1} = \{ hN : h \in H \}.$ > By definition, $Q=gN$ for some $g \in G$. Hence $\begin{align} > Q\{ hN: h \in H \}Q^{-1} = & gN\{ hN:h \in H \}g^{-1}N \\ > = & \{ \overbrace{ghg^{-1}}^{\in H}N : h \in H\} \text{ ($N$ is normal)}\\ > = & \{ h' N : h' \in H \} . > \end{align}$ > $h'$ is just a dummy variable used to make clear that we are not claiming $ghg^{-1} = h$, just $ghg^{-1} \in H$— the final set obtained is identical to $\{ hN : h \in H \}$ and thus we conclude the image of $H$ under the [[bijection]] is a [[normal subgroup]] of $G / N$. > > # 2 > $N \trianglelefteq H$: This is immediate because $N \trianglelefteq G$; $hN=Nh$ for all $h \in H$ because $h \in G$ for all $h \in H$. > $H / N \trianglelefteq G / N$: This follows from $(1)$ since $H$ is a [[normal subgroup]] of $G$ containing $N$ whose image under the [[bijection]] is $H / N$, and the [[bijection]] maps [[normal subgroup]]s of $G$ to [[normal subgroup]]s of $G / N$. > > Recall that the [[first isomorphism theorem]] guarantees the existence of the following commutative diagram: > ![[CleanShot 2023-09-18 at 20.50.10.jpg]] > This tells us that we want to construct a [[surjection]] $f:G / N \to G / H$ with $\ker f= H / N$. For $gN \in G / N$, define $f(gN):= gH$. Then $f$ is a [[surjection]], for letting $g'H \in G / H$ it is clear that $g'N \mapsto g'H$. Moreover, $\begin{align} > \ker f= & \{ gN \in G / N : f(gN)=e_{G / H} \} \\ > = & \{ gN \in G / N : f(gN)=H \} \\ > = & \{ gN \in G / N : gH = H \}\ \\ > = & \{ gN \in G / N : g \in H \} \\ > = & H / N > \end{align}$ > as desired. ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```