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\documentclass{amsart}
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\usepackage{sagetex}
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\usepackage{amssymb}
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\renewcommand{\dot}{\bullet}
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\newtheorem{define}{Definition}
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\newcommand{\aut}[1]{\mathbf{Aut}{#1}}
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\newcommand{\conj}[2][g]{{#1}^{-1}{#2}{#1}}
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\newcommand{\normal}{\triangleleft}
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\newcommand{\Aut}{\operatorname{Aut}}
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\newcommand{\Inn}{\operatorname{Inn}}
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\newcommand{\Out}{\operatorname{Out}}
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\newcommand{\gens}[1]{\left\langle{#1}\right\rangle}
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\newcommand{\epi}{\twoheadrightarrow}
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\newcommand{\epic}{\twoheadrightarrow}
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\newcommand{\mono}{\hookrightarrow}
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\newcommand{\Z}{\mathbf{Z}}
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\newcommand{\C}{\mathbf{C}}
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\newcommand{\R}{\mathbf{R}}
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\newcommand{\F}{\mathbf{F}}
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\newcommand{\N}{\mathbf{N}}
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\newcommand{\Q}{\mathbf{Q}}
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\newcommand{\ndef}{\overset{\text{def}}=}
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\def\smat{[\begin{smallmatrix} b&c\\ a&b \end{smallmatrix}]}
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\def\acts{\curvearrowright}
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\newcommand{\legendre}[1]{\left(\frac{ #1 }{p}\right)}
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\newcommand*\colvec[3][]{
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\begin{pmatrix}\ifx\relax#1\relax\else#1\\\fi#2\\#3\end{pmatrix}
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}
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\begin{document}
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\subsubsection*{sage filler}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%
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\begin{sagesilent}
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GF2 = GF(2)
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O = matrix(2, lambda i,j: i-j);
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M = matrix(2, lambda i,j: (-1+2*i-i*j));
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m1 = matrix(ZZ, 3, [[0,0,1],[0,1,0],[1,0,0]])
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m2 = matrix(ZZ, 3, [[0,1,0],[0,0,1],[1,0,0]])
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SSS = MatrixGroup(m1, m2)
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S3 = SymmetricGroup(3)
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U = GL(3, GF(5)).random_element()
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V = GL(3, GF(3)).random_element()
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GL3 = MatrixSpace(GF(7),3)
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\end{sagesilent}
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some groups: $C_n,S_n,A_n,D_n;GF(p), Aut(E/F),$ \\
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classic Lie type over $\R,\C,\bold{H}: GL,SO,SL, PSL,Sp,SU $ \\
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$$\mbox{Dic}_n := \langle a,x \mid a^{2n} = 1,\ x^2 = a^n,\ x^{-1}ax = a^{-1}\rangle$$ alt def by exact sequence $
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1 \to C_{2n} \to \mbox{Dic}_n \to C_2 \to 1$ \\
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G-actions \\
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$f:H \leq G \acts G; \ell_h(x) \ndef hx \in G $ \\
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other way may not work? $ G \acts H; ah \not\in H$ \\
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however $N \normal G, G \acts N; c_g(x) \ndef gxg^{-1}$ \\
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$G \acts G/H, f(g,aH) := (ga)H$ \\
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\section{permutations \& cycles}
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$\sigma \in Aut \N$, e.g.,
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\begin{itemize}
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\item $n \mapsto n$
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\item $\sigma \in \Z_p:n \mapsto n^2$
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\item $n \mapsto n+1$
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\end{itemize}
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These are really permutations, an \emph{$N$-cycle} satisfies
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\begin{enumerate} \item $o(\sigma) = N$, possibly infinity \end{enumerate} Remember, permutations form a symmetric group, with function composition as the binary operation. Thus for \emph{any} permutations, $\sigma, \tau$, you can compose and invert them $\tau \circ \sigma^{-1}$ . Whem $\sigma \circ \sigma' = \sigma' \circ \sigma$, we say they are \emph{disjoint} and write $\sigma \coprod \sigma'$
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Consider the strictly nondecreasing 2-cycles $\sigma_i \geq \sigma_j$ equality iff equal etcetc.
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\section*{milne}
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\subsubsection*{PROPOSITION 1.25} Let H be a subgroup of a group G. \\
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(a) An element $a$ of G lies in a left coset C of H if and only
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if $C = aH$ \\
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(b) Two left cosets are either disjoint or equal. \\
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(c) $aH = bH$ if and only if $a^{-1} b \in H$ \\
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(d) Any two left cosets have the same number of elements \\
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(possibly infinite).
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\begin{proof}
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Recall $a \in H \implies aH = H$. If this is the case, any left coset of $H$ via elements of $H$ will be $aH (=H)$. \\
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Otherwise for $a \ne c \in G - H: a \in cH \implies aH \subset cH$ (since $\exists h \in H: a = ch, h = c^{-1}a$) \\
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Using this, $cH = chH = c c^{-1}aH = aH \\
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\implies$ (a). \\
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and since, by (a), any intersection implies equality of cosets, only disjunct cosets remain \\
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$\implies$ (b) \\
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Going the other way, if $aH = cH$: $$ca^{-1}H = H, ca^{-1} \in H $$ $\implies$ (c) \\
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Define as follows
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$$\phi: aH \to bH: x \mapsto (ba^{-1})x, \phi^{-1}:x \mapsto ab^{-1}c, \phi\phi^{-1} =\phi^{-1}\phi=1 $$
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This defines a inverse system between $aH,bH$, thus iso ie same cardinality \\ $\implies $ (d)
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\end{proof}
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[Note 1: Cosets partition $G$, since $a \in aH$ along with 1.25(b).]
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[Note 2:In regards to right cosets, if we modify the function in 1.25(c) slightly
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$$\phi: aH \to Hb: ah \mapsto (ah)^{-1}ab = h^{-1}b \in Hb$$
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$$\phi^{-1}: Hb \to aH: hb \mapsto ab(hb)^{-1} = ah^{-1} \in aH$$
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we get a isomorphism between left and right cosets.]
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\begin{define} The \textbf{index} $(G:H)$ is the number of left (equivalently right) $H$-cosets in $G$ \end{define} %index
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\subsection*{THEOREM 1.26 (LAGRANGE)} If $G$ is finite, then
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$$(G:1) = (G:H)(H:1)$$
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In particular, the order of every subgroup of a finite group divides the order of the group.
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\begin{proof}
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$(G,1) = \sum_{a \in G}$ Cosets partition $G$ so $\sum_{aH \subset G}|aH| = |G| = (G:1)$ \\
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Cosets have equal cardinality relative to a given subgroup (which is itself the coset $eH$), and since the (left) multiplier determines the number of unique cosets
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\[ (G:1) = |G| = \sum_{a \in G}|aH| = (G:H)(H:1)
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\]
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\end{proof}
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\subsubsection*{COROLLARY 1.27} The order of each element of a finite group divides the order of the group.
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\begin{proof}
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Set $H = \gens{a}$, then $|H| = o(a)$
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\end{proof}
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One of the Sylow theorems is a partial converse of Lagrange for prime-powers $p^n$: \\
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\subsection*{THEOREM 5.2 (SYLOW I)} Let $G \in \bold{FinGp}$ be a finite group, and let $p$
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be prime. If $p^n | (G:1)$, then $G$ has a subgroup of order $p^n$
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\begin{proof}
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TBA
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\end{proof}
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Toshow: nexted subgroups $K<H<G: (G:K) = (G:H) (H:K)$
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\section*{Normal subgroups, conjugacy}
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\begin{define} $N \normal G$ if $\forall g \in G, gNg^{-1} = N$
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\end{define}
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What about if $g^{-1}Ng = N$? \\
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\[
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g^{-1}Ng = N \iff Ng = gN \iff gNg^{-1} = N
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\] So it doesn't matter how you conjugate. \\
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\subsubsection*{REMARK 1.32} suffices to show $gNg^{-1} \subset N (\forall g \in G)$
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\begin{proof}
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$$gNg^{-1} \subset N \implies N \subset g^{-1}Ng = gNg^{-1}$$ $\therefore gNg^{-1} = N$.
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\end{proof}
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\subsubsection*{important note:} this is only the case if \emph{every} $g \in G$ fulfills this criterion, and in the following example $G = GL(2,\Q), g = \left(\begin{smallmatrix}5&0\\0&1\end{smallmatrix}\right)$ and $H = \{\left(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right)\}_{n \in \Z} \cong \Z$
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\begin{sageblock}
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C= matrix(SR, 2, [1,'n',1,0])
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\end{sageblock}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%SCGE%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%SCGE%%%%%%%%%%%%%%%
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\[
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\sage{matrix([[5,0],[0,1]])}\sage{C}\sage{matrix([[5**-1,0],[0,1]])}
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= \sage{matrix([[5,0],[0,1]])*C*matrix([[5**-1,0],[0,1]])} \cong 5\Z
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\]
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\[
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\sage{matrix([[5**-1,0],[0,1]])}\sage{C}\sage{matrix([[5,0],[0,1]])}
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= \sage{matrix([[5**-1,0],[0,1]])*C*matrix([[5,0],[0,1]])}
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\]
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In the first case conjugation yields a proper subset, but conjugation by the inverse isn't even in $H$.
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%$\left(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right), \gens{\sage{C}}
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\subsubsection*{EXAMPLE 1.36} (a) Every subgroup of index two is normal. \\
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(b) Dihedral group, its cyclic and translational symmetries, $n=2, n>2$. \\
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(c) Subgroups in $\bold{Ab}$ are normal, but converse not true, e.g., $Q$
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\begin{proof}
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(a) The subgroup $H$ of index 2 has two cosets: namely itself $aH = H$, and $gH: g \in G-H$. In other words, $G$ is partitioned into $H \coprod gH$.
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Then by exclusion, $gH = Hg$ \\
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(b) by commutivity, $C_n \normal D_n \forall n$, but
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\end{proof}
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\begin{define} When $1 \trianglelefteq N$ is the only series of normal subgroups, $N$ is \textbf{simple}\end{define}
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\subsubsection*{PROPOSITION 1.37} If $H$ and $N$ are subgroups of $G$ and $N$ is normal, then $HN$ is a subgroup of $G$. If $H$ is also normal, then $HN$ is a normal subgroup of $G$.
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\begin{proof}
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(1) First the case of mutual normalcy: $H,N \normal G$,
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\[ gHNg^{-1} = g(g^{-1}Hg)(g^{-1}Ng)g^-1 = HN\]
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(2) Relaxing the normalcy condition on $H$:
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\[ (HN)^{-1} = NH = HN
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\]
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\end{proof}
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Moreover, if we let $X = N \cap N'$
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\begin{define} $\gens{X}_{N \normal G} {\ndef} \bigcap_{X \subset N}{N}$, the \textbf{normal subgroup generated by} $X$, is the intersection of normal subgroups containing $X$. As we will see this is equivalent to the \textbf{normal closure} $\conj{X}$.
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\end{define}
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\subsubsection*{LEMMA 1.38} If X is normal, then the subgroup $\gens{X}$ generated by it is also normal.
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\begin{proof}
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Say elements of $\gens{X}$ are of the form $a = a_1 \dots a_n$, then
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\end{proof}
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\subsubsection*{LEMMA 1.39} $\bigcup_{g \in G}{gXg^{-1}}$ is the smallest normal set containing $X$.
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\begin{proof}
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TBA
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\end{proof}
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\subsubsection*{PROPOSITION 1.40} $\gens{X} = \gens{\bigcup_{g \in G}{gXg^{-1}}} \normal G$
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\begin{proof}
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TBA
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\end{proof}
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\section*{Ch1 exercises}
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%sage: P = PermutationGroup(['(1,2,3)(4,5)'])
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%sage: p = P.gen(0)
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%sage: p.matrix()
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\begin{sageblock}
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var('q w r k')
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P = PermutationGroup(['(1,2,3)','(2,3)'])
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p = P.gen(0)
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IsoZ = matrix(SR, 2, [[1,'n'],[0,1]])
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a = matrix(SR, 3, 3, [[1, 'a', 'b'], [0, 1, 'c'], [0, 0, 1]])
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K = matrix(SR, [[q,w+r],[0,q^2*r]])
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A = matrix(2, [0,i, i,0])
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B = matrix(2, [0,1, -1,0])
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\end{sageblock}
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\subsubsection*{1-1} Using $$Q: = \langle A^2=B^2,A^4=1,A^3 = BAB^{-1} \rangle = \sage{MatrixGroup(A,B)}$$ Show $\forall H \leq Q, H \normal Q, \exists! a: oa=2, Q \notin \bold{Ab}$
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\begin{proof}
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\[
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\sage{A}\sage{B} = \sage{A*B} \not= \sage{B}\sage{A} = \sage{B*A}
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\] so $Q$ nonabelian.
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$AXA^{-1} = \sage{A*matrix(SR, 2, ['a','b','c','d'])*A**(-1)}$
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$BXB^{-1} = \sage{B*matrix(SR, 2, ['a','b','c','d'])*B^(-1)}$
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\end{proof}
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\subsubsection*{1-2} Using matrices in $GL(2,\Z[i])$, show $\gens{a,b|a^4 = b^3 =1} \notin \bold{FinGp}$
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\begin{proof}
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$ab = \sage{matrix([[1,1],[0,1]])}, \left\{(ab)^n = \sage{IsoZ}\right\} \cong \Z$
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\end{proof}
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\subsubsection*{1-3} Show $G:|G| \in 2\Z$ has an element $a:oa=2$.
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\begin{proof}
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oa | 2n,
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\end{proof}
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\subsubsection*{1-4} Let $n = \sum^r_1{n_i}$, use lagrange to show $\prod^r_1{n_i !} | n!$
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\begin{proof}
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Consider
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\end{proof}
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\subsubsection*{1-5} Let $N \normal G: (G:N)= n$. Show $g^n \in N$, and that in nonnnormal subgroups this may not be true.
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\begin{proof}
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TBA
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\end{proof}
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\subsubsection*{1-6} We say $m \in \N$ is the \textbf{exponent} of $G$ if it's the smallest annihilator of $G$. \\
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(a) Show $m=2 \implies G \in \bold{Ab}$ \\
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(b) Let $G$ be the following group. Verify that $m=p$ and $G \notin \bold{Ab}$ \\
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\[
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G :=\left\{\sage{a}\right\} \subset GL(3,{\mathbb{F}}_p)
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\]
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\begin{proof}
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(a) $abab=e, a^{-1}b^{-1} = ab=ba \qed$ \\
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(b) Show $a,b,c \in p\Z/$
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TBA
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\end{proof}
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\subsubsection*{1-7} Two subgroups $H$ and $H'$ of a group $G$ are said to be \textbf{commensurable} if $H \cap H'$ is of finite index in both $H$ and $H'$. Show that commensurability is an equivalence relation on subgroups of G.
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\subsubsection*{1-8} Show that a nonempty finite set with an associative binary
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operation satisfying the cancellation laws is a group.
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cancellation law: $an = am \implies n = m$
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\end{document}
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