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.
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 $\varphi(n)$.