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\begin{document}
\subsubsection*{sage filler}
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\begin{sagesilent}
GF2 = GF(2)
O = matrix(2, lambda i,j: i-j);
M = matrix(2, lambda i,j: (-1+2*i-i*j));
m1 = matrix(ZZ, 3, [[0,0,1],[0,1,0],[1,0,0]])
m2 = matrix(ZZ, 3, [[0,1,0],[0,0,1],[1,0,0]])
SSS = MatrixGroup(m1, m2)
S3 = SymmetricGroup(3)
U = GL(3, GF(5)).random_element()
V = GL(3, GF(3)).random_element()
GL3 = MatrixSpace(GF(7),3)
\end{sagesilent}
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some groups: $C_n,S_n,A_n,D_n;GF(p), Aut(E/F),$ \\
classic Lie type over $\R,\C,\bold{H}: GL,SO,SL, PSL,Sp,SU $ \\
$$\mbox{Dic}_n := \langle a,x \mid a^{2n} = 1,\ x^2 = a^n,\ x^{-1}ax = a^{-1}\rangle$$ alt def by exact sequence $
1 \to C_{2n} \to \mbox{Dic}_n \to C_2 \to 1$ \\
G-actions \\
$f:H \leq G \acts G; \ell_h(x) \ndef hx \in G $ \\
other way may not work? $ G \acts H; ah \not\in H$ \\
however $N \normal G, G \acts N; c_g(x) \ndef gxg^{-1}$ \\
$G \acts G/H, f(g,aH) := (ga)H$ \\
\section{permutations \& cycles}
$\sigma \in Aut \N$, e.g.,
\begin{itemize}
\item $n \mapsto n$
\item $\sigma \in \Z_p:n \mapsto n^2$
\item $n \mapsto n+1$
\end{itemize}
These are really permutations, an \emph{$N$-cycle} satisfies
\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 $\sigma \coprod \sigma'$
Consider the strictly nondecreasing 2-cycles $\sigma_i \geq \sigma_j$ equality iff equal etcetc.
\section*{milne}
\subsubsection*{PROPOSITION 1.25} Let H be a subgroup of a group G. \\
(a) An element $a$ of G lies in a left coset C of H if and only
if $C = aH$ \\
(b) Two left cosets are either disjoint or equal. \\
(c) $aH = bH$ if and only if $a^{-1} b \in H$ \\
(d) Any two left cosets have the same number of elements \\
(possibly infinite).
\begin{proof}
Recall $a \in H \implies aH = H$. If this is the case, any left coset of $H$ via elements of $H$ will be $aH (=H)$. \\
Otherwise for $a \ne c \in G - H: a \in cH \implies aH \subset cH$ (since $\exists h \in H: a = ch, h = c^{-1}a$) \\
Using this, $cH = chH = c c^{-1}aH = aH \\
\implies$ (a). \\
and since, by (a), any intersection implies equality of cosets, only disjunct cosets remain \\
$\implies$ (b) \\
Going the other way, if $aH = cH$: $$ca^{-1}H = H, ca^{-1} \in H $$ $\implies$ (c) \\
Define as follows
$$\phi: aH \to bH: x \mapsto (ba^{-1})x, \phi^{-1}:x \mapsto ab^{-1}c, \phi\phi^{-1} =\phi^{-1}\phi=1 $$
This defines a inverse system between $aH,bH$, thus iso ie same cardinality \\ $\implies $ (d)
\end{proof}
[Note 1: Cosets partition $G$, since $a \in aH$ along with 1.25(b).]
[Note 2:In regards to right cosets, if we modify the function in 1.25(c) slightly
$$\phi: aH \to Hb: ah \mapsto (ah)^{-1}ab = h^{-1}b \in Hb$$
$$\phi^{-1}: Hb \to aH: hb \mapsto ab(hb)^{-1} = ah^{-1} \in aH$$
we get a isomorphism between left and right cosets.]
\begin{define} The \textbf{index} $(G:H)$ is the number of left (equivalently right) $H$-cosets in $G$ \end{define} %index
\subsection*{THEOREM 1.26 (LAGRANGE)} If $G$ is finite, then
$$(G:1) = (G:H)(H:1)$$
In particular, the order of every subgroup of a finite group divides the order of the group.
\begin{proof}
$(G,1) = \sum_{a \in G}$ Cosets partition $G$ so $\sum_{aH \subset G}|aH| = |G| = (G:1)$ \\
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
\[ (G:1) = |G| = \sum_{a \in G}|aH| = (G:H)(H:1)
\]
\end{proof}
\subsubsection*{COROLLARY 1.27} The order of each element of a finite group divides the order of the group.
\begin{proof}
Set $H = \gens{a}$, then $|H| = o(a)$
\end{proof}
One of the Sylow theorems is a partial converse of Lagrange for prime-powers $p^n$: \\
\subsection*{THEOREM 5.2 (SYLOW I)} Let $G \in \bold{FinGp}$ be a finite group, and let $p$
be prime. If $p^n | (G:1)$, then $G$ has a subgroup of order $p^n$
\begin{proof}
TBA
\end{proof}
Toshow: nexted subgroups $K2$. \\
(c) Subgroups in $\bold{Ab}$ are normal, but converse not true, e.g., $Q$
\begin{proof}
(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 $H \coprod gH$.
Then by exclusion, $gH = Hg$ \\
(b) by commutivity, $C_n \normal D_n \forall n$, but
\end{proof}
\begin{define} When $1 \trianglelefteq N$ is the only series of normal subgroups, $N$ is \textbf{simple}\end{define}
\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 $G$.
\begin{proof}
(1) First the case of mutual normalcy: $H,N \normal G$,
\[ gHNg^{-1} = g(g^{-1}Hg)(g^{-1}Ng)g^-1 = HN\]
(2) Relaxing the normalcy condition on $H$:
\[ (HN)^{-1} = NH = HN
\]
\end{proof}
Moreover, if we let $X = N \cap N'$
\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} $\conj{X}$.
\end{define}
\subsubsection*{LEMMA 1.38} If X is normal, then the subgroup $\gens{X}$ generated by it is also normal.
\begin{proof}
Say elements of $\gens{X}$ are of the form $a = a_1 \dots a_n$, then
\end{proof}
\subsubsection*{LEMMA 1.39} $\bigcup_{g \in G}{gXg^{-1}}$ is the smallest normal set containing $X$.
\begin{proof}
TBA
\end{proof}
\subsubsection*{PROPOSITION 1.40} $\gens{X} = \gens{\bigcup_{g \in G}{gXg^{-1}}} \normal G$
\begin{proof}
TBA
\end{proof}
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\section*{Ch1 exercises}
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%sage: P = PermutationGroup(['(1,2,3)(4,5)'])
%sage: p = P.gen(0)
%sage: p.matrix()
\begin{sageblock}
var('q w r k')
P = PermutationGroup(['(1,2,3)','(2,3)'])
p = P.gen(0)
IsoZ = matrix(SR, 2, [[1,'n'],[0,1]])
a = matrix(SR, 3, 3, [[1, 'a', 'b'], [0, 1, 'c'], [0, 0, 1]])
K = matrix(SR, [[q,w+r],[0,q^2*r]])
A = matrix(2, [0,i, i,0])
B = matrix(2, [0,1, -1,0])
\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 $\forall H \leq Q, H \normal Q, \exists! a: oa=2, Q \notin \bold{Ab}$
\begin{proof}
\[
\sage{A}\sage{B} = \sage{A*B} \not= \sage{B}\sage{A} = \sage{B*A}
\] so $Q$ nonabelian.
$AXA^{-1} = \sage{A*matrix(SR, 2, ['a','b','c','d'])*A**(-1)}$
$BXB^{-1} = \sage{B*matrix(SR, 2, ['a','b','c','d'])*B^(-1)}$
\end{proof}
\subsubsection*{1-2} Using matrices in $GL(2,\Z[i])$, show $\gens{a,b|a^4 = b^3 =1} \notin \bold{FinGp}$
\begin{proof}
$ab = \sage{matrix([[1,1],[0,1]])}, \left\{(ab)^n = \sage{IsoZ}\right\} \cong \Z$
\end{proof}
\subsubsection*{1-3} Show $G:|G| \in 2\Z$ has an element $a:oa=2$.
\begin{proof}
oa | 2n,
\end{proof}
\subsubsection*{1-4} Let $n = \sum^r_1{n_i}$, use lagrange to show $\prod^r_1{n_i !} | n!$
\begin{proof}
Consider
\end{proof}
\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.
\begin{proof}
TBA
\end{proof}
\subsubsection*{1-6} We say $m \in \N$ is the \textbf{exponent} of $G$ if it's the smallest annihilator of $G$. \\
(a) Show $m=2 \implies G \in \bold{Ab}$ \\
(b) Let $G$ be the following group. Verify that $m=p$ and $G \notin \bold{Ab}$ \\
\[
G :=\left\{\sage{a}\right\} \subset GL(3,{\mathbb{F}}_p)
\]
\begin{proof}
(a) $abab=e, a^{-1}b^{-1} = ab=ba \qed$ \\
(b) Show $a,b,c \in p\Z/$
TBA
\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.
\subsubsection*{1-8} Show that a nonempty finite set with an associative binary
operation satisfying the cancellation laws is a group.
cancellation law: $an = am \implies n = m$
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\end{document}