Conclude that the multiplicative order modulo $ab$ of any $c$, $gcd(c,ab) = 1$ must be a proper divisor of $\phi(ab)$. 
a) Show that if $n = ab$ where $1 < a, b$ are odd and
  $gcd(a,b) = 1$, then $lcm(\phi(a),\phi(b)) < \phi(ab)$.
b) Conclude that the multiplicative order modulo $ab$ of any $c$, $gcd(c,ab) = 1$
  must be a proper divisor of $\phi(ab)$.  Hint: Chinese Remainder
  Theorem.  
c) Finally conclude that composite odd $n$ has no primitive root.

I've been having some trouble with this problem.  I solved part a) but I'm having some trouble with parts b) and c).  Any help would be useful.
 A: Since you have done a), we solve b), at one place using the result of a).
We have $c^{\varphi(a)}\equiv 1\pmod{a}$ and $c^{\varphi(b)}\equiv 1\pmod{b}$ by Euler's Theorem.
It follows that if $k$ is a multiple of $\varphi(a)$ then $c^k\equiv 1\pmod{a}$, and a similar result holds for $b$.
In particular, if $k$ is the lcm of $\varphi(a)$ and $\varphi(b)$, then $c^k\equiv 1\pmod{a}$ and $\pmod{b}$, and therefore $\pmod{ab}$, since $a$ and $b$ are relatively prime.
By part a), this lcm is less than $\varphi(a)\varphi(b)$, that is, less than $\varphi(ab)$.
Finally, we show that the order of $c$ modulo $ab$ divides this lcm $k$, and hence divides $\varphi(ab)$.
Let $e$ be the order of $c$ modulo $ab$. So $e$ is the smallest positive integer such that $c^e\equiv 1\pmod{ab}$.
We have
$$\varphi(ab)=qe+r,\tag{1}$$
for some integers $q$ and $r$ such that $0\le r\lt e$. We will show that $r=0$. From (1) we have
$$c^{\varphi(ab)}=(c^e)^q c^r.$$
But $c^{\varphi(ab)}\equiv 1\pmod{ab}$ and $(c^e)^q\equiv 1\pmod{ab}$. It follows that $c^r\equiv 1\pmod{ab}$. If $r\ne 0$, this contradicts the minimality of $e$. Thus $r=0$. This completes the proof of divisibility.
We need to modify c) slightly, to assert that if $n$ is the product of two relatively prime odd numbers, then $ab$ does not have a primitive root.  This will show that composite odd $n$ that are not prime powers do not have a primitive root. 
For we have shown that any $c$ relatively prime to $ab$ has order less than $\varphi(ab)$. A primitive root of $n$ has order exactly equal to $\varphi(n)$. 
