\documentclass[11pt]{article}
\usepackage{amsmath,amsfonts}
\usepackage[all]{xy}

\title{Solutions to Homework 5}
\author{}
\date{}

\newcommand \C {\mathbb C}
\newcommand \R {\mathbb R}
\newcommand \Z {\mathbb Z}

\DeclareMathOperator \Stab {Stab}
\DeclareMathOperator \GL {GL}

\renewcommand \phi \varphi

\begin{document}
\maketitle

\begin{enumerate}

\item Let $G$ be a group and $H$ a subgroup.  For $g, g' \in G$, show that the following are equivalent:
\begin{enumerate}
\item $g H = g' H$.
\item $g H \subset g' H$.
\item $g H \cap g' H \ne \emptyset$.
\item $g^{-1} g' \in H$.
\item $g' \in gH$.
\end{enumerate}
Warning: Do not assume that $G$ or $H$ is finite.

{\bf Solution:} (a) $\Rightarrow$ (b) and (b) $\Rightarrow$ (c) are clear.

For (c) $\Rightarrow$ (d), chose $h, h' \in H$ such that $gh = g'h'$.  Multiply on the left by $g^{-1}$ and on the right by $h'^{-1}$ to obtain $g^{-1} g' = h h'^{-1} \in H$.

For (d) $\Rightarrow$ (e), if $g^{-1} g' = h$ then $g' = gh \in gH$.

For (e) $\Rightarrow$ (a), write $g' = gh$.  For any $g'h' \in g'H$, we have $g'h' = (gh)h' = g(hh') \in gH$; thus $g'H \subset gH$.  For any $gh' \in gH$, we have $gh' = (g'h^{-1})h' = g'(h^{-1}h') \in g'H$; thus $gH \subset g'H$.


\item Let $G$ be a group and $H$ a subgroup.  Show that the following
are equivalent:
\begin{enumerate}
\item For all $g \in G$ one has $g H = H g$.
\item For all $g \in G$ one has $g H \subset H g$.
\item For all $g \in G$ one has $g H g^{-1} = H$.
\item For all $g \in G$ one has $g H g^{-1} \subset H$.
\end{enumerate}
Notice the difference between the statements
\[ \forall g \in G, (gH \subset Hg \Rightarrow gH = Hg) \]
and
\[ (\forall g \in G, gH \subset Hg) \Rightarrow (\forall g \in G, gH = Hg). \]
The implication (b) $\Rightarrow$ (a) is the latter.  The former holds if $H$ is finite.  Optional: Does it hold if $H$ is infinite?


{\bf Solution:} (a)~$\Rightarrow$~(b) and (c)~$\Rightarrow$~(d) are clear.

For (b)~$\Rightarrow$~(a), given $g \in G$ we have both $gH \subset Hg$ and $g^{-1} H \subset H g^{-1}$; multiplying by $g$ on the left and right we get $Hg \subset gH$, so $gH = Hg$.  More carefully, if $g^{-1} H \subset H g^{-1}$ then for every $h \in H$ there is an $h' \in H$ such that $g^{-1} h = h' g^{-1}$, so $hg = gh'$; thus $Hg \subset gH$, as desired.

The implication (d)~$\Rightarrow$~(c) is similar.

For (b)~$\Leftrightarrow$~(d), observe that for all $h \in H$ there is an $h' \in H$ with $gh = h'g$, if and only if for all $h \in H$ there is an $h' \in H$ with $ghg^{-1} = h'$.
\medskip

For the optional part, the answer is no.  One options is to let $G \subset \GL_n(\R)$ be the group of matrices of the form
\[ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}, \qquad a \ne 0, \]
and $H$ the subgroup of matrices of the form
\[ \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}, \qquad n \in \Z. \]
Take
\[ g = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}. \]
Then we have
\[ g \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} g^{-1} = \begin{pmatrix} 1 & 2n \\ 0 & 1 \end{pmatrix}, \]
so $gHg^{-1}$ is a proper subgroup of $H$.

Another option is to let $G$ be the free group on two generates $x,y$, that is, the group of strings
\[ x^{m_1} y^{n_1} x^{m_2} y^{n_2} \dotsm x^{m_k} y^{n_k}, \]
where $k$ is arbitrary, the exponents $m_i, n_i \in \Z$, and we impose only the most obvious relations: $x^0 = 1$, $x^a x^b = x^{a+b}$, and similarly with $y$.  Then let $H$ be the subgroup generated by $x^m y^n x^{-m}$, where $m \ge 0$ and $n$ is arbitrary: that is, elements where, as we read from left to right and keep a running sum of the exponents of $x$, the sum is always non-negative, and returns to zero by the end.  Then $xHx^{-1} \subset H$, but $y \notin xHx^{-1}$, because then we would have $x^{-1} y x \in H$.


\item
\begin{enumerate}
\item Let a group $G$ act on a set $X$ (on the left), and let $H \subset G$ be the stabilizer of a point $x \in X$.  Show that for $g \in G$, the stabilizer of $gx$ is $gHg^{-1}$.

{\bf Solution:} First, for any $ghg^{-1} \in gHg^{-1}$, we have $ghg^{-1} \cdot gx = ghx = gx$, so $ghg^{-1}$ is in the stabilizer of $gx$.

For the reverse inclusion, suppose that $k$ is in the stabilizer of $gx$.  Then  $kgx = gx$, so $g^{-1}k gx = x$, so $g^{-1} k g \in H$.  Writing $g^{-1} k g = h$, we then have $k = ghg^{-1} \in gHg^{-1}$.

Conversely, for $ghg^{-1} \in gHg^{-1}$, we have $ghg^{-1} \cdot gx = ghx = gx$, so $ghg^{-1}$ is in the stabilizer.

\item Give an explicit example where $G = D_4$.

{\bf Solution:} Let $G = D_4$ act on the four vertices of the square, numbered as shown,
\[ \xymatrix{
1 \ar@{-}[r] \ar@{-}[d] & 2 \ar@{-}[d] \\
4 \ar@{-}[r] & 3
} \]
and let $r$ act by rotating clockwise through $90^\circ$, and $s$ act by reflecting left-to-right.  We find that
\begin{align*}
\Stab(1) &= \{ 1, sr \} \\
\Stab(2) &= \{ 1, sr^3 \}
\end{align*}
We have $r \cdot 1 = 2$, and $r \cdot sr \cdot r^{-1} = rs = sr^3$, so indeed $r \Stab(1) r^{-1} = \Stab(2)$.

\end{enumerate}

\item Let $G$ be a group of order 8, and suppose that $g^2 = 1$ for all $g \in G$.  Show that $G \cong \Z_2 \times \Z_2 \times \Z_2$.

{\bf Solution:} From the midterm we know that $G$ is Abelian.  Since $G$ has only one element of order 1, it has seven elements of order 2.  Let $x, y \in G$ be two distinct elements.  Then $xy = yx$ is distinct from 1, $x$, $y$: indeed, if $xy = 1$ then $x = xy^2 = y$; if $xy = x$ then $y = 1$, and if $xy = y$ then $x = 1$.

Let $z \in G$ be distinct from $\{ 1, x, y, xy \}$.  Then $xz$, $yz$, and $xyz$ are distinct from those four, from $z$, and from one another by a similar argument.

Thus the map $\Z_2 \times \Z_2 \times \Z_2 \to G$ given by $(a,b,c) \mapsto x^a y^b z^c$ is an isomorphism.

\item
\begin{enumerate}
\item Let $\phi\colon G \to H$ be a surjective group homomorphism, and let $K = \ker \phi$.  Show that for all $h \in H$, the fiber
\[ \phi^{-1}(h) = \{ g \in G : \phi(g) = h \} \]
is a coset of $K$.  (Left or right doesn't matter, because $K$ is normal.  As always, beware that $\phi^{-1}$ is not a map $H \to G$.)

{\bf Solution:} Since $\phi$ is surjective, we can choose a $g \in \phi^{-1}(h)$.  I claim that $gK = \phi^{-1}(h)$.  First, if $gk \in gK$ then $\phi(gk) = \phi(g) \phi(k) = h \cdot 1 = h$, so $gk \in \phi^{-1}(h)$.  Second, if $g' \in \phi^{-1}(h)$ then $\phi(g^{-1}g') = \phi(g)^{-1} \phi(g') = h^{-1} h = 1$, so $g^{-1} g' \in K$, so $g' \in gK$ by problem 1.

\item For the homomorphism $\phi\colon \C^\times \to \R$ given by $\phi(z) = \log |z|$, describe the kernel and its cosets geometrically.

{\bf Solution:} They are circles centered at the (missing) origin.  The coset of $z \in \C^\times$ is the circle of radius $|z|$.  Alternatively, the fiber of $\phi$ over $x \in \R$ is the circle of radius $e^x$.
\vfill

\item For the homomorphism $\phi\colon \R \times \R \to \R$ given by $\phi(x,y) = x + y$, describe the kernel and its cosets geometrically.

{\bf Solution:} They are lines of slope $-1$.  The coset of a point $(x,y)$ is the line of slope $-1$ passing through the point.  Alternatively, the fiber of $\phi$ over $a \in \R$ is the line $x+y=a$.
\vfill

\end{enumerate}

\newpage
\item Optional: Show that
\begin{enumerate}
\item $D_6 \cong D_3 \times \Z_2$.

{\bf Solution:} Define a map $\phi\colon D_3 \times \Z_2 \to D_6$ by
\[ \phi(s^a r^b, c) = s^a r^{2b+3c}. \]
This is well-defined because $r^6 = 1$ in $D_6$.  To see that $\phi$ is a homomorphism, we consider two cases:
\begin{align*}
\phi (s^a r^{b_1}, c_1) \cdot \phi(r^{b_2}, c_2)
&= s^a r^{2b_1 + 3c_1} r^{2b_2 + 3c_2} \\
&= s^a r^{2(b_1 + b_2) + 3(c_1 + c_2)} \\
&= \phi(s^a r^{b_1 + b_2}, c_1 + c_2) \\[1em]
\phi (s^a r^{b_1}, c_1) \cdot \phi(s r^{b_2}, c_2)
&= s^a r^{2b_1 + 3c_1} s r^{2b_2 + 3c_2} \\
&= s^a s r^{-(2b_1 + 3c_1)} r^{2b_2 + 3c_2} \\
&= s^{a + 1} r^{2(-b_1+b_2) + 3(-c_1 + c_2)} \\
&= \phi(s^{a+1} r^{-b_1 + b_2}, -c_1 + c_2) \\
&= \phi(s^a s r^{-b_1} r^{b_2}, c_1 + c_2) \\
&= \phi(s^a r^{b_1} s r^{b_2}, c_1 + c_2).
\end{align*}
To see that $\phi$ is surjective, we observe that any integer can be written in the form $2b + 3c$ for some $b, c \in \Z$ because 3 is odd.

\item $D_8 \not\cong D_4 \times \Z_2$.

{\bf Solution:} First we establish that for $n \ge 2$, the center
\[ Z(D_{2n}) = \{ 1, r^n \}. \]
For a rotation $r^m \in D_{2n}$, we have $r^m \in Z(D_n)$ if and only if $sr^m = r^ms$, which is true if and only if $sr^m = sr^{-m}$, which is true if and only if $m \equiv 0$ or $n \pmod{2n}$.  For a reflection $sr^m \in D_{2n}$, we have $sr^m \cdot r = r \cdot sr^m = sr^{m-1}$, so if $sr^m \in Z(D_{2n})$ then $m+1 \equiv m-1 \pmod{2n}$, which is impossible because $2n \ge 4$.

Thus the center of $D_8$ has order 2.  But
\[ Z(D_4 \times \Z_2) \cong Z(D_4) \times Z(\Z_2), \]
as we saw on the optional part of homework 1, problem 5, and this has order 4.
\item Generalize to $D_{2n}$.

{\bf Solution:} If $n$ is odd then the map $\phi\colon D_n \times \Z_2 \to D_{2n}$ given by
\[ \phi(s^a r^b, c) = s^a r^{2b+nc} \]
is an isomorphism.  If $n$ is even then $Z(D_{2n})$ has order 2 while $Z(D_n \times \Z_2)$ has order 4.

Geometrically, we can consider an $n$-gon inscribed in a $2n$-gon; then the antipodal map (= rotation through $180^\circ$) preserves the $n$-gon if $n$ is even, and does not preserve it if $n$ is odd.

\end{enumerate}


\end{enumerate}

\end{document}