From db667eff2268ecf752d4e55b0bff863eff4af8d4 Mon Sep 17 00:00:00 2001 From: Hearot Date: Tue, 17 Oct 2023 14:15:43 +0200 Subject: [PATCH] fix(algebra1): riordina i teoremi di isomorfismo secondo l'ordine del corso --- .../9. I teoremi di isomorfismo/main.pdf | Bin 177187 -> 177183 bytes .../9. I teoremi di isomorfismo/main.tex | 64 +++++++++--------- 2 files changed, 32 insertions(+), 32 deletions(-) diff --git a/Secondo anno/Algebra 1/9. I teoremi di isomorfismo/main.pdf b/Secondo anno/Algebra 1/9. 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I teoremi di isomorfismo/main.tex b/Secondo anno/Algebra 1/9. I teoremi di isomorfismo/main.tex index 8c839f9..8cc1365 100644 --- a/Secondo anno/Algebra 1/9. I teoremi di isomorfismo/main.tex +++ b/Secondo anno/Algebra 1/9. I teoremi di isomorfismo/main.tex @@ -60,7 +60,38 @@ \[ \Ker f = \{ gN \mid \varphi(g) = e \} = \{ gN \mid g \in \Ker \varphi \} = \Ker \varphi \quot N. \] - \begin{theorem}[Secondo teorema di isomorfismo, o teorema del diamante] + \begin{theorem}[Secondo teorema di isomorfismo] + Siano $H$ e $N$ due sottogruppi normali di $G$ e sia + $N \leq H$. Allora\footnote{ + Ci sono più modi per vedere che $H \quot N$ è + normale in $G \quot N$. Un modo di vederlo si + ottiene dalla dimostrazione stessa del teorema, + dal momento che si ottiene che $H \quot N$ è + il kernel dell'omomorfismo $\varphi$. Altrimenti, + se $hN \in H \quot N$, $gN hN g\inv N = (ghg\inv)N$, + e poiché $H$ è normale in $G$, $ghg\inv \in H$, da + cui $(ghg\inv)N \in H \quot N$. + }: + \[ \frac{G \quot N}{H \quot N} \cong G \quot H. \] + \end{theorem} + + \begin{proof} + Si costruisce l'omomorfismo $\varphi : G \quot N \to G \quot H$ tale per cui $gN \mapsto gH$. Si verifica innanzitutto + che la mappa $\varphi$ è ben definita: + \[ gnH = gH \impliedby N \subseteq H. \] + Inoltre $\varphi$ è effettivamente un omomorfismo dal momento + che: + \[ \varphi(gkN) = gkH = gH \, kH = \varphi(gN) \varphi(kN). \] + Chiaramente $\varphi$ è una mappa surgettiva e quindi + $\Im \varphi = G \quot H$. + Allora, se $g \in \Ker \varphi$, $\varphi(gN) = gH = H$, e quindi $g \in H$. Pertanto $\Ker \varphi = \{ + gN \mid g \in H + \} = H \quot N$. Si conclude allora, per il Primo teorema + di isomorfismo, che: + \[ \frac{G \quot N}{H \quot N} \cong G \quot H. \] + \end{proof} + + \begin{theorem}[Terzo teorema di isomorfismo, o teorema del diamante] Siano $H$, $N \leq G$ con $N \nsgeq G$. Allora\footnote{ Si osserva che effettivamente $H \cap N$ è normale in $H$. Infatti se $g \in H \cap N$, allora, se @@ -101,35 +132,4 @@ la tesi: \[ H \quot (H \cap N) \cong HN \quot N. \] \end{proof} - - \begin{theorem}[Terzo teorema di isomorfismo] - Siano $H$ e $N$ due sottogruppi normali di $G$ e sia - $N \leq H$. Allora\footnote{ - Ci sono più modi per vedere che $H \quot N$ è - normale in $G \quot N$. Un modo di vederlo si - ottiene dalla dimostrazione stessa del teorema, - dal momento che si ottiene che $H \quot N$ è - il kernel dell'omomorfismo $\varphi$. Altrimenti, - se $hN \in H \quot N$, $gN hN g\inv N = (ghg\inv)N$, - e poiché $H$ è normale in $G$, $ghg\inv \in H$, da - cui $(ghg\inv)N \in H \quot N$. - }: - \[ \frac{G \quot N}{H \quot N} \cong G \quot H. \] - \end{theorem} - - \begin{proof} - Si costruisce l'omomorfismo $\varphi : G \quot N \to G \quot H$ tale per cui $gN \mapsto gH$. Si verifica innanzitutto - che la mappa $\varphi$ è ben definita: - \[ gnH = gH \impliedby N \subseteq H. \] - Inoltre $\varphi$ è effettivamente un omomorfismo dal momento - che: - \[ \varphi(gkN) = gkH = gH \, kH = \varphi(gN) \varphi(kN). \] - Chiaramente $\varphi$ è una mappa surgettiva e quindi - $\Im \varphi = G \quot H$. - Allora, se $g \in \Ker \varphi$, $\varphi(gN) = gH = H$, e quindi $g \in H$. Pertanto $\Ker \varphi = \{ - gN \mid g \in H - \} = H \quot N$. Si conclude allora, per il Primo teorema - di isomorfismo, che: - \[ \frac{G \quot N}{H \quot N} \cong G \quot H. \] - \end{proof} \end{document} \ No newline at end of file