---- - [ ] they're all the same, should really try to just collect all into one spot > [!theorem] Theorem. ([[first isomorphism theorem]]) > Let $\phi:G \to H$ be any [[group homomorphism|group homomorphism]]. Then there exists a unique map $\psi:G / \ker \phi \to H$ such that $\phi=\psi \circ \pi$ and $\psi$ is an [[group isomorphism|isomorphism]] onto its [[image]]. [^1] > > ```tikz \usepackage{tikz-cd} \usepackage{amsmath} \begin{document} \begin{tikzcd}[] G \arrow[r, "\phi"] \arrow[d, two heads, "\pi"'] & H \\ G/\ker(\phi) \arrow[ru, hook, dashed, "\psi"'] & \end{tikzcd} \end{document}``` > > > [!proposition] Corollary. > For any [[surjection|surjective]] [[group homomorphism]] $\phi:G \to H$ we have $G / \text{ker } \phi \cong H,$ > since $\psi$ in the diagram above then becomes a [[surjection]]. ^proposition > [!justification] Motivation. > Recall the [[the canonical decomposition of a set function#^intuition|motivation]] for [[the canonical decomposition of a set function]]: we wanted to use the [[universal property of quotient sets]] and manufactured conditions to satisfy it. Indeed, on a purely set-theoretic basis we go ahead and declare $g_{1} \sim g_{2} \iff \phi(g_{1})=\phi(g_{2})$, and in turn the [[characterization of quotienting a group|universal property of quotient groups (scroll down)]] promises a [[group embedding]] $\overline{\phi}:G /{\sim} \hookrightarrow H$. But we can say more: the [[characterization of quotienting a group]] also says the [[equivalence class|equivalence classes]] comprising $G /{\sim}$ are uniquely determined to be the [[coset|cosets]] of a [[normal subgroup]] $N=[e_{G}]_{\sim}$, and $g \in [e_{G}]_{\sim} \iff \phi(g)=\phi(e_{G})=e_{H} \iff g \in \text{ker }\phi$. So $N=\ker \phi$. ^justification > [!proof]+ Proof. ([[first isomorphism theorem]]) > Recall the [[the canonical decomposition of a set function#^intuition|motivation]] for [[the canonical decomposition of a set function]]: we wanted to use the [[universal property of quotient sets]] and manufactured conditions to satisfy it. Indeed, on a purely set-theoretic basis we go ahead and declare $g_{1} \sim g_{2} \iff \phi(g_{1})=\phi(g_{2})$, and in turn the [[characterization of quotienting a group|universal property of quotient groups (scroll down)]] promises a [[group embedding]] $\overline{\phi}:G / {\sim} \to H$. But we can say more: the [[characterization of quotienting a group]] also says the [[equivalence class|equivalence classes]] of $G /{\sim}$ are uniquely determined to be the [[coset|cosets]] of a [[normal subgroup]] $N=[e_{G}]_{\sim}$, and $g \in [e_{G}]_{\sim} \iff \phi(g)=\phi(e_{G})=e_{H} \iff g \in \text{ker }\phi$. So $N=\ker \phi$. > >Alternatively, we can argue right from the definition of [[normal subgroup]] using the lemmas below. In particlar, > [[first isomorphism theorem#^1728d1|Proposition 1]] shows that $\psi$ exists, is a [[group homomorphism]], and is unique. [[first isomorphism theorem#^fa135f|Proposition 2]] shows that it is [[injection|injective]] (thus bijective onto its [[image]]). This is the proof. [^1]: Here, $\pi:G \to G / K$ denotes the [[kernel iff normal subgroup|natural projection homomorphism]]. > [!proposition] Proposition 1. (Uniquely factorizing a homomorphism using the natural projection homomorphism) > Let $\phi:{}G_{} \to G_{}'$ be any [[group homomorphism]] between [[group]]s $G,G_{1}$ for which$N \subset \ker \phi,$ > where $N \trianglelefteq G$. > \ > Then there exists a unique [[group homomorphism|homomorphism]] $\psi:G / N \to G_{1}$ such that $\phi=\psi \circ \pi$, where $\pi$ denotes the [[kernel iff normal subgroup|natural projection homomorphism]]. ^1728d1 ```tikz \begin{document} \begin{tikzpicture} G \arrow{d}[swap]{\varphi} \arrow{dr}{\varphi_1} & \\ G/N \arrow{r}[swap]{\psi} & G_1 \end{tikzpicture} \end{document} ``` ```tikz ``` > [!proof] Proof of Proposition 1. > *If* such a $\psi$ existed, we would need $\psi(aN)=\varphi_{1}(a)$ for all $a \in G$ and so we in fact tentatively define $\psi(aN)=\phi_{1}(a)$ with uniqueness being immediate. But whenever you define a function using [[coset|coset representatives]] you must check it is well-defined. That is, we need to check that if $aN=bN$, then $\varphi_{1}(a)=\varphi_{1}(b)$. Well, to say $aN=bN$ is to say that $a=bn$ for some $n\in N$.[^2] So we can argue $\varphi_{1}(a)=\phi_{1}(bn)=\varphi _{1}(b)\varphi_{1}(n)=\varphi_{1}(b),$ > where in the last step we used that $N \subset \ker \varphi$. This shows that our choice of $\psi$ works. > [!proposition] Proposition 2. (Characterizing injectivity of $\psi$) > The [[group homomorphism|homomorphism]] $\psi: G / N \to G_{1}$ is an [[injection]] *if and only if* $N=\ker \varphi$. ^fa135f > [!proof] Proof of Proposition 2. > Set $K:=\ker \varphi _{1}$. Note that $\begin{align} \ker \psi = & \{ aN : \varphi_{1}(a)=e_{G_{1}} \} \\ = & \{ aN : a \in K \} \\ = & K / N \ \ (\text{ recall that we assumed } K \supset N) \end{align}$ Now, $\begin{align} \psi \text{ is injective} \iff & \ker \psi = \{e_{G / N} \} \\ \iff & K / N = \{e_{G / N}\} \\ \iff & K / N = N \\ \iff & K=N. \end{align}$ > [!intuition] > If you have a [[group homomorphism|homomorphism]] with some (say, nontrival) [[kernel]] and you want to make the map [[injection|injective]] somehow, you can do so by 'killing (modding by) the entire [[kernel]]'... actually, this is the *only* way to do it. [^2]: One way of seeing this is that $a=ae \in aN=bN$ and since [[Lagrange's Theorem|cosets are either identical or disjoint]] the inclusion goes the other way, hence the use of 'is to say'. ---- #### ----- #### 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 > ```