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\begin{document}

\title[252 Worksheet \#3]{Math 101: Algebra I \\ Worksheet, Day \#3}
\date{Wednesday, 14 September 2016}

\newcommand{\shortblank}{\underline{\hspace{1in}}}
\newcommand{\longblank}{\underline{\hspace{2in}}}
\newcommand{\Longblank}{\underline{\hspace{2.5in}}}
\newcommand{\twolongblank}{\underline{\hspace{3.5in}}}
\newcommand{\prooflongblank}{\underline{\hspace{4.5in}}}
\newcommand{\fulllongblank}{\underline{\hspace{\textwidth}}}

\maketitle

Fill in the blanks as we finish our first pass on prerequisites of group theory.

\section{Subgroups, cosets}

Let $G$ be a group.  Recall that a subgroup $H \leq G$ is a subset that is a group under the binary operation of $G$.  The \defi{subgroup criterion} says that $H \subseteq G$ is a subgroup if and only if \twolongblank.  It shows that if $\phi:G \to G'$ is a group homomorphism, then $\phi(G) \leq G'$ is a subgroup: {\underline{\hspace{3in}}}.

Let $H \leq G$ be a subgroup.  Define an equivalence relation on $G$ by $a \sim b$ if and only if $a^{-1}b \in H$.  This is an equivalence relation because {\underline{\hspace{3in}}} \\ \fulllongblank. The equivalence classes are $a H = \{ah : h \in H\}$ for $a \in G$; note $a H = b H$ if and only if $H = a^{-1} b H$.  Let $G/H$ be the set of equivalence classes.

We similarly define equivalence on the other side, with classes $Ha$ for $a \in G$, and write $H \backslash G$ for the set of equivalence classes.

\defi{Lagrange's theorem} states: {\underline{\hspace{4.4in}}} \\ \fulllongblank.

It implies \defi{Fermat's little theorem}: {\underline{\hspace{4in}}} \\ \prooflongblank\ because {\underline{\hspace{1.33in}}} \\
\fulllongblank.

\vfill\newpage

\section{Centralizer, normalizer, stabilizer}

Let $A \subseteq G$ be a subset.  The \defi{centralizer} of $A$ in $G$ is $C_A(G)=$ \longblank,
consisting of the elements of $G$ \twolongblank.  The centralizer ${C_A(G) \leq G}$ is a subgroup: 
{\underline{\hspace{4in}}}.  The \defi{center} of $G$ is $Z(G)=C_G(G)$. 

Similarly, the \defi{normalizer} of $A$ in $G$ is $N_A(G)=$ \longblank.  Note that $C_G(A) \leq N_G(A)$.  The normalizer $N_A(G) \leq G$ is a subgroup.

Let $G \circlearrowright X$ act on a set $X$.  Let $x \in X$.  The \defi{stabilizer} of $x \in G$ is the set $\mathrm{Stab}_G(x) = G_x = \{g \in G : gx=x\}$.  We verify that $G_x \leq G$ is a subgroup.  Recall that the kernel of the action is the subgroup \twolongblank.  

\begin{enumerate}
\item Let $G \circlearrowright X=G$ act on itself by conjugation, so that \longblank.  Then the centralizer of $A$ in terms of the group action is \longblank \\
\twolongblank.  

\item  Let $X=\calP(G)$ be the \defi{power set} of $G$, consisting of \Longblank \\ \longblank.  Then $G \circlearrowright X$ acts on $X$ by conjugation via $(g,A) \mapsto g A g^{-1}=\{gag^{-1} : a \in A\}$.  Under this action, $N_G(A)$ is \longblank.

\item   Let the group $N_G(A) \circlearrowright A$ act on $A$ by conjugation.  Then $C_G(A) \leq N_G(A)$ is \longblank.
\end{enumerate}

\newpage

\section{Cyclic groups}

A group $G$ is \defi{cyclic} if \twolongblank, and we write \longblank.  In this case, we have an isomorphism 
\begin{center}
\longblank \\
\longblank
\end{center}
Consequently, any two cyclic groups of the same order are isomorphic.  

\begin{lem}
Let $G$ be a cyclic group, generated by $x$, and let $a \in \Z$.  
\begin{enumalph}
\item If $x$ has infinite order, then $G=\la x^a \ra$ if and only if \shortblank.
\item If $x$ has order $n<\infty$, then $G=\la x^a \ra$ if and only if \shortblank; in particular, the number of generators of $G$ is \longblank.
\end{enumalph}
\end{lem}

Every subgroup of a cyclic group is cyclic: {\underline{\hspace{3.3in}}} \\ \fulllongblank \\ \fulllongblank.  
In fact, the subgroups of a finite cyclic group correspond bijectively with \\ 
\longblank via \twolongblank.

\newpage

\section{Generating subgroups}

Let $G$ be a group.  Given any nonempty collection $\{H_i\}_{i \in I}$ of subgroups of $G$, the intersection $\bigcap_{i \in I} H_i \leq G$ is a subgroup.  

Let $A \subseteq G$.  The \defi{subgroup generated by $A$}, written $\langle A \rangle \subseteq G$, is
\[ \langle A \rangle = \bigcap_{\substack{A \subseteq H \\ H \leq G }} H. \]
The subgroup generated by $A$ is the smallest subgroup of $G$ containing $A$ in the sense that \fulllongblank.  A more concrete way of thinking about $\langle A \rangle$ is that it consists of all elements of $G$ that can be written as \defi{words} in elements of $A \cup A^{-1}$, where $A^{-1}=\{a^{-1} : a \in A\}$: this means 
\fulllongblank.
A subgroup $H \leq G$ is \defi{finitely generated} if \\
\prooflongblank.

\begin{exm}
The subgroup of $\GL_2(\C)$ generated by 
\[ A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},\quad  B=\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \]
is {\underline{\hspace{6.35in}}} \\
\fulllongblank.
\end{exm}

\begin{exm}
The subgroup of $\GL_2(\Q)$ generated by
\[ S=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},\quad T=\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix} \]
is infinite even though $S,T$ have orders \longblank because
{\underline{\hspace{1.15in}}} \\
\fulllongblank.
\end{exm}

\section{Kernels, normal subgroups, quotient groups}

Let $\phi:G \to G'$ be a group homomorphism.  The \defi{kernel} of $\phi$ is the subgroup
\begin{center}
$\ker \phi=$\longblank\  $\leq G$.
\end{center}

\begin{lem}
$\phi$ is injective if and only if $\ker\phi=\{1\}$.
\end{lem}

\begin{proof}
{\underline{\hspace{6in}}} \\
\prooflongblank.
\end{proof}

\begin{lem}
Let $K \leq G$ be a subgroup.  Then the following conditions are equivalent:
\begin{enumroman}
\item For all $a \in G$, we have $aK=Ka$;
\item For all $a \in G$, we have $aKa^{-1} \subseteq K$; and
\item For all $a \in G$, we have $aKa^{-1}=K$.
\end{enumroman}
\end{lem}

\begin{proof}
Proves itself.
\end{proof}

A subgroup $K \leq G$ is \defi{normal} if the equivalent conditions (i)--(iii) hold, and we write $K \trianglelefteq G$.  

\begin{exm}
Every subgroup of an abelian group is normal.
\end{exm}

\begin{prop}
If $K \leq G$, then $G/K$ is a group under $(aK)(bK)=abK$ if and only if $K \trianglelefteq G$.
\end{prop}

\begin{proof}
{\underline{\hspace{6in}}} \\
\fulllongblank.
\end{proof}

If $K \trianglelefteq G$, then the group $G/K$ is the \defi{quotient group}.  Kernels of surjective group homomorphisms are the same as normal subgroups in the sense that \Longblank \\ \fulllongblank.  In general, if $\phi: G \to G'$ is a group homomorphism, then it factors:
\vspace{2in}

\begin{thm}[First isomorphism theorem]\  \\
\fulllongblank
\end{thm}

There are also three more isomorphism theorems, concerning intersections, quotients, and lattices.

\begin{exm}
The group $\Q/\Z$ looks like \\
\fulllongblank.
\end{exm}

\newpage

\section{Composition series}

A (finite or infinite) group is \defi{simple} if $\#G > 1$ and {\underline{\hspace{2.7in}}} \\
\longblank.

A \defi{composition series} for $G$ is a sequence of subgroups
\[ 1 = N_0 \leq N_1 \leq \dots \leq N_{k-1} \leq N_k = G \]
such that $N_i \trianglelefteq N_{i+1}$ and $N_{i+1}/N_i$ is a simple group for all $0 \leq i \leq k-1$; we then call the set of groups $N_{i+1}/N_i$ the \defi{composition factors}.  

\begin{exm}
A composition series for $S_3$ is: \\
\fulllongblank

A composition series for $D_8$ is: \\
\fulllongblank

A composition series for $A_4$ is: \\
\fulllongblank
\end{exm}

\begin{exm}
Let $F$ be a field with $\#F=p$.  A composition series for the Heisenberg group $H(F)$ is: {\underline{\hspace{6.3in}}}
\end{exm}

\begin{thm}[Jordan--H\"older]
Every nontrivial finite group has a composition series and the composition factors in any composition series are unique (up to reordering).
\end{thm}

To classify all finite groups, we classify all finite simple groups and then try to ``put them back together'' to get all finite groups.

\begin{thm}
Every finite simple group is isomorphic to either one in a list of $18$ infinite families of simple groups or one of $26$ \emph{sporadic} simple groups.
\end{thm}

Examples of simple groups include:
\begin{itemize}
\item \prooflongblank
\item \prooflongblank
\item \prooflongblank
\end{itemize}

\end{document}


