\documentclass[11pt]{article}
\usepackage{amsmath}
\pagestyle{empty}
\newcommand{\sig}{\sigma}
\newcommand{\ra}{\rightarrow}
\newcommand{\eps}{\epsilon}
\newcommand{\ah}{\alpha}
\newcommand{\lam}{\lambda}
\newcommand{\gam}{\gamma}
\newcommand{\ol}{\overline}
\newcommand{\So}{\\ {\tt Solution:}}
\newcommand{\ben}{\begin{enumerate}}
\newcommand{\een}{\end{enumerate}}
\newcommand{\hsone}{\hspace*{1cm}}
\newcommand{\hstwo}{\hspace*{2cm}}
\begin{document}
\begin{center} {\Large\bf Algebra , Assignment 6 \\ Solutions}
\end{center}
\ben
\item Let $\sig: G\ra G$ be an automorphism of $G$. Let
$x,y\in G$ with $y=\sig(x)$.
\ben
\item Assume $x^s=e$. Prove $y^s=e$.
\So $y^s=\sig(x)^s=\sig(x^s)=\sig(e)=e$
\item Assume $x^s\neq e$. Prove $y^s\neq e$.
\So As above, $y^s=\sig(x^s)$. An automorphism
is injective (and surjective, for that matter) so if
$y^s=e$ then $x^s=e$. The required statement is the
contrapositive.
\item Show $o(x)=o(y)$ (You can assume they are both finite.)
\So : $o(x)$ is the {\em least} $s$ for which $x^s=e$ and
$o(y)$ is the {\em least} $s$ for which $y^s=e$. Since the
values $s$ are the same for both $x,y$ the least value $s$ is
the same for both.
\een
\item Let $\sig\in S_n$ be given by $\sig(i)=n+1-i$ (so $1\cdots n$
is ``reversed.") When $n=403$ is $\sig$ even or odd? For which
$n$ is $\sig$ even?
\So We flip $1,403$, $2,402$ all the way to $201,202$ so
there are $201$ flips so $\sig$ is odd. More generally it depends
on whether $n$ is even or odd. When $n=2k+1$ there are $k$ flips
so it is even when $k$ is even otherwise odd. When $n=2k$ there are $k$ flips
so it is even when $k$ is even otherwise odd.
\item Here we write perutations in terms of their disjoint cycles.
\ben
\item Given $\sig=(12)(34)$ and $\gam=(56)(13)$ find $\tau\in S_6$ with
$\tau^{-1}\sig\tau=\gam$
\So We send $1234$ into $5613$. The other $56$ go into the other $24$ in either
order:
\[ \tau=
\begin{pmatrix} 1&2&3&4&5&6\\ 5 & 6 & 1 & 3 & 2 & 4
\end{pmatrix}
\mbox{ or }
\begin{pmatrix} 1&2&3&4&5&6\\ 5 & 6 & 1 & 3 & 4 & 2
\end{pmatrix}
\]
\item Prove that with $\sig=(123)$ and $\gam=(13)(578)$ there is no $\tau$ with
$\tau^{-1}\sig\tau=\gam$
\So Such a $\tau$ exists if and only if $\sig,\gam$ have the same cycle sizes with
the same multiplicity: these are $3$ for $\sig$ and $2,3$ for $\gam$.
\item Set $\sig=(12)(34)$. In $S_8$ precisely how many $\gam$ have the property
that
$\tau^{-1}\sig\tau=\gam$ for some $\tau$?
\So We must have $\gam=(ab)(cd)$. So there are $8\cdot 7\cdot 6\cdot 5$ choices for
$a,b,c,d$ except we are overcounting by an eight factor as we can flip $ab$, flip $cd$,
flip $ab$ with $cd$. So the number is $7\cdot 6\cdot 5= 210$.
\een
\item Consider the symmetries of the square. Use the table and
notation on the website file. For each symmetry $v$ find the
conjugacy class $C(v)$ and the Normalizer $N(v)$ as defined
in \S 2.11. Check by calculation that the class equation,
the equation at the top of Page 85, holds.
\So Call this grouup $G$. Copying solution 1, the
elements are $I,R,S,T,V,H,D,A$ and the
table is
\begin{center}\begin{tabular}{r|rrrrrrrr}
* & I & R & S & T & V & H & D & A \\
\hline
I & I & R & S & T & V & H & D & A \\
R & R & S & T & I & D & A & H & V \\
S & S & T & I & R & H & V & A & D \\
T & T & I & R & S & A & D & V & H \\
V & V & A & H & D & I & S & T & R \\
H & H & D & V & A & S & I & R & T \\
D & D & V & A & H & R & T & I & S \\
A & A & H & D & V & T & R & S & I
\end{tabular}
\end{center}
For the normalizer $N(g)$ we look at which $x$ have $gx=xg$. We see:
\\ $N(I)=G$ (this is always the case with the identity)
\\ $N(S)=G$
\\ $N(R)=\{I,R,S,T\}=N(T)$
\\ $N(V)=\{I,S,V,H\}$
\\ $N(H)=\{I,S,V,H\}=N(V)$
\\ $N(D)=\{I,S,D,A\}=N(S)$
\\ The normalizers are always subgroups and so must have $1,2,4$ or $8$ elements.
\\ For the conjugacy classes we first note $C(I)=\{I\}$ as this is always the
case with the identity. Looking at the eight values $g^{-1}Rg$ we find
$R=IRI$, $T=VRV$ and no more so $C(R)=\{R,T\}$. We see that $S$ commutes
with everything so $C(S)=\{S\}$. (The center of the group, denoted $Z[G]$,
is those elements that commute with everything, here $Z(G)=\{I,S\}$.
Looking at the eight values $g^{-1}Vg$ we find
$V=IVI$, $H=RVT$ and no more so $C(V)=\{V,H\}$.
Looking at the eight values $g^{-1}Dg$ we find
$D=IDI$, $A=RDT$ and no more so $C(D)=\{D,A\}$. To summarize:
\\ $C(I)=\{I\}$, $C(S)=\{S\}$
\\ $C(R)=C(T)=\{R,T\}$
\\ $C(V)=C(H)=\{V,H\}$
\\ $C(D)=C(A)=\{D,A\}$
Note that for each symmetry $g$, $8=o(G)=o(C(x))\cdot o(N(x))$.
For the Class Equation take one from each equivalence class -- say $I,S,R,V,D$.
We get
\[ 8=o(G) =
\frac{o(G)}{o(N(I)}+
\frac{o(G)}{o(N(S)}+
\frac{o(G)}{o(N(R)}+
\frac{o(G)}{o(N(V)}+
\frac{o(G)}{o(N(V)}= 1+1+2+2+2 \]
\item In $S_4$ set $\sig=(12)(34)$, $\tau=(13)(24)$, $\gam=(13)(24)$, $e$ the
indentity. Let $H=\{e,\sig,\tau,\gam\}$. Give the multiplication table for
$H$. Show that $H$ is a subgroup of $S_4$. Show that $H$ is a subgroup of
$A_4$. Show that $H$ is a normal subgroup of $A_4$. Show that
$H$ is a normal subgroup of $S_4$. Set $G_1=(Z_4,+)$ and
$G_2=(Z_2\times Z_2,+)$. (We call $G_2$ the viergr\"uppe.) Which of
$H_1,H_2$ is $H$ isomorphic to? Given an explicit (there are many
correct answers) isomorphism $\phi$ from $H$ to whichever ($G_1$ or $G_2$) it is
\So $H$ is the viergr\"uppe, $\sig\tau=\gam$, $\sig\gam=\tau$, $\tau\gam=\sig$ and
everthing squared is $e$. (In $(Z_4,+)$ only $2$ is of order two while in the
viergr\"uppe all three elements (other than $e$ are of order two. We have $|H|=4$,
$|A_4|=4!/2 = 12$, $|S_4|=4!=24$. All $e,\sig,\tau,\gam$ are even permutations
so $H$ is a subgroup of $A_4$. For any $\kappa\in S_4$, $\kappa^{-1}H\kappa$ consists
the identity plus three permutations consisting of two two cycles, and so
$\kappa^{-1}H\kappa=H$. Thus $H$ is a normal subgroup of $S_4$. In general, if
$H\subset K\subset G$, all groups, and $H$ is a normal subgroup of $G$ then
(just examine the definition!) $H$ is tautologically a normal subgroup of $K$.
So in this case $H$ is a normal subgroup of $A_4$.
\item Let $GL_n(R)$ be, as usual, the $n\times n$ nonsingular
matrices over the Reals, under multiplication. Suppose
$A\in GL_n(R)$ has real eigenvectors $\vec{v_1},\ldots,\vec{v_n}$
with distinct real eigenvalues $\lam_1,\ldots,\lam_n$, so that
$A\vec{v_i}=\lam_i\vec{v_i}$ for $1\leq i\leq n$. Recall from
Linear Algebra that this implies that the vectors $\vec{v_i}$
form a basis for $R^n$. Let $\sim$
be the conjugacy relation defined in \S 2.11
\ben
\item Suppose $B\sim A$. Prove that $B$
has the same eigenvalues $\lam_1,\ldots,\lam_n$. That is, show
that there are real eigenvectors $\vec{w_1}\ldots,\vec{w_n}$ with
$B\vec{w_i}=\lam_i\vec{w_i}$ for $1\leq i\leq n$.
\So We have $B=P^{-1}AP$ with $P$ nonsingular so set
$\vec{w_i}$ to be that vector with $P\vec{w_i}=\vec{v_i}$ Then
\[ P^{-1}AP\vec{w_i}= P^{-1}A\vec{v_i} = P^{-1}\lam_i\vec{v_i} = \lam_iP^{-1}\vec{v_i} = \lam_i\vec{w_i} \]
\item (*) Conversely, Suppose $B$ has real eigenvectors $\vec{w_1},\ldots,\vec{w_n}$
with distinct real eigenvalues $\lam_1,\ldots,\lam_n$, the same values
as for $A$. Prove that $B\sim A$.
\So As the $\vec{v_i}$ and the $\vec{w_i}$ both form bases there is a matrix $P$
with $P\vec{w_i}=\vec{v_i}$ and then $P^{-1}AP\vec{w_i}=B\vec{w_i}$ for all $i$ so that
$P^{-1}AP=B$.
\een
\een
\end{document}