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\centerline{Mathematics 243, section 1 -- Algebraic Structures}
\centerline{Solutions for Exam 3}
\centerline{December 8, 2003}
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\noindent
I.  A) To show $h_a$ is one-to-one, assume that $h_a(x) = h_a(y)$
for some $x,y\in G$.  Then:
$$xa = ya \Rightarrow (xa)a^{-1} = (ya)a^{-1} \Rightarrow x = y$$
Hence $h_a$ is one-to-one.  To see that $h_a$ is onto,
let $y \in G$ be a general element and consider $x = ya^{-1}$.  Then
we have 
$$h_a(x) = xa = (ya)a^{-1} = y$$
Hence every element of $G$ is in the image of $h_a$, so 
$h_a$ is onto.

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B)  The result from part A says that in the group operation
table for $G$, the column for $a\in G$ contains each element
of $G$ exactly once.
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\noindent
II.  A) {\it Comment: Too many people were thinking in a very
sloppy way in working on this question.  The problem does not
ask for the generators of ${\bf Z}_{24}$, it asks for the generators
of the cyclic subgroup $H = \langle [15] \rangle \subset {\bf Z}_{24}$.}

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\noindent
{\it Best solution:}  
Recall the general fact in the proofs to know for this exam:
In a finite cyclic group $G = \langle a\rangle$ of order $n$, 
$\langle a^k\rangle = 
\langle x^d \rangle$ where $d = {\rm gcd}(n,k)$. Here 
${\bf Z}_{24}$ is cyclic (generator $[1]$ under addition)
$n = 24$, and ${\rm gcd}(15,24) = 3$.  So the same subgroup
is generated by any $[k]$ with ${\rm gcd}(k,24) = 3$.  This gives
$$k = [3], [9], [15],\ {\rm or}\ [21].$$

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\noindent
{\it Also correct, but more ``brute force'' approach}:  Compute the
elements of 
$$H = \langle [15]\rangle = \{[0],[15],[6],[21],[12],[3],[18],[9]\}.$$
Then check one-by-one to see which of them generate all of 
$H$.  Only $[3],[9],[15],[21]$ do.  For example,
$$\langle [18] \rangle = \{[0],[18],[12],[6]\}$$
only has order 4, so $[18]$ is not a generator of $H$.  But
$$\langle [3] = \{[0],[3],[6],[9],[12],[15],[18],[21]\} = H,$$
so $[3]$ is also one of the generators of $H$.

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B)  By direct computation, 
$$\sigma = (1532)(24368)(12479) = (14795368)(2) = (14795368)$$
which has order 8, since it is an 8-cycle.

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\noindent
III.  A) Since $\phi$ is a group homomorphism,
$$\phi(e_G) = \phi(e_G\cdot e_G) = \phi(e_G)\cdot \phi(e_G)$$
hence, after multiplying both sides by $\phi(e_G)^{-1}$ in $H$,
we have 
$$\eqalign{& \phi(e_G)^{-1}\phi(e_G) = \phi(e_G)^{-1}\phi(e_G)\phi(e_G)\cr
 \Rightarrow\ & e_H = \phi(e_G)\cr}$$

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\noindent
B)  Recall the definition:
$$\ker(\phi) = \{ x \in G\ |\ \phi(x) = e_H\}.$$ 
First,  $\ker(\phi)$ is nonempty since by part A, $e_G \in \ker(\phi)$.
Next, if $x,y\in \ker(\phi)$, then 
$$\phi(xy) = \phi(x) \phi(y) = e_H e_H =e_H$$
Hence $xy \in \ker(\phi)$.  Finally, if
$x\in \ker(\phi)$, then by a general property of homomorphisms
that we proved in class,  
$$\phi(x^{-1}) = (\phi(x))^{-1} = e_H^{-1} = e_H$$
Hence $x^{-1} \in \ker(\phi)$.  We have shown the three 
properties needed to see that $\ker(\phi)$ is a subgroup of $G$.

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\noindent
IV.  A)  Let $A = \pmatrix{a & b\cr -b & a\cr}$ and 
$B = \pmatrix{c & d\cr -d & c\cr}$ be {\it two general matrices in $G$ -- 
it is not sufficient to consider only the product $A^2$}.  Then
$$AB = \pmatrix{a & b\cr -b & a\cr}\pmatrix{c & d\cr -d & c\cr} 
=\pmatrix{ac-bd & ad + bc\cr -(ad+bc) & ac - bd\cr}$$
Since the product $AB$ has the same form (that is, the diagonal
entries are equal and the off-diagonal entries are negatives
of each other), the product will be in $G$ as long as we
know that at least one of these is nonzero.  Note that the 
$\det(A) = a^2 + b^2 \ne 0$ and $\det(B) = c^2 + d^2 \ne 0$.
Hence $\det(AB) = \det(A)\det(B) = (a^2 + b^2)(c^2+d^2) = 
(ac - bd)^2 + (ad + bc)^2 \ne 0$.  It follows that at least
one of $ac - bd$ and $ad + bc$ is nonzero, and $G$ is closed
under matrix products.

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\noindent
B)  The general formula for matrix inverses for $2\times 2$ matrices
shows if $A = \pmatrix{a & b\cr -b & a\cr}$, then
$$A^{-1} = \pmatrix{{a\over a^2 + b^2} &  {-b\over a^2 + b^2}\cr
{b\over a^2 + b^2} & {a\over a^2 + b^2}\cr}$$
Again, the two diagonal entries are the same, and the off-diagonal
entries are negatives.  So $G$ is closed under matrix inverses.

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{\it Note:  It follows from the two parts of this problem that
$G$ is a subgroup of $GL_2({\bf R})$, since $G$ is clearly not empty.}

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\noindent
{\it Extra Credit.} To do this in any reasonable way, you 
have to remember the $2\times 2$ rotation matrices we
saw on one of the problem sets: Let  $A = \pmatrix{\cos\theta &-\sin\theta\cr
\sin\theta & \cos\theta\cr}$.  If 
$\theta = 2\pi/7$, then $A$ will have order $7$:
$$A^7 =\pmatrix{\cos(7\theta) &-\sin(7\theta)\cr
                \sin(7\theta) &\cos(7\theta)\cr} = 
 \pmatrix{\cos(2\pi) &-\sin(2\pi)\cr
          \sin(2\pi) &\cos(2\pi)\cr} = \pmatrix{1&0\cr 0&1\cr}.$$

\bye  
