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\pages{1-7}
    \atxy{115}{172}{\parbox{\textwidth}{infinite cyclic group $C_\infty=\{\ldots,a^{-1},1,a,\ldots\}\cong\mathbb{Z}$\\$a^n\leftrightarrow n$\\$a^ma^n\leftrightarrow m+n$}}
\pages{8-9}
    \atxy{117}{35}{$=\left<x,y|x^2=y^2=xy=1\right>$}
\pages{10-11}
    \atxy{106}{169}{Sheet1 Q8}
    \atxy{76}{184}{\parbox{\textwidth}{\href{http://en.wikipedia.org/wiki/Generating_set_of_a_group\#Finitely_generated_group}{Wikipedia}: the free group $F_2$ contains a subgroup generated by $y^n x y^{-n}, n \ge 1$, which is free on countably many generators.}}
\pages{12-13}
    \atxy{20}{18}{Diamond lemma}
\pages{14}
    \atxy{127}{167}{$\phi:x_1^{m_1}x_2^{m_2}\dots x_n^{m_n}\mapsto f(x_1)^{m_1}f(x_2)^{m_2}\dots f(x_n)^{m_n}$}
\pages{15-17}
    \atxy{100.5}{104.3}{\color{white}\rule{2mm}{2mm}}
    \atxy{100.5}{104.3}{\scriptsize2}
    \atxy{106.4}{104.3}{\color{white}\rule{2mm}{2mm}}
    \atxy{106.4}{104.3}{\scriptsize\it n}
    \atxy{120}{154.2}{\parbox{\textwidth}{Dicyclic group $\mathrm{Dic}_{4n}=\left<a,b\mid a^{2n}=1,a^n=b^2,bab^{-1}=a^{-1}\right>$\\Note this implies $b^4=1$\\$n=2$: quaternion group $Q_8=\{\pm1,\pm i,\pm j,\pm k\}$\\$\{\pm1\}$ is the center\\smallest group that is neither abelian nor dihedral}}
\pages{18}
    \atxy{19.4}{200}{\parbox[t]{\textwidth}{{\bf Example }$D_8\overset{d}\to C_2\times C_2\cong V_4=\{1,g,h,gh\}$\\$a\mapsto g,b\mapsto h$ defines a homomorphism $\phi$.\\[1ex]{\bf Proof. }By Lemma 31 we verify:\begin{align*}a^4&\mapsto g^4=1\\b^2&\mapsto h^2=1\\bab^{-1}a&\mapsto hgh^{-1}g=1\qquad\square\end{align*}$\phi$ is onto $V_4$ as $\operatorname{Im}\phi$ contains the generators $g$ and $h$.\\$\ker\phi=\{1,a^2\}\cong C_2$, by first isomorphism theorem,\[D_8/C_2\cong V_4\]This is an example that $G$ is (solvable) not abelian but $N,G/N$ are abelian.}}
\pages{19-20}
    \atxy{125}{164}{\parbox{\textwidth}{In general, let $H,K$ be subgroups of a group $G$.\\$HK$ is a subgroup of $G$ if and only if $HK=KH$.}}
\pages{21}
    \atxy{117}{73.5}{Similarly, $A_4/V_4\cong C_3$.}
    \atxy{20}{183}{\href{https://proofwiki.org/wiki/Correspondence_Theorem_(Group_Theory)}{Correspondence Theorem}}
\pages{22}
    \atxy{40}{52.5}{$G/K$ is simple$\iff K$ is a maximal normal subgroup of $G$. \small Proof: $\begin{aligned}\nexists\bar N:{}&1\subsetneq\bar N\subsetneq G/K\\\nexists N:{}&K\subsetneq N\subsetneq G\end{aligned}$}
    \atxy{19}{217}{\parbox[t]{136mm}{\href{https://planetmath.org/conjugacyinan}{conjugacy in $A_n$}: A conjugacy class in $S_n$ splits into two distinct conjugacy classes under the action of $A_n$ if and only if its cycle type consists of distinct odd integers.}}
    \atxy{159}{156}{\href{https://groupprops.subwiki.org/wiki/Alternating_groups_are_simple}{Proof of $A_n,n>5$ are simple}}
\pages{23}
    \atxy{77}{183}{Composition series of $S_5$ is $1\lhd A_5\lhd S_5$}
\pages{24}
    \atxy{20}{75}{\parbox[t]{\textwidth}{$|GL(2,7)|=(7^2-1)(7^2-7)$\\$\det:GL(2,7)\to\mathbb F_7^\times$ is onto homomorphism, kernel is $SL(2,7)$\\by first isomorphism theorem $|SL(2,7)|=\frac{|GL(2,7)|}{7-1}$\\$A_5$ is smallest nonabelian simple group, $PSL(2,7)$ is second smallest.\\$|PSL(2,7)|=\frac{|SL(2,7)|}2$\\Another example:\\$|PSL(2,\mathbb F_5)|=\frac{5^3-5}2=60,|A_5|=\frac{5!}2=60$. In fact, $PSL(2,\mathbb F_5)\cong A_5$}}
    \atxy{30}{58}{$SL(2,7)$ is 2$\times$2 matrices in $\mathbb F_7$ of determinant 1. $PSL(2,7)=SL(2,7)/\{\pm I\}$ is simple.}
\pages{25}
    \atxy{106.6}{118.3}{\color{white}\rule{2mm}{3.4mm}}
    \atxy{106.6}{118.3}{\large``}
\pages{26-27}
    \atxy{72}{211}{Proof: Third isomorphism theorem.}
\pages{28}
    \atxy{20}{80}{\parbox[t]{\textwidth}{\href{https://math.stackexchange.com/questions/2879557}{How many} distinct composition series does the group $D_{12}$ have?\begin{gather*}1\lhd C_2\lhd C_6\lhd D_{12}\\1\lhd C_3\lhd C_6\lhd D_{12}\end{gather*}In general, $D_{2n}$ has factors $C_{p_i}$ for various primes.\\[1em]For $n\ge5$, $A_n$ is simple, so the composition series for $S_n$ is\[1\lhd A_n\lhd S_n\]}}
\pages{29}
    \atxy{180}{161}{\href{https://groupprops.subwiki.org/wiki/A3_in_S3}{$A_3$ in $S_3$}}
    \atxy{110}{117}{\parbox{84mm}{\begin{flushright}\href{https://math.stackexchange.com/questions/255274/}{Normal subgroup is not transitive}: $V_4\lhd A_4\lhd S_4$, but $V_4\leqslant S_4$ is not normal.\end{flushright}}}
\pages{30}
    \atxy{20}{39.5}{Abelian property is preserved under taking {\bf subgroups, quotients} but not {\bf extensions}: $C_2\cong D_{2n}/C_n$ but $D_{2n}$ nonAbelian}
    \atxy{75}{43}{Solvable property is preserved under taking {\bf subgroups, quotients} and {\bf extensions}.}
    \atxy{90}{154}{($H_{i+1}\cap G_i=H\cap G_{i+1}\cap G_i=H\cap G_i=H_i$)}
\pages{31}
    \atxy{126}{73}{\parbox[t]{\textwidth}{If $f:G\to H$ is homomorphism, $H$ abelian,\\then $G'\leqslant\ker f$.}}
    \atxy{114}{82}{\parbox[t]{10cm}{Example. $F_2=\text{free group on 2 letters}=\langle a,b|\emptyset\rangle$\\$(F_2)_{\text{ab}}=\langle x,y|xy=yx\rangle\cong\mathbb Z^2$}}
    \atxy{69}{91}{\href{https://math.stackexchange.com/questions/1510680}{For $n\ge5$, show that $[A_n,A_n]=A_n$}}
\pages{32}
    \atxy{181.4}{100}{\color{white}\rule{5mm}{4mm}}
    \atxy{181.7}{99.5}{$\Rightarrow$}
    \atxy{73.2}{113.6}{\color{white}\rule{2mm}{2mm}}
    \atxy{73}{115.7}{\small\it i}
    \atxy{26}{178.2}{\textbullet\quad The derived series for $D_{2n}$ is $D_{2n}\rhd\langle a^2\rangle\rhd\{e\}$ where $\langle a^2\rangle\cong C_{2n}$ when $n$ is odd}\atxy{121}{183}{$\cong C_n$ when $n$ is even}
\pages{33}
    \atxy{126}{124}{\href{https://groupprops.subwiki.org/wiki/Product_formula}{Order of Subgroup Product}}
    \atxy{76}{191}{\href{https://groupprops.subwiki.org/wiki/Permutable_complement}{Permutable complements}}
    \atxy{76}{195}{\href{https://groupprops.subwiki.org/wiki/Normal_complement}{Normal complement} $G=HN=NH$.}
\pages{34}
    \atxy{78}{28}{$H$ acts on $N$ by conjugation, but this action is nontrivial, so this is not {\it direct} product.}
    \atxy{87}{44}{\href{https://groupprops.subwiki.org/wiki/S3_in_S4}{$S_3$ in $S_4$}}
\pages{35}
    \atxy{147.5}{98}{\color{green}{\rule{2.3cm}{.5mm}}}
    \atxy{153}{101}{\parbox[t]{\textwidth}{$\phi(1_H)=\text{id}_N$, so\\$(n_1,1)\circ(n_2,1)=(n_1n_2,1)$}}
    \atxy{130}{167.3}{,}
    \atxy{20}{221}{\parbox[t]{20cm}{{\bf Example.} (Affine group) semidirect product $\mathbb R^n\rtimes\text{GL}(n,\mathbb R)$\\$H=\{x\mapsto Ax:A\in\text{GL}(n,\mathbb R)\}$ group of invertible linear maps\\$N=\{x\mapsto x+b|b\in\mathbb R^n\}$ group of translation\\$\text{Aff}(n,\mathbb R)=\{x\mapsto Ax+b|A\in\text{GL}(n,\mathbb R),b\in\mathbb R^n\}$ group of affine transform with multiplication\[A'(Ax+b)=A'Ax+(Ab+b')\]{\bf Note.} $b\mapsto Ab$ is an automorphism of $N$}}
    \atxy{134.2}{97.6}{\color{white}\rule{1.2mm}{4.6mm}}
\pages{36}
    \atxy{110}{35}{in language of rings: $\text{Aut}(\mathbb Z_n)=U(\mathbb Z_n),\psi(g)=g^i,\gcd(i,n)=1$}
    \atxy{142}{181.5}{$=\text{Dic}_{12}$\quad dicyclic group}
    \atxy{40.7}{206}{\scriptsize2}
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    \atxy{157.4}{206.2}{\scriptsize4}
    \atxy{150}{189}{\scriptsize2}
    \atxy{1}{65}{\href{https://users.metu.edu.tr/sozkap/461/The number of homomorphisms from Zn to Zm.pdf}{The number of homomorphisms from $\mathbb Z^n$ to $\mathbb Z^m$ is $\gcd(m,n)$}}
\pages{37}
    \atxy{187}{163}{\color{white}\rule{2mm}{6mm}}
\pages{38-39}
    \atxy{117}{121}{\color{green}$\text{Aut}(\mathbb Z_2)$ is trivial, since any automorphism fixes $0$ so fixes $1$}
    \atxy{29.5}{130}{$S_3$ as an extension of $A_3$ by $C_2$ does not split \href{https://en.wikipedia.org/wiki/Splitting_lemma\#Counterexample}{Splitting lemma--Counterexample}}
    \atxy{97}{51}{\setlength{\unitlength}{8pt}
    \begin{picture}(6,4)\put(4,1.1){\oval(3,1)[b]}\put(3.8,.6){\vector(-1,0){0}}\put(3.8,-.3){\scriptsize\it j}\end{picture}}
    \atxy{139}{43}{\parbox[t]{74mm}{{\bf Step 1: }$i(A)\cap j(B)=\{e\}$\\
    {\it Proof: }Let $g\in i(A)\cap j(B)$.\[g\in i(A)\Rightarrow\pi(g)=e\]$g\in j(B)\Rightarrow\exists b\in B:g=j(b)$\[\pi\circ j=\text{id}_B\Rightarrow\pi(g)=\pi(j(b))=b\]Comparing two equations, $b=e$, so $g=e.\quad\blacksquare$}}
    \atxy{102}{85}{\parbox[t]{\textwidth}{{\bf Step 2: }$G=i(A)\cdot j(B)$\\{\it Proof: }For any $g\in G,g=gh^{-1}\cdot h$, where $h=j(\pi(g))\in j(B)$.\\It remains to prove $gh^{-1}\in i(A)$.\begin{align*}\pi(gh^{-1})&=\pi(g)\pi(h^{-1})=\pi(g)\pi(j(\pi(g^{-1})))\\&=\pi(g)\pi(g^{-1})=e\\&\Rightarrow gh^{-1}\in\ker\pi=i(A).\quad\blacksquare\end{align*}}}
\pages{40}
    \atxy{72}{112}{\color{green}$\Omega=\{(x_1,\dots,x_p)\mid x_i\in G,x_1x_2\dots x_p=1\}$}
    \atxy{148}{106}{\color{green}$x_1x_2\dots x_p=1,C_p$ acts on $\Omega$ by cyclic perm}
    \atxy{148}{110}{\color{green}conjugate by $x_1\Rightarrow x_2\dots x_px_1=1$}
    \atxy{138}{114}{\color{green}orbits are of size 1 or $p\Rightarrow\exists x\ne1,(x,\dots,x)\in\Omega$}
    \atxy{195}{118}{\color{green}$\Rightarrow x^p=1$}
    \atxy{20}{186}{\color{green}Any subgroup conjugate to Sylow $p$-subgroup is a Sylow $p$-subgroup}
    \atxy{20}{215}{\href{https://math.stackexchange.com/questions/4238356}{A normal $p$-subgroup is contained in every Sylow $p$-subgroup}}
    \atxy{45}{201}{\color{white}\rule{9mm}{2.5ex}}
    \atxy{45}{201}{\href{https://math.stackexchange.com/questions/150534}{MSE}}
\pages{41-42}
    \atxy{77}{145}{\color{green}we don't know $H$ is normal in $G$, so right cosets don't form a group.}
    \atxy{191}{158}{\color{green}\#size-1-orbits}
    \atxy{191}{162}{\color{green}$\equiv|\Sigma|\pmod p$}
\pages{43}
    \atxy{30}{220}{\color{green}$\text{Cauchy}\Rightarrow\exists\text{index-2 subgroup}$}
    \atxy{107}{202}{\begin{tabular}{|c|l|}\hline
        \bf Order&\bf Abelian Group\\\hline
        1     &$Z_1 = S_1 = A_2$\\\hline
        2     &$Z_2 = S_2 = D_2$\\\hline
        \end{tabular}}
    \atxy{154}{235}{\begin{tabular}{|c|l|}\hline
        3     &$Z_3 = A_3$\\\hline
        4     &$Z_4 = Dic_1$\\\hline
        4     &$Z_2^2 = K_4 = D_4$\\\hline
        5    &$Z_5 $\\ \hline
        6    &$Z_6 = Z_3\times Z_2$\\ \hline
        7    &$Z_7 $\\ \hline
        8    &$Z_8 $\\ \hline
        8    &$Z_4\times Z_2$\\ \hline
        8    &$Z_2^3$\\ \hline
        9    &$Z_9 $\\ \hline
        9    &$Z_3^2$\\ \hline
        10  &$Z_{10} = Z_5\times Z_2$\\ \hline
        11  &$Z_{11}$\\\hline
        12  &$Z_{12} = Z_4\times Z_3$\\\hline
        12  &$Z_6\times Z_2 = Z_3\times Z_2^2$\\\hline
        13  &$Z_{13} $\\\hline
        14  &$Z_{14} = Z_7\times Z_2$\\\hline
        15  &$Z_{15} = Z_5\times Z_3$\\\hline
        \end{tabular}}
    \atxy{83}{250}{\begin{tabular}{|c|l|} \hline
        \bf Order&\bf Non-Abelian Group\\\hline
        6&$D_6=S_3=Z_3\rtimes Z_2$\\\hline
        8&$D_8$\\\hline
        8&$Q_8$\\\hline
        10&$D_{10}$\\\hline
        12&$Q_{12}=Z_3\rtimes Z_4$\\\hline
        12&$A_4=K_4\rtimes Z_3=(Z_2\times Z_2)\rtimes Z_3$\\\hline
        12&$D_{12}=D_6\times Z_2$\\\hline
        14&$D_{14}$\\\hline
        \end{tabular}}
\pages{44}
    \atxy{138}{22}{\color{green}$Z(G)\leqslant G$}
    \atxy{35}{70}{\href{https://proofwiki.org/wiki/Group_of_Order_p_q_is_Cyclic}{ProofWiki}}
\pages{45}
    \atxy{58.6}{169}{\color{green}\rule{61mm}{1px}}
    \atxy{125.6}{170.6}{\color{green}$V_4=\{1,(12)(34),(13)(24),(14)(23)\}\lhd A_4$}
    \atxy{112}{197}{\color{green}$G=\{1,a,a^2,a^3,a^4,a^5,b,ba,ba^2,ba^3,ba^4,ba^5\}$}
    \atxy{45}{27}{\href{https://math.stackexchange.com/questions/6078}{MSE}}
\pages{46}
\pages{47}
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