\magnification=\magstep1
\def\rotmat #1{\pmatrix{\cos #1 & -\sin #1\cr
                        \sin #1 & \cos #1\cr}}
\centerline{Mathematics 243 -- Algebraic Structures, section 1}
\centerline{Selected Solutions -- Problem Set 9}
\centerline{November 20, 2006}
\bigskip
\noindent
3.2/11.  Let $H$ be a subgroup of a group $G$, let $a \in G$ be
a fixed element, and let 
$$K = \{x \in G | x = aha^{-1},\ {\rm some}\ h \in H\}$$
We must show $K$ is a subgroup of $G$.
First, $e \in K$, since $e = aea^{-1}$, and $e \in H$ because
$H$ is a subgroup.  Next, if $x,y \in K$, we must show 
that $xy \in K$.  Saying $x\in K$ means $x = aha^{-1}$ for 
some $h \in H$.  Similarly $y \in K$ means $y = aka^{-1}$
for some $k \in H$.  Then 
$$xy = (aha^{-1})(aka^{-1}) = ah(a^{-1}a)ka^{-1} = aheka^{-1} = ahka^{-1}$$
Since $H$ is a subgroup and $h,k \in H$, we have $hk \in H$.  Therefore
$xy \in K$.  Finally, we must show that if $x \in H$, then 
$x^{-1} \in H$ too.  But note by the ``reverse order law'' (see Theorem 3.4
part d on page 122 of the text), 
$$x^{-1} = (aha^{-1})^{-1} = (a^{-1})^{-1}h^{-1}a^{-1}= ah^{-1}a^{-1}$$
Since $H$ is a subgroup of $G$ and $h \in H$, $h^{-1} \in H$.
This says $x^{-1} \in K$.  We have verified all three of the conditions
of Theorem 3.9 in the text, so $K$ is a subgroup of $G$.
\bigskip
\noindent
3.2/12.  Let $G$ be an abelian group (i.e. the group operation
is commutative), and let $H = \{x \in G | h^{-1} = h\}$.
In other words, $H$ is the set of elements in $G$ that {\it 
are their own inverses}.  We want to show that $H$ is a subgroup 
of $G$.  First, since $e^{-1} = e$ for the identity $e$ in $G$, 
$e \in H$, so $H$ is nonempty.  Next, if $h, k \in H$, so 
$h = h^{-1}$ and $k = k^{-1}$, then by the reverse order law and commutativity,
$$hk = h^{-1}k^{-1} = (kh)^{-1} = (hk)^{-1}.$$
Hence $hk \in H$ (because it is its own inverse).  Finally,
since $h^{-1} = h$ for all $h \in H$, every element in $H$ has an
inverse in $H$ (itself!!).  We have verified all three of the conditions
of Theorem 3.9 in the text, so $H$ is a subgroup of $G$.
\bigskip
\noindent
3.2/14b.  Let
$$H = \left\{\pmatrix{a & -b\cr b & a\cr} : a^2 + b^2 = 1\right\}$$
$H$ is not empty since the identity matrix $\pmatrix{1 & 0 \cr 0 & 1\cr} \in H$
(take $a = 1, b = 0$).  Next, if we take two matrices in $H$ and multiply
$$\pmatrix{a & -b\cr b & a\cr}\pmatrix{c & -d\cr d & c\cr} = \pmatrix{ac - bd & -(ad+bc)\cr ad+bc &ac - bd\cr}$$
The product matrix satisfies 
$$\eqalign{(ac - bd)^2 + (ad+bc)^2 &= a^2c^2 - 2acbd + b^2d^2 + a^2d^2 + 2adbc + b^2c^2\cr
&=(a^2 + b^2)(c^2 + d^2)\cr
&=1\cr}$$
Hence $H$ is closed under products.  Similarly, since $a^2 + b^2 = 1$,
$$\pmatrix{a & -b\cr b & a\cr}^{-1} = \pmatrix{a & b\cr -b & a\cr} 
= \pmatrix{a & -(-b)\cr -b & a\cr}$$
This is a matrix of the same form, with $b$ replaced by $-b$.  Hence $H$
is closed under taking inverses too.  This shows that $H$ is a subgroup 
of $G = {\rm GL}_2({\bf R})$ by Theorem 3.9.  

The proof of part d is similar.  Here if we take $a = 0, b = 1$ we get
$I_2 \in H$.  Then if we take two general matrices in $H$ and multiply
$$\pmatrix{1 & a \cr 0 & b} \pmatrix{1 & c\cr 0 & d\cr} = \pmatrix{1 & c + ad\cr 0 & bd}$$
This is of the right form to be in $H$ since the zero is in the lower right entry,
and $b \ne 0, d\ne 0$ imply $bd \ne 0$.  This shows $H$ is closed under products.
Then, 
$$\pmatrix{1 & a \cr 0 & b}^{-1} = \pmatrix{1 & -a/b \cr 0 & 1/b}$$
which is also in $H$.  Hence $H$ is a subgroup of $G = {\rm GL}_2({\bf R})$ by Theorem 3.9.
\bigskip
\noindent
3.2/17.  Let $G$ be a group,  and let 
$$Z(G) = \{a \in G | ax = xa,\ {\rm all}\ a \in G\}$$
We must show $Z(G)$ is a subgroup of $G$.
First, $e \in Z(G)$, since $ex = xe = x$.  
Next, if $a,b \in Z(G)$, we must show 
that $ab \in Z(G)$.  Saying $a\in Z(G)$ means $ax = xa$ for 
all $x\in G$.  Similarly $b \in Z(G)$ means $bx = xb$
for all $x \in G$.  Then using associativity and these properties
we have for any $x \in G$:
$$(ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab)$$
Hence $ab$ also commutes with every $x$ in $G$, so $ab \in Z(G)$.
Finally, we must show that if $a \in Z(G)$, then 
$a^{-1} \in Z(G)$ too.  From the equation $ax = xa$, if we multiply
on both sides on the left and right by $a^{-1}$ we have 
$a^{-1}(ax)a^{-1} = a^{-1}(xa)a^{-1}$.  Regrouping by 
associativity, and cancelling when we can this gives
$xa^{-1} = a^{-1} x$ for all $x$ in $G$.  Hence $a^{-1} \in Z(G)$.
We have verified all three of the conditions
of Theorem 3.9 in the text, so $Z(G)$ is a subgroup of $G$.
\bigskip
\noindent
3.3/13a.  First, we show by induction on $n$ that 
$$\rotmat{\theta}^n = \rotmat{n\theta}$$
for all $n \ge 1$.  The base case is completely obvious (nothing 
to prove).  So assume that 
$$\rotmat{\theta}^k = \rotmat{k\theta}$$
and consider
$$\rotmat{\theta}^{k+1} =\rotmat{\theta}^k\rotmat{\theta} $$
By the induction hypothesis, this is 
$$\rotmat{k\theta}\rotmat{\theta} =$$ 
$$\pmatrix{\cos k\theta \cos\theta - \sin k \theta\sin\theta &
-\cos k\theta \sin\theta - \sin k \theta\cos\theta\cr
\cos k\theta \cos\theta + \sin k \theta\sin\theta&
\cos k\theta \cos\theta - \sin k \theta\sin\theta\cr}$$
Using the trig identities
$$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B), \quad
\sin(A+B) = \sin(A)\cos(B) + \sin(B) \cos(A)$$
this last matrix equals
$$\rotmat{(k\theta + \theta)} = \rotmat{(k+1)\theta}$$
which is what we wanted to show.  
To finish part a, we must also show 
that the negative powers of $\rotmat{\theta}$ have the same 
form.  This can be done by a second induction argument (base case $n = -1$,
induction step $n = -k$ implies $n = -(k+1)$).  Here is another way to 
do the same thing.  Note by our general formula for $2\times 2$
inverse matrices, since $\det \rotmat{n\theta} = \cos^2(n\theta) + \sin^2(n\theta) = 1$,
we have 
$$\rotmat{n\theta}^{-1} = \pmatrix{\cos(n\theta) & \sin(n\theta)\cr
                          -\sin(n\theta) & \cos(n\theta)\cr}
                       =\rotmat{(-n \theta)}$$
because of the identities $\sin(-n \theta)= -\sin(n \theta)$ and
$\cos(-n \theta) = \cos(n \theta)$.
Hence the negative powers of $\rotmat{\theta}$ will 
come from the matrices of the same form with negative 
integer multiples of $\theta$.

b.  If $\theta = 90^\circ = \pi/2$ radians, then 
$$\rotmat{ \pi/2 } = \pmatrix{0 & -1\cr 1 & 0},$$ and 
$$H = \left\{\pmatrix{0 & -1\cr 1 & 0}^n : n \in {\bf Z}\right\} = 
\left\{I,  \pmatrix{0 & -1\cr 1 & 0}, \pmatrix{-1 & 0\cr 0 & -1}, \pmatrix{0 & 1\cr -1 & 0}\right\}$$
(since $\pmatrix{0 & -1\cr 1 & 0}^4 = I$

\bigskip
c.  If $\theta = 120^\circ = 2\pi/3$ radians, then 
$$\rotmat{ 2\pi/3 } = \pmatrix{ -1/2 & -\sqrt{3}/2 \cr \sqrt{3}/2 & -1/2\cr},$$
and 
$$H = \left\{\pmatrix{ -1/2 & -\sqrt{3}/2 \cr \sqrt{3}/2 & -1/2\cr}^n : n \in {\bf Z}\right\}
= \left\{I, \pmatrix{ -1/2 & \sqrt{3}/2 \cr -\sqrt{3}/2 & -1/2\cr}, 
\pmatrix{ -1/2 & -\sqrt{3}/2 \cr \sqrt{3}/2 & -1/2\cr} \right\}$$
(since $\pmatrix{ -1/2 & -\sqrt{3}/2 \cr \sqrt{3}/2 & -1/2\cr}^3 = I$).
\bigskip
\noindent
3.3/14.  We can show this by 
showing that the set $G$ of all $[a]$ in ${\bf Z}_n$ 
with ${\rm gcd}(a,n) = 1$ is a group under multiplication 
mod $n$.  We know that multiplication mod $n$ is associative
in all cases, so it is associative on the subset of classes
which have multiplicative inverses.  We are including exactly
the elements in ${\bf Z}_n$ that do have multiplicative
inverses, so every element of $G$ has an inverse for 
multiplication and that inverse is also an element of $G$.  
The element $[1] \in G$ is an identity for multiplication.
So the only thing that must be proved is that $G$ is closed
under multiplication.  If ${\rm gcd}(a,n) = {\rm gcd}(b,n) = 1$,
then we know there exist equations $am + n\ell = 1 = bp + nq$
for some integeres $m,\ell,p,q$.  If we multiply, we get
$$1 = (am+n\ell)(bp + nq) = (ab)pm + n(\ell p + qa + n \ell q)$$
This implies that ${\rm gcd}(ab,n) = 1$ also.  Hence
$[a][b] = [ab]$ also has a multiplicative inverse in ${\bf Z}_n$. 
\bye
