5.3: Teorema del isomorfismo
- Page ID
- 111059
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Hemos observado algunos casos ahora en los que: 1. Definir un homomorfismo\(\rho: G\rightarrow H\), and then 2. Notice that \(G/\mathord K \sim H\), where \(K\) is the kernel of \(\rho\). This isn't an accident!
La prueba es solo construir una correspondencia entre los cosets del kernel\(gK\) and elements of the image \(I\). Indeed, in any coset \(gK\) all elements map to the same element of the image. \(\rho(gk)=\rho(g)\rho(k)=\rho(g)1=\rho(g)\) for any \(k\in K\).
Esto sugiere un homomorfismo desde el conjunto de cosets a la imagen: set\(\phi(gK)=\rho(g)\). This is a homomorphism, since \(\phi(ghK)=\rho(gh)=\rho(g)\rho(h)=\phi(gK)\phi(hK)\).
El mapa\(\phi\) is also one-to-one: if \(\phi(gK)=\phi(hK)\), we have \(\rho(g)=\rho(h)\), so that \(1=\rho(g^{-1}h)\), meaning \(g^{-1}h\in K\). Then \(h=g(g^{-1}h)\in gK\), which tells us that \(gK=hK\), since cosets are either equal or disjoint.
El mapa\(\phi\) is onto, since any element in the image may be written as \(\rho(g)\) for some \(g\), which is also the image of \(gK\) under \(\phi\). Therefore, the map \(\phi\) is an isomorphism.
TODO: ¡Fotos!
A este teorema se le suele llamar el “primer teorema del isomorfismo”. Hay tres teoremas de isomorfismo, todos los cuales tratan sobre las relaciones entre los grupos de cocientes. El tercer teorema del isomorfismo tiene una declaración particularmente agradable:\((G/\mathord N)/\mathord (H/\mathord N) \sim G/\mathord H\), which one can relate to the the numerical identity
\[\frac{ \frac{n}{m} }{ \frac{p}{m} }=\frac{n}{p}.\]