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\title{Math 430 -- Problem Set 5 Solutions}
\date{Due March 18, 2016}

\begin{document}
\maketitle


\begin{enumerate}

\item[10.4.] Let $T$ be the group of nonsingular upper triangular $2 \times 2$ matrices with entries in $\RR$; that is, matrices of the form
\[
\begin{pmatrix} a & b \\ 0 & c \end{pmatrix},
\]
where $a, b, c \in \RR$ and $ac \ne 0$. Let $U$ consist of matrices of the form
\[
\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix},
\]
where $x \in \RR$.
\begin{enumerate}
\item Show that $U$ is a subgroup of $T$.
\begin{solution}
Taking $x = 0$, we see that the identity matrix is in $U$.  The inverse of $\begin{psmallmatrix} 1 & x \\ 0 & 1 \end{psmallmatrix}$ is $\begin{psmallmatrix} 1 & -x \\ 0 & 1 \end{psmallmatrix}$, which is also in $U$.  Finally,
\[
\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & y \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & x+y \\ 0 & 1 \end{pmatrix},
\]
which is in $U$.
\end{solution}
\item Prove that $U$ is abelian.
\begin{solution}
This follows from the formula for multiplication of elements of $U$ given above, together with the commutativity of addition in $\RR$.
\end{solution}
\item Prove that $U$ is normal in $T$.
\begin{solution}
\begin{align*}
\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ 0 & c \end{pmatrix}^{-1} &= \begin{pmatrix} a & ax+b \\ 0 & c \end{pmatrix} \begin{pmatrix} 1/a & -b/(ac) \\ 0 & 1/c \end{pmatrix} \\
&= \begin{pmatrix} 1 & ax/c \\ 0 & 1 \end{pmatrix}
\end{align*}
\end{solution}
\item Show that $T/U$ is abelian.
\begin{solution}
Note that
\[
\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & c \end{pmatrix} \begin{pmatrix} 1 & b/a \\ 0 & 1 \end{pmatrix},
\]
so every coset in $T/U$ has a representative that is a diagonal matrices.  Since diagonal matrices commute with each other, $T/U$ is commutative.

Alternatively, note that
\begin{align*}
\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \begin{pmatrix} a' & b' \\ 0 & c' \end{pmatrix} \begin{pmatrix} a & b \\ 0 & c \end{pmatrix}^{-1} \begin{pmatrix} a' & b' \\ 0 & c' \end{pmatrix}^{-1} &= \begin{pmatrix} aa' & ab' + bc' \\ 0 & cc' \end{pmatrix} \begin{pmatrix} 1/(aa') & -(b'c+bc')/(aca'c') \\ 0 & 1/(cc') \end{pmatrix} \\
&= \begin{pmatrix} 1 & (ab' -b'c)/(cc') \\ 0 & 1 \end{pmatrix}.
\end{align*}
Since $U$ contains the commutator subgroup of $T$, $T/U$ is abelian by 10.14.
\end{solution}
\item Is $T$ normal in $\GL_2(\RR)$?
\begin{solution}
No.  For example,
\[
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}.
\]
\end{solution}
\end{enumerate}
\item[10.7.] Prove or disprove: If $H$ is a normal subgroup of $G$ such that $H$ and $G/H$ are abelian, then $G$ is abelian.
\begin{solution}
$U \triangleleft T$ from the previous problem provides a counterexample, as does $A_3 \triangleleft S_3$.
\end{solution}
\item[10.9.] Prove or disprove: If $H$ and $G/H$ are cyclic, then $G$ is cyclic.
\begin{solution}
$A_3 \triangleleft S_3$ provides a counterexample, as does $\ZZ_2 \triangleleft \ZZ_2 \times \ZZ_2$.
\end{solution}
\item[10.14.] Let $G$ be a group and let $G' = \langle aba^{-1}b^{-1} \rangle$; that is, $G'$ is the subgroup of all finite products of elements in $G$ of the form $aba^{-1}b^{-1}$. The subgroup $G'$ is called the commutator subgroup of $G$.
\begin{enumerate}
\item Show that $G'$ is a normal subgroup of $G$.
\begin{solution}
Suppose $\gamma = aba^{-1}b^{-1}$ is a generator of $G'$.  Since $g\gamma g^{-1} = (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}$, we have that $g\gamma g^{-1} \in G'$.  Since conjugation by $g$ is a homomorphism, every product of such elements will also be an element of $G'$.  Thus $G'$ is normal.

Alternatively, note that $g\gamma g^{-1} = g\gamma g^{-1} \gamma^{-1} \gamma \in G'$ since $\gamma \in G'$ and $g\gamma g^{-1} \gamma^{-1}$ is a commutator.
\end{solution}
\item Let $N$ be a normal subgroup of $G$.  Prove that $G/N$ is abelian if and only if $N$ contains the commutator subgroup of $G$.
\begin{solution}
Suppose $a, b \in G$.  Then
\begin{align*}
(aN)(bN) = (bN)(aN) &\Leftrightarrow Nab = Nba \\
&\Leftrightarrow Naba^{-1}b^{-1} = N \\
&\Leftrightarrow aba^{-1}b^{-1} \in N.
\end{align*}
So 
\begin{align*}
\mbox{$G/N$ is abelian} &\Leftrightarrow \mbox{$(aN)(bN) = (bN)(aN)$ for all $a, b \in G$} \\
&\Leftrightarrow \mbox{$aba^{-1}b^{-1} \in N$ for all $a, b \in G$} \\
&\Leftrightarrow G' \subseteq N.
\end{align*}
\end{solution}
\end{enumerate}
\item[11.2.] Which of the following maps are homomorphisms? If the map is a homomorphism, what is the kernel?
\begin{enumerate}
\item $\phi : \RR^* \to \GL_2(\RR)$ defined by 
\[
\phi(a) = \begin{pmatrix} 1 & 0 \\ 0 & a \end{pmatrix}
\]
\begin{solution}
This is a homomorphism since $\begin{psmallmatrix} 1 & 0 \\ 0 & a \end{psmallmatrix} \begin{psmallmatrix} 1 & 0 \\ 0 & b \end{psmallmatrix} = \begin{psmallmatrix} 1 & 0 \\ 0 & ab \end{psmallmatrix}$.  The kernel is $\{1\} \subset \RR^*$.
\end{solution}
\item $\phi : \RR \to \GL_2(\RR)$ defined by 
\[
\phi(a) = \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}
\]
\begin{solution}
This is a homomorphism since $\begin{psmallmatrix} 1 & 0 \\ a & 1 \end{psmallmatrix} \begin{psmallmatrix} 1 & 0 \\ b & a \end{psmallmatrix} = \begin{psmallmatrix} 1 & 0 \\ a+b & 1 \end{psmallmatrix}$.  The kernel is $\{0\} \subset \RR$.
\end{solution}
\item $\phi : \GL_2(\RR) \to \RR$ defined by 
\[
\phi\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = a + d
\]
\begin{solution}
This is not a homomorphism since it maps the identity to $2$, which is not the identity in $\RR$.
\end{solution}
\item $\phi : \GL_2(\RR) \to \RR^*$ defined by 
\[
\phi\left(\begin{psmallmatrix} a & b \\ c & d \end{psmallmatrix}\right) = ad - bc
\]
\begin{solution}
This is a homomorphism, since
\begin{align*}
\phi\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}\right) &= \phi\left(\begin{pmatrix} aa'+bc' & ab'+bd' \\ ca'+dc' & cb'+dd' \end{pmatrix}\right) \\
&= (aa'+bc')(cb'+dd') - (ab'+bd')(ca'+dc') \\
&= ada'd' + bcb'c' - adb'c' - bca'd' \\
&= (ad - bc)(a'd' - b'c') \\
&= \phi\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) \phi\left(\begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}\right).
\end{align*}
The kernel is $\SL_2(\RR)$, the subgroup of $\GL_2(\RR)$ consisting of matrices of determinant $1$.
\end{solution}
\item $\phi : \mathbb{M}_2(\RR) \to \RR$ defined by 
\[
\phi\left(\begin{psmallmatrix} a & b \\ c & d \end{psmallmatrix}\right) = b,
\]
where $\mathbb{M}_2(\RR)$ is the additive group of $2 \times 2$ matrices with entries in $\RR$.
\begin{solution}
This is a homomorphism, since
\[
\phi\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}\right) = b + b' = \phi\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) + \phi\left(\begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}\right).
\]
The kernel is the group (under addition) of lower triangular matrices:
\[
\left\{\begin{pmatrix} a & 0 \\ b & c \end{pmatrix} : a, b, c \in \RR\right\}.
\]
\end{solution}
\end{enumerate}
\item[11.9.] If $\phi : G \to H$ is a group homomorphism and $G$ is abelian, prove that $\phi(G)$ is abelian.
\begin{solution}
If $x, y \in \phi(G)$ then there exist $a,b \in G$ with $x = \phi(a)$ and $y = \phi(b)$.  Then $xy = \phi(a)\phi(b) = \phi(ab) = \phi(ba) = \phi(b)\phi(a) = yx$, so $\phi(G)$ is abelian.
\end{solution}
\end{enumerate}

\end{document}