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galois-notes.tex: detour, isomorphism theorems
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@@ -37,6 +37,21 @@
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note = {\url{https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf}},
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url = {https://scipp.ucsc.edu/~haber/ph251/Dn_subgroups.pdf}
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}
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@misc{judson,
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author={Thomas W. Judson},
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title={Abstract Algebra: theory and applications},
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year={1994},
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note={Available at \url{http://abstract.ups.edu/download.html} },
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pages={438}
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}
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@misc{dummitfoote,
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author={David S. Dummit and Richard M. Foote},
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title={Abstract Algebra (Third Edition)},
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year={2004},
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pages={945}
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}
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\end{filecontents}
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\nocite{*}
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@@ -87,7 +102,8 @@
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\tableofcontents
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\section{Recap on the degree of field extensions}
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\section{Galois Theory notes}
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\subsection{Chapters 4-6}
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(Definitions, theorems, lemmas, corollaries and examples enumeration follows from Ian Stewart's book \cite{ianstewart}).
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\begin{defn}{4.10}
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@@ -176,6 +192,201 @@
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[...] TODO: pending to add key parts up to Chapter 15.
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\subsection{Detour: Isomorphism Theorems}
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\begin{thm}{}(\emph{First Isomorphism Theorem}) \label{1stisothm}
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\begin{minipage}{0.75\textwidth}
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If $\psi: G \longrightarrow H$ a group homomorphism, then $ker(\psi) \triangleleft G$.\\
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Let $\phi: G \longrightarrow G/ker(\psi)$ be the canonical homomorphism.\\
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Then $\exists$ unique isomorphism $\eta: G/ker(\psi) \longrightarrow \psi(G)$ such that $\psi = \eta \phi$.\\
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$\Longleftrightarrow$ ie. $G/ker(\psi) \cong \psi(G)$.
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\end{minipage}
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\hfill
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\begin{minipage}{0.2\textwidth}
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\begin{tikzpicture}[node distance=2.5cm, auto]
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\node (G) {$G$};
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\node (H) [right of=G] {$H$};
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\node (GmodK) [below of=G, xshift=1.25cm] {$G/\ker(\psi)$};
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\draw[->] (G) to node {$\psi$} (H);
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\draw[->] (G) to node [swap] {$\phi$} (GmodK);
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\draw[->] (GmodK) to node [swap] {$\eta$} (H);
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\end{tikzpicture}
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\end{minipage}
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\end{thm}
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\begin{proof}
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\emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
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Let $K=ker(\psi)$.
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Since
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$$\eta: G/K \longrightarrow \psi(G)$$
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let
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$$\eta: gK \longrightarrow \psi(g)$$
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ie. $\eta(gK)=\psi(g)$.
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\begin{enumerate}[i.]
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\item show that $\eta$ is a \emph{well defined} map:
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if $g_1 K=g_2 K$, then for some $k \in K$, $g_1 k =g_2$, so
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$$\eta(g_1K)=\psi(g_1) = \psi(g_1)\psi(k) = \psi(g_1 k) = \psi(g_2) = \eta(g_2 k)$$
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Thus, $\eta$ does not depend on the choice of coset representatives, and
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the map $\eta: G/ker(\psi) \longrightarrow \psi(G)$ is uniquely defined
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since $\psi=\eta\phi$.
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\item show that $\eta$ is a homomorphism:
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Observe:
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$$\eta(g_1 K g_2 K) = \eta(g_1 g_2 K) = \psi(g_1 g_2) = \psi(g_1) \psi(g_2) = \eta(g_1 K) \eta(g_2 K)$$
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$\Longrightarrow$ so $\eta$ is a homomorphism.
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\item show that $\eta$ is an isomorphism:
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Since each element of $H=\psi(G)$ has at least a preimage, then $\eta$ is \emph{surjective} (onto $\psi(G)$).
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Show that it is also \emph{injective} (onet-to-one):
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Suppose 2 different preimatges lead to the same image in $\psi(G)$, ie.
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$\eta(g_1 K) = \eta(g_2 K)$
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then,
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$$\psi(g_1) = \psi(g_2)$$
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which implies $\psi(g_1^{-1} g_2) = e$, ie. $g_1^{-1} g_2 \in ker(\psi)$,
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hence
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$$g_1^{-1} g_2 K = K$$
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$$g_1 K = g_2 K$$
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so $\eta$ is injective.
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\end{enumerate}
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Since $\eta$ is injective and surjective $\Longrightarrow$ $\eta$ is a bijective homomorphism,\\
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ie. $\eta$ is an \emph{isomorphism}.
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\end{proof}
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\begin{thm}{}(\emph{Second Isomorphism Theorem}) \label{2ndisothm}
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Let $H \subseteq G$, $N \triangleleft G$. Then
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\begin{enumerate}[i.]
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\item $HN \subseteq G$
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\item $H \cap N \triangleleft H$
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\item $\frac{H}{H \cap N} \cong \frac{HN}{N}$
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\end{enumerate}
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\end{thm}
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\begin{proof}
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\emph{(proof from Thomas W. Judson book "Abstract Algebra" \cite{judson})}\\
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\begin{enumerate}[i.]
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\item show $HN \subseteq G$:\\
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Note that $HN = \{ hn : h\in H, n\in N \}$. Let $h_1 n_1, h_2 n_2 \in HN$.
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Since $N$ normal $\Longrightarrow~ h_2^{-1} n_1 h_2 \in N$, so
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$$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \in HN$$
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[Recall: since $N \triangleleft G$, $gN=Ng ~\forall g \in G$ $\Longrightarrow gn=n'g$ for some $n' \in N$.]
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To see that $(hn)^{-1} \in HN$:\\
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since $(hn)^{-1} = n^{-1} h^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
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Thus $HN \subseteq G$.
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In fact, $$HN = \bigcup_{h \in H} hN$$
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(TODO: diagram)
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\item show that $H \cap N \triangleleft H$:\\
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Let $h \in H,~ n \in H \cap N$ (recall: $H \cap N \subseteq H$).\\
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Then $h^{-1}nh \in H$ $\longleftarrow$ since $h^{-1}, n, h \in H$.\\
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Since $N \triangleleft G$, $h^{-1} n h \in N$.\\
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Therefore, $h^{-1}nh \in H \cap N$ $\Longrightarrow~ H \cap N \triangleleft H$
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\item show that $\frac{H}{H \cap N} \cong \frac{HN}{N}$:\\
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Define a map
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\begin{align*}
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\phi: &H \longrightarrow \frac{HN}{N}\\
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\text{by}~ \phi: &h \longmapsto hN
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\end{align*}
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$\phi$ is surjective (onto), since any coset $hnN=hN$ is the image of $h \in H$, ie. $\phi(h)$
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$\phi$ is a homomorphism, since
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$$\phi(h h') = h h' N = hN h'N = \phi(h)\phi(h')$$
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By the First Isomorphism Theorem \ref{1stisothm},
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$$\frac{HN}{N} \cong \frac{H}{ker(\phi)}$$
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and since
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\begin{align*}
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&ker(\phi) = \{ h \in H : h \in N\}\\
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&\text{then}~~ker(\phi) = H \cap N
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\end{align*}
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so then,
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$$\frac{HN}{N} = \phi(H) \cong \frac{H}{ker(\phi)} = \frac{H}{H \cap N}$$
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thus
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$$\frac{HN}{N} \cong \frac{H}{H \cap N}$$
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\end{enumerate}
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\end{proof}
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\begin{thm}{}(\emph{Third Isomorphism Theorem}) \label{2ndisothm}\\
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Let $H \subseteq K$ and $K \triangleleft G,~ H \triangleleft G$.\\
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Then
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$\frac{K}{H} \triangleleft \frac{G}{H}$
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and
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$$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
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\end{thm}
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\begin{proof}
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\emph{(proof from Dummit and Foote book ``Abstract Algebra" \cite{dummitfoote})}\\
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Easy to see that $\frac{K}{H} \triangleleft \frac{G}{H}$.
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Define
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\begin{align*}
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\psi: &\frac{G}{H} \longrightarrow \frac{G}{K}\\
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\text{by}~ \psi: &gH \longmapsto gK
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\end{align*}
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To show that $\psi$ is \emph{well defined}:\\
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suppose $g_1 H = g_2 H$, then $g_1 = g_2 h$ for some $h \in H$.\\
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Since $H \subseteq K \Longrightarrow~ h \in K$, hence $g_1 K = g_2 K$,\\
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ie. $\psi(g_1 H) = \psi(g_2 H)$, which shows that $\psi$ is well defined.
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Since $g \in G$ may be chosen arbitrarily in $G$, $\psi$ is a surjective homomorphism.
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Finally,
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\begin{align*}
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ker(\psi) &= \{ gH \in \frac{G}{H} \mid \psi(gH)=1K \}\\
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&= \{ gH \in \frac{G}{H} \mid gK =1K \}\\
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&= \{ gH \in \frac{G}{H} \mid g \in K \}\\
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&= \frac{K}{H}
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\end{align*}
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By the First Isomorphism Theorem (\ref{1stisothm}),
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\begin{tikzpicture}[node distance=2.5cm, auto]
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\node (GmodH) {$\frac{G}{H}$};
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\node (GmodK) [right of=GmodH] {$\frac{G}{K}$};
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\node (GmodHmodKer) [below of=G, xshift=1.25cm] {$\frac{G/H}{\ker(\psi)} = \frac{G/H}{K/H}$};
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\draw[->] (GmodH) to node {$\psi$} (GmodK);
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\draw[->] (GmodH) to node [swap] {$\phi$} (GmodHmodKer);
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\draw[->] (GmodHmodKer) to node [swap] {$\eta$} (GmodK);
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\end{tikzpicture}
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So, by
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$$\eta: \frac{G/H}{K/H} \longrightarrow \frac{G}{K}$$
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since $\eta$ is
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bijective (we know it by the First Isomorphism Theorem), $\eta$ it is the isomorphism:
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$$\frac{ G/H }{ K/H } \cong \frac{G}{K}$$
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\end{proof}
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\subsection{Chapter 14}
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\newpage
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@@ -296,7 +507,7 @@ $$t^{12}-1 = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12}$$
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\begin{defn}{21.5}
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The polynomial $\Phi_d(t)$ defined by
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$$\Phi_n(t) = \prod_{a\in \mathbb{Z}_n,(a,n)=1} (t- \zeta^a)$$
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is the $n$-th \emph{cyclotomic polynomial} over \mathbb{C}.
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is the $n$-th \emph{cyclotomic polynomial} over $\mathbb{C}$.
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\end{defn}
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\begin{cor}{21.6}
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@@ -305,14 +516,15 @@ $$t^{12}-1 = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_{12}$$
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\begin{thm}{21.9}
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\begin{enumerate}
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\item The Galois group $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ consists of the
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$\mathbb{Q}$-automorphisms $\psi_j$ defined by
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$$\psi_j(\zeta)=\zeta^j$$
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where $0 \leq j \leq n-1$ and $j$ is prime to $n$.
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\item The Galois group $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ consists of the
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$\mathbb{Q}$-automorphisms $\psi_j$ defined by
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$$\psi_j(\zeta)=\zeta^j$$
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where $0 \leq j \leq n-1$ and $j$ is prime to $n$.
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\item $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q}) \stackrel{iso}{\cong} \mathbb{Z}_n^*$, and is an abelian group.
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\item its order is $\phi(n)$
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\item if $n$ is prime, $\mathbb{Z}_n^*$ is cyclic
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\item $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q}) \stackrel{iso}{\cong} \mathbb{Z}_n^*$, and is an abelian group.
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\item its order is $\phi(n)$
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\item if $n$ is prime, $\mathbb{Z}_n^*$ is cyclic
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\end{enumerate}
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\end{thm}
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