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\centerline{Mathematics 243, section 3 -- Algebraic Structures}
\centerline{Solutions for Problem Set 8}
\centerline{November 16, 2012}
\vskip 10pt
\noindent
{\it `A' Section}
\bigskip
\begin{enumerate}
\item[1.]  Consider the $2\times 2$ matrices $I_2,S,X,Y,D,T$ defined by 
\begin{align*} 
I_2 = \begin{pmatrix} 1&0 \\ 0&1\end{pmatrix}, &\quad S = \begin{pmatrix} 0&-1 \\ 1&0\end{pmatrix}\\
X = \begin{pmatrix} 1&0 \\ 0&-1\end{pmatrix}, &\quad Y = \begin{pmatrix} -1&0 \\ 0&1\end{pmatrix}\\
D = \begin{pmatrix} 0&1 \\ 1&0\end{pmatrix}, &\quad T = \begin{pmatrix} 0&-1 \\ -1&0\end{pmatrix}
\end{align*}
Let $G = \{I_2,S,S^2,S^3,X,Y,D,T\}$.  Construct the operation table for matrix multiplication
on this set and verify that $G$ is a group under this operation.
\item[] {\it Solution:}  The table looks like this (taking the products in order
row $\times$ column in all cases):
\[
\begin{array}{llllllllll}
          \cdot & \vline & I_2 & S   & S^2 & S^3 & X   & Y   & D   & T \\ \hline
            I_2 & \vline & I_2 & S   & S^2 & S^3 & X   & Y   & D   & T \\
              S & \vline & S   & S^2 & S^3 & I_2 & D   & T   & Y   & X  \\
            S^2 & \vline & S^2 & S^3 & I_2 & S   & Y   & X   & T   & D  \\
            S^3 & \vline & S^3 & I_2 & S   & S^2 & T   & D   & X   & Y  \\
              X & \vline & X   & T   & Y   & D   & I_2 & S^2 & S^3 & S \\
              Y & \vline & Y   & D   & X   & T   & S^2 & I_2 & S   & S^3 \\
              D & \vline & D   & X   & T   & Y   & S   & S^3 & I_2 & S^2 \\
              T & \vline & T   & Y   & D   & X   & S^3 & S   & S^2 & I_2
\end{array}
\]
The table shows that $G$ is closed under matrix multiplication, so this is 
a binary operation.  Matrix multiplication is always associative as we saw
before, so this operation is associative.  The element $I_2$ is an identity.
Finally, each element has an inverse for matrix multiplication.  $I,S^2,X,Y,T,D$ 
are their own inverses, while $S^{-1} = S^3$ and $(S^3)^{-1} = S$.  
\item[2.]   
\begin{enumerate}
\item[a.] Let $G = \mathbb{Z}_{12}$ under the operation of \emph{addition} modulo 12.  Determine
the cyclic subgroups $\langle [a]\rangle$ for all $[a] \in \mathbb{Z}_{12}$.    
\item[] {\it Solution:}  We have 
\begin{align*}
\langle [0]\rangle &= \{[0]\}\\
\langle [1]\rangle &= \mathbb{Z}_{12} = \langle [5] \rangle = \langle [7] \rangle  = \langle [11] \rangle\\
\langle [2]\rangle &= \{[0],[2],[4],[6],[8],[10]\} = \langle [10]\rangle\\
\langle [3]\rangle &= \{[0],[3],[6],[9]\} = \langle [9]\rangle\\
\langle [4]\rangle &= \{[0],[4],[8]\} = \langle [8]\rangle\\
\langle [6]\rangle &= \{[0],[6]\}
\end{align*}
\item[b.] Conjecture a general formula for the number of elements in $\langle [a]\rangle$ in 
terms of the integers $a$ and $12$.  Check out your conjecture on the corresponding list of 
cyclic subgroups of $\mathbb{Z}_{20}$ constructed in class on November 14. 
\item[] {\it Solution:}  The pattern that fits all of these examples is that the number
of elements in $\langle [a]\rangle \subseteq \mathbb{Z}_n$ is equal to $n/\gcd(a,n)$.  
For instance, with $n = 12$ and $a = 10$, we have $\gcd(10,12) = 2$, and $\langle [10]\rangle$
has $12/2 = 6$ elements.  Moreover, all the $a$ with the same $\gcd(a,n)$ apparently generate
the same cyclic subgroup.  \emph{Comment:}  These patterns do hold in general and we will 
prove them shortly. 
\end{enumerate}
\end{enumerate}
\vskip 10pt
\noindent
{\it `B' Section}
\vskip 10pt
\begin{enumerate}
\item[1.]  Let 
$$G = \left\{\begin{pmatrix} 0&a&b\\ 0&0&c\\ 0&0&0\end{pmatrix} \mid a,b,c \in \mathbb{R}\right\}$$
Show that $G$ is a group under the operation of matrix addition.
\item[] {\it Solution:}  We know that $M_{3\times 3}(\mathbb{R})$ is a group under
matrix addition.  So a sneaky method here is to apply the ``shortcut method'' 
of Theorem 3.10 in the text to show that $G$ is a subgroup of $M_{3\times 3}(\mathbb{R})$.  $G$
is definitely nonempty since it contains elements for all triples $a,b,c$ of real numbers.  Next, if
$$X = \begin{pmatrix} 0&a&b\\ 0&0&c\\ 0&0&0\end{pmatrix} \qquad \text{and} \qquad
Y = \begin{pmatrix} 0&a'&b'\\ 0&0&c'\\ 0&0&0\end{pmatrix}$$
are in $G$, then 
$$X - Y = \begin{pmatrix} 0&a - a' &b - b'\\ 0&0&c - c'\\ 0&0&0\end{pmatrix} \in G$$
as well.  Therefore $G$ is a subgroup of $M_{3\times 3}(\mathbb{R})$, hence a group.  
\item[2.]  Let 
$$G = \left\{\begin{pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1\end{pmatrix} \mid a,b,c \in \mathbb{R}\right\}$$
Is $G$ is a group under the operation of matrix multiplication.  If so, say why; if not say which 
properties in the definition fail.  
\item[] {\it Solution:}  We know that $GL_3(\mathbb{R})$ is a group under
matrix multiplication.  So a sneaky method here is to apply the ``shortcut method'' 
of Theorem 3.10 in the text to show that $G$ is a subgroup of $GL_3(\mathbb{R})$.  $G$
is definitely nonempty since it contains elements for all triples $a,b,c$ of real numbers.  Next, if
$$X = \begin{pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1\end{pmatrix} \qquad \text{and} \qquad
Y = \begin{pmatrix} 1&a'&b'\\ 0&1&c'\\ 0&0&1\end{pmatrix}$$
are in $G$, then 
$$Y^{-1} = \begin{pmatrix} 1&-a' & a'c' - b' \\ 0&1& -c'\\ 0&0&1\end{pmatrix}.$$
Hence
$$X Y^{-1} = \begin{pmatrix} 1&a - a'& a'c' - b' - ac' + b\\ 0&1&c - c'\\ 0&0&1\end{pmatrix} \in G$$
as well.  Therefore $G$ is a subgroup of $GL_3(\mathbb{R})$, hence a group.  
\item[3.]  Let $G$ be a group, and consider the relation $R$ on $G$ defined by 
$xRy \leftrightarrow y = axa^{-1}$ for some $a\in G$.  (If $xRy$ is true, the $a$ that works
in the equation will depend on $x,y$.)   $R$ is called the \emph{conjugacy} relation on $G$.
\begin{enumerate}
\item[a.] Show that conjugacy is an equivalence relation on $G$.
\item[] {\it Solution:}  For all $x\in G$, we have $xRx$ since $x = exe^{-1}$.  Therefore $R$
is reflexive.  If $xRy$, then $y = axa^{-1}$.  If we multiply both sides of this equality 
by $a^{-1}$ on the left and $a$ on the right, we get $x = a^{-1}ya = a^{-1}y(a^{-1})^{-1}$.  
It follows that $yRx$ is also true, so $R$ is symmetric.  Finally, if $xRy$ and $yRz$, then 
$y = axa^{-1}$ and $z = byb^{-1}$ for some $a,b \in G$.  Substituting, we get
$z = b(axa^{-1})b^{-1} = (ba)x(ba)^{-1}$ by associativity and the reverse order law for inverses.
Therefore $R$ is transitive. 
\item[b.] Show that $G$ is an abelian group if and only if the equivalence classes
for the conjugacy relation satisfy $[x] = \{x\}$ for all $x \in G$.
\item[] {\it Solution:}  We have $[x] = \{x\}$ if and only if $x = axa^{-1}$ for all $a \in G$, 
or equivalently if and only if $xa = ax$.  This is true for all classes $[x]$ if and only if 
$xa = ax$ for all $a$ and all $x$ in $G$. That is the same as saying the operation in $G$
is commutative, or that $G$ is an abelian group.
\item[c.]  More generally, show that if $[x] = \{x\}$ for some element $x \in G$, then 
$xa = ax$ for all $a \in G$ and conversely.  The set of such elements $x$ is called the \emph{center}
of $G$: The center is the subset of $G$ defined by:
$$Z(G) = \{x \in G \mid xa = ax \text{ for all a in } G\}.$$
\item[] {\it Solution:}  One direction is essentially the same as the argument for part of the proof of part b, 
except we are no longer assuming that $[x] = \{x\}$ holds for all $x\in G$, just for some 
particular $x \in G$.  This proof is ``reversible,'' so it can be phrased like this:
\begin{align*}
[x] = \{x\} &\Leftrightarrow x = axa^{-1} \text{ for all } a \in G\\
&\Leftrightarrow xa = axa^{-1}a \text{ for all } a \in G\\
&\Leftrightarrow xa = ax(a^{-1}a) \text{ for all } a \in G,\text{ by associativity}\\
&\Leftrightarrow xa = axe \text{ for all } a \in G, \text{ by definition of inverse}\\
&\Leftrightarrow xa = ax \text{ for all } a \in G, \text{ by definition of identity element}.
\end{align*}
Hence $[x] = \{x\}$ if and only if $x \in Z(G)$, the center of $G$.   
\item[d.]  Show that the center of $G$ (as defined in part c) is a subgroup of $G$.
\item[] {\it Solution:}  We again use the criterion of Theorem 3.10:  $Z(G)$ is not
empty since it always contains the identity element $e \in G$ -- recall $ex = x = xe$
for all $x \in G$.   Next, let $x,y \in Z(G)$, then $xa = ax$ and $ya = ay$ for all $a \in G$.
But then $y^{-1}yay^{-1} = y^{-1}ayy^{-1}$ as well so $ay^{-1} = y^{-1}a$ for all $a \in G$.
Hence by associativity and this last observation, for all $a \in G$:
$$(xy^{-1})a = x(y^{-1}a) = x(ay^{-1}) = (xa)y^{-1} = (ax)y^{-1} = a(xy^{-1}).$$
This shows $xy^{-1} \in Z(G)$, so $Z(G)$ is a subgroup of $G$.  
\item[e.]  Let $x \in G$ be a fixed element and define $C_x = \{a \in G \mid x = axa^{-1}\}$.  Show that
$C_x$ is a subgroup of $G$ ($C_x$ is called the \emph{centralizer of $x$}).
\item[] {\it Solution:}  The idea of the proof is similar to that of the proof of part d.  
First, $C_x$ is not empty since $C_x$ contains at least the identity element.
Let $a,b \in C_x$.  Then $x = axa^{-1} = bxb^{-1}$.  It also follows by multiplying both
sides of the equality $x = bxb^{-1}$ by $b^{-1}$ on the left and $b$ on the right
that $b^{-1}xb = x$.  But then by the reverse order law and associativity,
$$(ab^{-1})x(ab^{-1})^{-1} = ab^{-1}xba^{-1} = a(b^{-1}xb)a^{-1} = axa^{-1} = x.$$
It follows that $ab^{-1} \in C_x$, so $C_x$ is a subgroup by Theorem 3.10.
\item[f.] Let $G = \mathcal{S}(A)$ be the group of permutations of $A = \{a,b,c\}$.  Using the names for
the elements of this group we introduced in Problem Set 3, find all of the equivalence classes for the 
conjugacy relation on $G$ (there are three of them), determine the centralizer of each element of
$G$, and the center of $G$.  How are the sizes of the equivalence class of $x$ and the number of elements of the 
centralizer of $x$ related in each case? 
\item[] {\it Solution:}  Recall that $\mathcal{S}(A) = \{I_A,R_a,R_b,R_c,C_1,C_2\}$.  Computing we find
that there are exactly three conjugacy classes:
\begin{align*}
[I_A] &= \{I_A\}\\
[R_a] &= \{R_a,R_b,R_c\} = [R_b] = [R_a]\\
[C_1] &= \{C_1,C_2\}
\end{align*}
 To start, since $I_A$ is the identity element of this group, which commutes
with every element, it follows from part c above, that $[I_A] = \{I_A\}$. 
The rest can be read off from the operation table you derived in Problem Set 3. For example, 
$$R_b = R_c \circ R_a \circ R_c^{-1},$$
so $R_b \in [R_a]$.  Similarly 
$$R_c = R_b \circ R_a \circ R_b^{-1},$$
so $R_c \in [R_a]$.  On the other hand conjugating $C_1$ or $C_2$ by any element of 
$\mathcal{S}(A)$ always yields either $C_1$ or $C_2$, so those elements form 
another conjugacy class.  The centralizers are as follows:
\begin{align*}  
C_{I_A} &= \mathcal{S}(A)\\
C_{C_1} = C_{C_2} &= \{I_A,C_1,C_2\}\\
C_{R_a} &= \{I_A,R_a\}\\
C_{R_b} &= \{I_A,R_b\}\\
C_{R_c} &= \{I_A,R_c\}.
\end{align*}
(For instance, from the group table for $\mathcal{S}(A)$, we see $I,C_1,C_2$
all commute with $C_1$, but $R_a \circ C_1 = R_b \ne R_c = C_1 \circ R_a$, 
$R_b \circ C_1 = R_c \ne R_a = C_1 \circ R_b$, and $R_c \circ C_1 = R_a \ne R_b = C_1 \circ R_c$.
These computations show that none of $R_a,R_b,R_c$ are in the centralizer of $C_1$.) 
In each case, the product of the size of the conjugacy class times the order of
the centralizer equals $6 = |\mathcal{S}(A)|$.  (Equivalently, the size of the conjugacy 
class of $x$ is $\frac{6}{|C_x|}$ in each case.)
\end{enumerate}
\item[4.]  Let $H$ and $K$ be subgroups of a group $G$.
\begin{enumerate}
\item[a.]  Show that $H \cap K$ is also subgroup of $G$.
\item[] {\it Solution:}  Both $H$ and $K$ contain $e$, so $H \cap K \ne \emptyset$.
Next, let  $x,y\in H\cap K$, then $xy^{-1} \in H$ since $H$ is a subgroup.  Similarly
$xy^{-1}\in K$ since $K$ is a subgroup.  Thus $xy^{-1} \in H\cap K$.  It follows
that $H \cap K$ is a subgroup of $G$ by Theorem 3.10.
\item[b.]  Find an example where $H\cup K$ \emph{is} a subgroup of $G$ and one where $H\cup K$ \emph{is not}
a subgroup of $G$. 
\item[]  {\it Solution:}  Consider the subgroups $H = \{I_A,R_a\}$ and $K = \{I_A,R_b\}$
of $\mathcal{S}(A)$ from question 3f above.  $H\cup K$ is not a subgroup of $\mathcal{S}(A)$
because $H\cup K$ is not closed under composition $R_a \circ R_b =  C_1 \notin H\cup K$.  
On the other hand if $H = K$, then $H \cup K = H = K$ is a subgroup. 
\end{enumerate}
\end{enumerate}
\end{document}
