\documentclass{amsart}1\usepackage{sagetex}2\usepackage{amssymb}3%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%4\renewcommand{\dot}{\bullet}5\newtheorem{define}{Definition}6\newcommand{\aut}[1]{\mathbf{Aut}{#1}}7\newcommand{\conj}[2][g]{{#1}^{-1}{#2}{#1}}8\newcommand{\normal}{\triangleleft}9\newcommand{\Aut}{\operatorname{Aut}}10\newcommand{\Inn}{\operatorname{Inn}}11\newcommand{\Out}{\operatorname{Out}}12\newcommand{\gens}[1]{\left\langle{#1}\right\rangle}13\newcommand{\epi}{\twoheadrightarrow}14\newcommand{\epic}{\twoheadrightarrow}15\newcommand{\mono}{\hookrightarrow}16\newcommand{\Z}{\mathbf{Z}}17\newcommand{\C}{\mathbf{C}}18\newcommand{\R}{\mathbf{R}}19\newcommand{\F}{\mathbf{F}}20\newcommand{\N}{\mathbf{N}}21\newcommand{\Q}{\mathbf{Q}}22\newcommand{\ndef}{\overset{\text{def}}=}23\def\smat{[\begin{smallmatrix} b&c\\ a&b \end{smallmatrix}]}24\def\acts{\curvearrowright}25\newcommand{\legendre}[1]{\left(\frac{ #1 }{p}\right)}26\newcommand*\colvec[3][]{27\begin{pmatrix}\ifx\relax#1\relax\else#1\\\fi#2\\#3\end{pmatrix}28}29%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%30\begin{document}31\subsubsection*{sage filler}32%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%33%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%34%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%35\begin{sagesilent}36GF2 = GF(2)37O = matrix(2, lambda i,j: i-j);38M = matrix(2, lambda i,j: (-1+2*i-i*j));39m1 = matrix(ZZ, 3, [[0,0,1],[0,1,0],[1,0,0]])40m2 = matrix(ZZ, 3, [[0,1,0],[0,0,1],[1,0,0]])41SSS = MatrixGroup(m1, m2)42S3 = SymmetricGroup(3)43U = GL(3, GF(5)).random_element()44V = GL(3, GF(3)).random_element()45GL3 = MatrixSpace(GF(7),3)46\end{sagesilent}47%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%48%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%49some groups: $C_n,S_n,A_n,D_n;GF(p), Aut(E/F),$ \\50classic Lie type over $\R,\C,\bold{H}: GL,SO,SL, PSL,Sp,SU $ \\51$$\mbox{Dic}_n := \langle a,x \mid a^{2n} = 1,\ x^2 = a^n,\ x^{-1}ax = a^{-1}\rangle$$ alt def by exact sequence $521 \to C_{2n} \to \mbox{Dic}_n \to C_2 \to 1$ \\53G-actions \\54$f:H \leq G \acts G; \ell_h(x) \ndef hx \in G $ \\55other way may not work? $ G \acts H; ah \not\in H$ \\56however $N \normal G, G \acts N; c_g(x) \ndef gxg^{-1}$ \\57$G \acts G/H, f(g,aH) := (ga)H$ \\58\section{permutations \& cycles}59$\sigma \in Aut \N$, e.g.,60\begin{itemize}61\item $n \mapsto n$62\item $\sigma \in \Z_p:n \mapsto n^2$63\item $n \mapsto n+1$64\end{itemize}65These are really permutations, an \emph{$N$-cycle} satisfies66\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'$6768Consider the strictly nondecreasing 2-cycles $\sigma_i \geq \sigma_j$ equality iff equal etcetc.6970\section*{milne}71\subsubsection*{PROPOSITION 1.25} Let H be a subgroup of a group G. \\72(a) An element $a$ of G lies in a left coset C of H if and only73if $C = aH$ \\74(b) Two left cosets are either disjoint or equal. \\75(c) $aH = bH$ if and only if $a^{-1} b \in H$ \\76(d) Any two left cosets have the same number of elements \\77(possibly infinite).78\begin{proof}79Recall $a \in H \implies aH = H$. If this is the case, any left coset of $H$ via elements of $H$ will be $aH (=H)$. \\80Otherwise for $a \ne c \in G - H: a \in cH \implies aH \subset cH$ (since $\exists h \in H: a = ch, h = c^{-1}a$) \\81Using this, $cH = chH = c c^{-1}aH = aH \\82\implies$ (a). \\83and since, by (a), any intersection implies equality of cosets, only disjunct cosets remain \\84$\implies$ (b) \\85Going the other way, if $aH = cH$: $$ca^{-1}H = H, ca^{-1} \in H $$ $\implies$ (c) \\86Define as follows87$$\phi: aH \to bH: x \mapsto (ba^{-1})x, \phi^{-1}:x \mapsto ab^{-1}c, \phi\phi^{-1} =\phi^{-1}\phi=1 $$88This defines a inverse system between $aH,bH$, thus iso ie same cardinality \\ $\implies $ (d)89\end{proof}90[Note 1: Cosets partition $G$, since $a \in aH$ along with 1.25(b).]9192[Note 2:In regards to right cosets, if we modify the function in 1.25(c) slightly93$$\phi: aH \to Hb: ah \mapsto (ah)^{-1}ab = h^{-1}b \in Hb$$94$$\phi^{-1}: Hb \to aH: hb \mapsto ab(hb)^{-1} = ah^{-1} \in aH$$95we get a isomorphism between left and right cosets.]9697\begin{define} The \textbf{index} $(G:H)$ is the number of left (equivalently right) $H$-cosets in $G$ \end{define} %index98\subsection*{THEOREM 1.26 (LAGRANGE)} If $G$ is finite, then99$$(G:1) = (G:H)(H:1)$$100In particular, the order of every subgroup of a finite group divides the order of the group.101\begin{proof}102$(G,1) = \sum_{a \in G}$ Cosets partition $G$ so $\sum_{aH \subset G}|aH| = |G| = (G:1)$ \\103Cosets have equal cardinality relative to a given subgroup (which is itself the coset $eH$), and since the (left) multiplier determines the number of unique cosets104\[ (G:1) = |G| = \sum_{a \in G}|aH| = (G:H)(H:1)105\]106\end{proof}107108\subsubsection*{COROLLARY 1.27} The order of each element of a finite group divides the order of the group.109\begin{proof}110Set $H = \gens{a}$, then $|H| = o(a)$111\end{proof}112One of the Sylow theorems is a partial converse of Lagrange for prime-powers $p^n$: \\113\subsection*{THEOREM 5.2 (SYLOW I)} Let $G \in \bold{FinGp}$ be a finite group, and let $p$114be prime. If $p^n | (G:1)$, then $G$ has a subgroup of order $p^n$115\begin{proof}116TBA117\end{proof}118Toshow: nexted subgroups $K<H<G: (G:K) = (G:H) (H:K)$119120\section*{Normal subgroups, conjugacy}121\begin{define} $N \normal G$ if $\forall g \in G, gNg^{-1} = N$122\end{define}123What about if $g^{-1}Ng = N$? \\124\[125g^{-1}Ng = N \iff Ng = gN \iff gNg^{-1} = N126\] So it doesn't matter how you conjugate. \\127\subsubsection*{REMARK 1.32} suffices to show $gNg^{-1} \subset N (\forall g \in G)$128\begin{proof}129$$gNg^{-1} \subset N \implies N \subset g^{-1}Ng = gNg^{-1}$$ $\therefore gNg^{-1} = N$.130\end{proof}131\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$132%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%133%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%134%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%135\begin{sageblock}136C= matrix(SR, 2, [1,'n',1,0])137\end{sageblock}138%%%%%%%%%%%%%%%%%%%%%%%%%%%%SCGE%%%%%%%%%%%%%%%139%%%%%%%%%%%%%%%%%%%%%%%%%%%%SCGE%%%%%%%%%%%%%%%140\[141\sage{matrix([[5,0],[0,1]])}\sage{C}\sage{matrix([[5**-1,0],[0,1]])}142= \sage{matrix([[5,0],[0,1]])*C*matrix([[5**-1,0],[0,1]])} \cong 5\Z143\]144\[145\sage{matrix([[5**-1,0],[0,1]])}\sage{C}\sage{matrix([[5,0],[0,1]])}146= \sage{matrix([[5**-1,0],[0,1]])*C*matrix([[5,0],[0,1]])}147\]148In the first case conjugation yields a proper subset, but conjugation by the inverse isn't even in $H$.149150%$\left(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right), \gens{\sage{C}}151\subsubsection*{EXAMPLE 1.36} (a) Every subgroup of index two is normal. \\152(b) Dihedral group, its cyclic and translational symmetries, $n=2, n>2$. \\153(c) Subgroups in $\bold{Ab}$ are normal, but converse not true, e.g., $Q$154\begin{proof}155(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 $H \coprod gH$.156Then by exclusion, $gH = Hg$ \\157(b) by commutivity, $C_n \normal D_n \forall n$, but158\end{proof}159\begin{define} When $1 \trianglelefteq N$ is the only series of normal subgroups, $N$ is \textbf{simple}\end{define}160161\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 $G$.162\begin{proof}163(1) First the case of mutual normalcy: $H,N \normal G$,164\[ gHNg^{-1} = g(g^{-1}Hg)(g^{-1}Ng)g^-1 = HN\]165(2) Relaxing the normalcy condition on $H$:166\[ (HN)^{-1} = NH = HN167\]168\end{proof}169Moreover, if we let $X = N \cap N'$170\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} $\conj{X}$.171\end{define}172\subsubsection*{LEMMA 1.38} If X is normal, then the subgroup $\gens{X}$ generated by it is also normal.173\begin{proof}174Say elements of $\gens{X}$ are of the form $a = a_1 \dots a_n$, then175\end{proof}176\subsubsection*{LEMMA 1.39} $\bigcup_{g \in G}{gXg^{-1}}$ is the smallest normal set containing $X$.177\begin{proof}178TBA179\end{proof}180\subsubsection*{PROPOSITION 1.40} $\gens{X} = \gens{\bigcup_{g \in G}{gXg^{-1}}} \normal G$181\begin{proof}182TBA183\end{proof}184%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%185%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%186%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%187\section*{Ch1 exercises}188%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%189%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%190%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%191%sage: P = PermutationGroup(['(1,2,3)(4,5)'])192%sage: p = P.gen(0)193%sage: p.matrix()194195\begin{sageblock}196var('q w r k')197P = PermutationGroup(['(1,2,3)','(2,3)'])198p = P.gen(0)199IsoZ = matrix(SR, 2, [[1,'n'],[0,1]])200a = matrix(SR, 3, 3, [[1, 'a', 'b'], [0, 1, 'c'], [0, 0, 1]])201K = matrix(SR, [[q,w+r],[0,q^2*r]])202A = matrix(2, [0,i, i,0])203B = matrix(2, [0,1, -1,0])204\end{sageblock}205206%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%207%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%208\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}$209\begin{proof}210\[211\sage{A}\sage{B} = \sage{A*B} \not= \sage{B}\sage{A} = \sage{B*A}212\] so $Q$ nonabelian.213$AXA^{-1} = \sage{A*matrix(SR, 2, ['a','b','c','d'])*A**(-1)}$214$BXB^{-1} = \sage{B*matrix(SR, 2, ['a','b','c','d'])*B^(-1)}$215216\end{proof}217\subsubsection*{1-2} Using matrices in $GL(2,\Z[i])$, show $\gens{a,b|a^4 = b^3 =1} \notin \bold{FinGp}$218\begin{proof}219$ab = \sage{matrix([[1,1],[0,1]])}, \left\{(ab)^n = \sage{IsoZ}\right\} \cong \Z$220\end{proof}221\subsubsection*{1-3} Show $G:|G| \in 2\Z$ has an element $a:oa=2$.222\begin{proof}223oa | 2n,224\end{proof}225\subsubsection*{1-4} Let $n = \sum^r_1{n_i}$, use lagrange to show $\prod^r_1{n_i !} | n!$226\begin{proof}227Consider228\end{proof}229\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.230\begin{proof}231TBA232\end{proof}233234\subsubsection*{1-6} We say $m \in \N$ is the \textbf{exponent} of $G$ if it's the smallest annihilator of $G$. \\235(a) Show $m=2 \implies G \in \bold{Ab}$ \\236(b) Let $G$ be the following group. Verify that $m=p$ and $G \notin \bold{Ab}$ \\237\[238G :=\left\{\sage{a}\right\} \subset GL(3,{\mathbb{F}}_p)239\]240\begin{proof}241(a) $abab=e, a^{-1}b^{-1} = ab=ba \qed$ \\242(b) Show $a,b,c \in p\Z/$243244245TBA246\end{proof}247%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%248249\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.250\subsubsection*{1-8} Show that a nonempty finite set with an associative binary251operation satisfying the cancellation laws is a group.252cancellation law: $an = am \implies n = m$253%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%254%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%255%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%256257\end{document}258259