\magnification=\magstep1
\noindent
Patrick,
\vskip 10pt
The problems you are asking about that I put on the review
sheet are for your ``general mathematical education'' and a way to prepare for the exam
by looking at problems that are somewhat {\it harder} than the ones
you'll actually be seeing.  In other words, if you can do those,
you'll not have any difficulty with the exam because you'll really
understand everything.  The actual exam will be closer to the 
practice problems.  
\vskip 10pt
\noindent
On 4.10 A, what you are saying is not quite correct.  (You need to review what it means for something
to be a subgroup of the integers under addition -- see Definition 4.18.) The set $\langle a \rangle \cap \langle b \rangle$
is the set of integers that are simultaneously multiples of $a$ and multiples of $b$.  That means it is 
the set of all integers $x$ such that $x = s a = t b$ for some integers $s,t$.  So you had the definition 
more or less correct. However, to show that that is a subgroup of ${\bf Z}$, you need to show first that if 
you take $x_1,x_2 \in \langle a \rangle \cap \langle b \rangle$, then the sum 
$x_1 + x_2 \in \langle a \rangle \cap \langle b \rangle$ as well.  This is true because 
you have $x_1 = s_1 a = t_1 b$ and $x_2 = s_2 a = t_2b$ for some integers $s_1,t_1,s_2,t_2$.  So
$$\eqalign{
x_1 + x_2 &= s_1 a + s_2 a = (s_1 + s_2) a\quad {\rm and}\cr
x_1 + x_2 &= t_1 b + t_2 b = (t_1 + t_2) b.\cr}$$
This shows $x_1 + x_2 \in \langle a \rangle \cap \langle b \rangle$.  Then, you also need to show
that $-x_1 \in \langle a \rangle \cap \langle b\rangle$.  But this follows too since 
$-x_1 = (-s_1) a = (-t_1) b$, so $-x_1$ is also a multiple of both $a$ and $b$.  
\bigskip
\noindent
Now, by Theorem 4.20, we know that $\langle a \rangle \cap \langle b\rangle = m {\bf Z}$ for some
(unique) positive integer $m$.  I'm calling it $m$ rather than $d$ in the theorem because this $m$
is going to be the least common multiple of $a,b$.  
\bigskip
\noindent
Then for part B what you need to show is:
\item{1.}  $m$ is a common multiple of $a$ and $b$ -- that is, $a|m$ and $b|m$
\item{2.}  If $n$ is any other common multiple of $a$ and $b$, then $m | n$  (so $m$ is the smallest
number that is a common multiple of $a,b$).
\bigskip
\noindent
Statement 1 follows directly from the definition of $\langle a\rangle \cap \langle b\rangle$.  
$m$ is in that subgroup so it is a multiple of $a$ and a multiple of $b$.  
Statement 2 is proved like this:
If $a|n$ and $b|n$, then $n = sa = tb$ for some integers $s,t$.  
But that is equivalent to saying that $n \in \langle a\rangle \cap \langle b\rangle$.  
Hence $m|n$ since $m$ divides everything in 
$\langle a\rangle \cap \langle b\rangle$.
\bigskip
\noindent
For part C, use the suggestion:  Assume $ab = md$ (for any two integers $m,d$ -- 
not necessarily the lcm and gcd yet).  We want to show that $d$ is a common
divisor of $a$ and $b$ if and only if $m$ is a common multiple of $a$ and $b$.  
\bigskip
\noindent
$\Rightarrow:$ If $d|a$ and $d|b$ then we can write $a = ds$ and $b = dt$
for some integers $s,t$.  But then $(ds)(dt) = md$, so $dst = m$.  This shows $(ds) = a$ divides $m$
and $(dt) = b$ divides $m$.  Hence $m$ is a common multiple of $a,b$.  
\bigskip 
\noindent
$\Leftarrow$:
Conversely, if $m$ is a common multiple
of $a,b$, then we have $m = as$ and $m = bt$ for some integers $s,t$.  Substituting the first one into the equation $ab = md$ we get $ab = asd$ so $b = sd$ and $d|b$.  Similarly,
if we substitute the second one in to $ab = md$, we get $ab = btd$ so $a = td$, which
says $d|a$.  
\bigskip
\noindent
Now to prove that $ab = \gcd(a,b){\rm lcm}(a,b)$, we can argue like this.  $ab$ is clearly a
common multiple of $a,b$, so it is in the subgroup $\langle a\rangle \cap \langle b\rangle$.
That shows that ${\rm lcm}(a,b)$ divides the product $ab$ and we have an equation $ab = {\rm lcm}(a,b)\cdot q$
for some integer $q$.  But what we proved above (the ``suggestion'') shows that $q$ must then be a common 
divisor of $a,b$.  Arguing by contradiction, suppose that $q \ne d = \gcd(a,b)$.  Then
from the properties of $d = \gcd(a,b)$ we know that $q|d$, or $d = qs$ for some integer $s > 1$.  
If we multiply $s$ on both sides of the equation $ab = {\rm lcm}(a,b)\cdot q$, we get
$abs = {\rm lcm}(a,b) d$.  But $d$ divides $a$ so $a = dr$ for some integer $r$ and we get 
$(dr)bs = {\rm lcm}(a,b) d$ so $rbs = {\rm lcm}(a,b)$.  Similarly $b = dt$ for some $t$ so we get
$a(dt)s = {\rm lcm}(a,b)d$, so $ats = {\rm lcm}(a,b)$.  But now we have a contradiction because
of the factor $s > 1$ on the left side of both of these equations.  The number $rb = at$ is
also a common multiple of $a,b$ and it's smaller than $rbs = ats$ since $s > 1$.  Hence the other
factor $q$ in the equation $ab = {\rm lcm}(a,b)\cdot q$ must equal $d = \gcd(a,b)$.  
\bye





