1\documentclass{amsart}
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31\begin{document}
32\subsubsection*{sage filler}
33%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%
34%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%
35%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%
36\begin{sagesilent}
37    GF2 = GF(2)
38    O = matrix(2, lambda i,j: i-j);
39    M = matrix(2, lambda i,j: (-1+2*i-i*j));
40    m1 = matrix(ZZ, 3, [[0,0,1],[0,1,0],[1,0,0]])
41    m2 = matrix(ZZ, 3, [[0,1,0],[0,0,1],[1,0,0]])
42    SSS = MatrixGroup(m1, m2)
43    S3 = SymmetricGroup(3)
44    U = GL(3, GF(5)).random_element()
45    V = GL(3, GF(3)).random_element()
46    GL3 = MatrixSpace(GF(7),3)
47\end{sagesilent}
48%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%%
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50some groups: $C_n,S_n,A_n,D_n;GF(p), Aut(E/F),$ \\
51classic Lie type over $\R,\C,\bold{H}: GL,SO,SL, PSL,Sp,SU$ \\
52$$\mbox{Dic}_n := \langle a,x \mid a^{2n} = 1,\ x^2 = a^n,\ x^{-1}ax = a^{-1}\rangle$$ alt def by exact sequence $53 1 \to C_{2n} \to \mbox{Dic}_n \to C_2 \to 1$ \\
54G-actions \\
55$f:H \leq G \acts G; \ell_h(x) \ndef hx \in G$ \\
56other way may not work? $G \acts H; ah \not\in H$ \\
57however $N \normal G, G \acts N; c_g(x) \ndef gxg^{-1}$ \\
58$G \acts G/H, f(g,aH) := (ga)H$ \\
59\section{permutations \& cycles}
60$\sigma \in Aut \N$, e.g.,
61\begin{itemize}
62\item $n \mapsto n$
63\item $\sigma \in \Z_p:n \mapsto n^2$
64\item $n \mapsto n+1$
65\end{itemize}
66These are really permutations, an \emph{$N$-cycle} satisfies
67\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 $\sigma \coprod \sigma'$
68
69Consider the strictly nondecreasing 2-cycles $\sigma_i \geq \sigma_j$ equality iff equal etcetc.
70
71\section*{milne}
72\subsubsection*{PROPOSITION 1.25} Let H be a subgroup of a group G. \\
73(a) An element $a$ of G lies in a left coset C of H if and only
74if $C = aH$ \\
75(b) Two left cosets are either disjoint or equal. \\
76(c) $aH = bH$ if and only if $a^{-1} b \in H$ \\
77(d) Any two left cosets have the same number of elements \\
78(possibly infinite).
79\begin{proof}
80    Recall $a \in H \implies aH = H$. If this is the case, any left coset of $H$ via elements of $H$ will be $aH (=H)$. \\
81    Otherwise for $a \ne c \in G - H: a \in cH \implies aH \subset cH$ (since $\exists h \in H: a = ch, h = c^{-1}a$) \\
82    Using this, $cH = chH = c c^{-1}aH = aH \\ 83 \implies$ (a). \\
84    and since,  by (a), any intersection implies equality of cosets, only disjunct cosets remain \\
85    $\implies$ (b) \\
86    Going the other way, if $aH = cH$: $$ca^{-1}H = H, ca^{-1} \in H$$ $\implies$ (c) \\
87    Define as follows
88    $$\phi: aH \to bH: x \mapsto (ba^{-1})x, \phi^{-1}:x \mapsto ab^{-1}c, \phi\phi^{-1} =\phi^{-1}\phi=1$$
89    This defines a inverse system between $aH,bH$, thus iso ie same cardinality \\ $\implies$ (d)
90    \end{proof}
91[Note 1: Cosets partition $G$, since $a \in aH$ along with 1.25(b).]
92
93[Note 2:In regards to right cosets, if we modify the function in 1.25(c) slightly
94$$\phi: aH \to Hb: ah \mapsto (ah)^{-1}ab = h^{-1}b \in Hb$$
95$$\phi^{-1}: Hb \to aH: hb \mapsto ab(hb)^{-1} = ah^{-1} \in aH$$
96we get a isomorphism between left and right cosets.]
97
98\begin{define} The \textbf{index} $(G:H)$ is the number of left (equivalently right) $H$-cosets in $G$ \end{define} %index
99\subsection*{THEOREM 1.26 (LAGRANGE)} If $G$ is finite, then
100$$(G:1) = (G:H)(H:1)$$
101In particular, the order of every subgroup of a finite group divides the order of the group.
102\begin{proof}
103    $(G,1) = \sum_{a \in G}$ Cosets partition $G$ so $\sum_{aH \subset G}|aH| = |G| = (G:1)$ \\
104    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
105    $(G:1) = |G| = \sum_{a \in G}|aH| = (G:H)(H:1) 106$
107    \end{proof}
108
109\subsubsection*{COROLLARY 1.27} The order of each element of a finite group divides the order of the group.
110\begin{proof}
111    Set $H = \gens{a}$, then $|H| = o(a)$
112    \end{proof}
113One of the Sylow theorems is a partial converse of Lagrange for prime-powers $p^n$: \\
114\subsection*{THEOREM 5.2 (SYLOW I)} Let $G \in \bold{FinGp}$ be a finite group, and let $p$
115be prime. If $p^n | (G:1)$, then $G$ has a subgroup of order $p^n$
116\begin{proof}
117TBA
118    \end{proof}
119Toshow: nexted subgroups $K<H<G: (G:K) = (G:H) (H:K)$
120
121\section*{Normal subgroups, conjugacy}
122\begin{define} $N \normal G$ if $\forall g \in G, gNg^{-1} = N$
123 \end{define}
124What about if $g^{-1}Ng = N$? \\
125$126 g^{-1}Ng = N \iff Ng = gN \iff gNg^{-1} = N 127$ So it doesn't matter how you conjugate. \\
128\subsubsection*{REMARK 1.32} suffices to show $gNg^{-1} \subset N (\forall g \in G)$
129\begin{proof}
130    $$gNg^{-1} \subset N \implies N \subset g^{-1}Ng = gNg^{-1}$$ $\therefore gNg^{-1} = N$.
131    \end{proof}
132\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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136\begin{sageblock}
137    C= matrix(SR, 2, [1,'n',1,0])
138    \end{sageblock}
139%%%%%%%%%%%%%%%%%%%%%%%%%%%%SCGE%%%%%%%%%%%%%%%
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141$142 \sage{matrix([[5,0],[0,1]])}\sage{C}\sage{matrix([[5**-1,0],[0,1]])} 143 = \sage{matrix([[5,0],[0,1]])*C*matrix([[5**-1,0],[0,1]])} \cong 5\Z 144$
145$146 \sage{matrix([[5**-1,0],[0,1]])}\sage{C}\sage{matrix([[5,0],[0,1]])} 147 = \sage{matrix([[5**-1,0],[0,1]])*C*matrix([[5,0],[0,1]])} 148$
149In the first case conjugation yields a proper subset, but conjugation by the inverse isn't even in $H$.
150
151%$\left(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right), \gens{\sage{C}} 152\subsubsection*{EXAMPLE 1.36} (a) Every subgroup of index two is normal. \\ 153(b) Dihedral group, its cyclic and translational symmetries,$n=2, n>2$. \\ 154(c) Subgroups in$\bold{Ab}$are normal, but converse not true, e.g.,$Q$155\begin{proof} 156 (a) The subgroup$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$. 157 Then by exclusion,$gH = Hg$\\ 158 (b) by commutivity,$C_n \normal D_n \forall n$, but 159 \end{proof} 160\begin{define} When$1 \trianglelefteq N$is the only series of normal subgroups,$N$is \textbf{simple}\end{define} 161 162\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$G$. 163\begin{proof} 164(1) First the case of mutual normalcy:$H,N \normal G$, 165$gHNg^{-1} = g(g^{-1}Hg)(g^{-1}Ng)g^-1 = HN$ 166(2) Relaxing the normalcy condition on$H$: 167$(HN)^{-1} = NH = HN 168$ 169 \end{proof} 170Moreover, if we let$X = N \cap N'$171\begin{define}$\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}$. 172 \end{define} 173\subsubsection*{LEMMA 1.38} If X is normal, then the subgroup$\gens{X}$generated by it is also normal. 174\begin{proof} 175Say elements of$\gens{X}$are of the form$a = a_1 \dots a_n$, then 176 \end{proof} 177\subsubsection*{LEMMA 1.39}$\bigcup_{g \in G}{gXg^{-1}}$is the smallest normal set containing$X$. 178\begin{proof} 179TBA 180 \end{proof} 181\subsubsection*{PROPOSITION 1.40}$\gens{X} = \gens{\bigcup_{g \in G}{gXg^{-1}}} \normal G$182\begin{proof} 183TBA 184 \end{proof} 185%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 186%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 187%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 188\section*{Ch1 exercises} 189%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%% 190%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%% 191%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%% 192%sage: P = PermutationGroup(['(1,2,3)(4,5)']) 193%sage: p = P.gen(0) 194%sage: p.matrix() 195 196\begin{sageblock} 197 var('q w r k') 198 P = PermutationGroup(['(1,2,3)','(2,3)']) 199 p = P.gen(0) 200 IsoZ = matrix(SR, 2, [[1,'n'],[0,1]]) 201 a = matrix(SR, 3, 3, [[1, 'a', 'b'], [0, 1, 'c'], [0, 0, 1]]) 202 K = matrix(SR, [[q,w+r],[0,q^2*r]]) 203 A = matrix(2, [0,i, i,0]) 204 B = matrix(2, [0,1, -1,0]) 205\end{sageblock} 206 207%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%% 208%%%%%%%%%%%%%%%%%%%%%%%%%%%%SAGE%%%%%%%%%%%%%%% 209\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}$210\begin{proof} 211$212\sage{A}\sage{B} = \sage{A*B} \not= \sage{B}\sage{A} = \sage{B*A} 213$ so$Q$nonabelian. 214$AXA^{-1} = \sage{A*matrix(SR, 2, ['a','b','c','d'])*A**(-1)}$215$BXB^{-1} = \sage{B*matrix(SR, 2, ['a','b','c','d'])*B^(-1)}$216 217 \end{proof} 218\subsubsection*{1-2} Using matrices in$GL(2,\Z[i])$, show$\gens{a,b|a^4 = b^3 =1} \notin \bold{FinGp}$219\begin{proof} 220$ab = \sage{matrix([[1,1],[0,1]])}, \left\{(ab)^n = \sage{IsoZ}\right\} \cong \Z$221 \end{proof} 222\subsubsection*{1-3} Show$G:|G| \in 2\Z$has an element$a:oa=2$. 223\begin{proof} 224 oa | 2n, 225\end{proof} 226\subsubsection*{1-4} Let$n = \sum^r_1{n_i}$, use lagrange to show$\prod^r_1{n_i !} | n!$227\begin{proof} 228Consider 229 \end{proof} 230\subsubsection*{1-5} Let$N \normal G: (G:N)= n$. Show$g^n \in N$, and that in nonnnormal subgroups this may not be true. 231\begin{proof} 232TBA 233 \end{proof} 234 235\subsubsection*{1-6} We say$m \in \N$is the \textbf{exponent} of$G$if it's the smallest annihilator of$G$. \\ 236(a) Show$m=2 \implies G \in \bold{Ab}$\\ 237(b) Let$G$be the following group. Verify that$m=p$and$G \notin \bold{Ab}$\\ 238$239 G :=\left\{\sage{a}\right\} \subset GL(3,{\mathbb{F}}_p) 240$ 241\begin{proof} 242(a)$abab=e, a^{-1}b^{-1} = ab=ba \qed$\\ 243(b) Show$a,b,c \in p\Z/$244 245 246TBA 247 \end{proof} 248%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 249 250\subsubsection*{1-7} Two subgroups$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'$. Show that commensurability is an equivalence relation on subgroups of G. 251\subsubsection*{1-8} Show that a nonempty finite set with an associative binary 252operation satisfying the cancellation laws is a group. 253cancellation law:$an = am \implies n = m\$
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258\end{document}
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