Order of product versus product of orders We are given that $ord_m a =x$, $ord_m b=y$ and $gcd(x,y)=1$. (The author of my text doesn't say so, but another book mentions that this requires $gcd(a,m)=gcd(b,m)=1$.) The problem is to show that $ord(ab)\equiv xy \pmod{\phi(m)}$.
$(a^x)^y\equiv 1^y \pmod{m} \equiv 1 \pmod{m}$ and $(b^y)^x\equiv 1^x \pmod{m}\equiv 1\pmod{m}$
So $(ab)^{xy}=(a^x)y(b^y)^x\equiv 1$.
The problem is to show that $(ab)^{xy-n}\not\equiv 1$ for $n>0$.
APPROACH 1
To use the condition that $gcd(x,y)=1$, I thought of the relation $xy=lcm(x,y)$.
One way to approach it would be to assume that $xy=k\phi(m)+r$ and show that $r=0$
$(ab)^{xy}\equiv (ab)^r \pmod{\phi(m)}$ or
$(ab)^{lcm(x,y)} = a^{lcm(x,y)}b^{lcm(x,y)}\equiv (ab)^r \pmod{\phi(m)}$
APPROACH 2
The text has a theorem which says that if r is a primitive root modulo m, then
$r^i\equiv r^j\pmod{m}$ implies $i\equiv j \pmod{\phi(m)}$ So if we can identify i with $ord(ab)$ and j with $xy$, we are done. These two conditions leads to
$r^{ord(ab)}\equiv r^{xy}\pmod{\phi(m)}$
a and b can be written as powers of r:
$a=r^p$, $b=r^q$ with $p,q\ge 1$ so we get
$r^{ord(r^{p+q})}\equiv r^{xy} \pmod{m}$
$ord(r^{p+q})$ is the minimum solution for t of $(r^{p+q})^t\equiv 1 \pmod{m}$. I don't know where to go from here.
 A: You have correctly shown
$$(ab)^{xy} \equiv 1 \pmod{m} \tag{1}\label{eq1A}$$
However, the problem, as stated, is not necessarily to show $(ab)^{xy - n} \not\equiv 1 \pmod{m}$ for $n \gt 0$ (and, implicitly, $n \lt xy$), although this ends up being the case.
Next, for simpler algebra, let
$$z = \operatorname{ord}_m(ab) \tag{2}\label{eq2A}$$
so you're asked to prove
$$z \equiv xy \pmod{\phi(m)} \tag{3}\label{eq3A}$$
As for your two proposed approaches, I don't believe either one will work in general. For the first one, you

... assume that $xy=k\phi(m)+r$ and show that $r=0$

However, \eqref{eq3A} gives $xy = j\phi(m) + z$, where $z \gt 0$, so your assumption can only be correct only if $z = \phi(m)$ giving $k = j + 1$, but this is not necessarily always true since $z \lt \phi(m)$ is possible.
With your second approach, as stated in primitive root modulo $n$, primitive roots only exist for $n$ being $2$, $4$, $p^k$ or $2p^k$ where $p^k$ is a power of an odd prime $p$. The question doesn't state $n$ has a primitive root, or is one of the required forms, so you can't assume this in general.
From \eqref{eq1A}, \eqref{eq2A} and the multiplicative order divides any positive integer power which gives a congruence of $1$, we have $z \mid xy$. Thus, $xy = ez$ for some $e \ge 1$. Assume $z \lt xy$, i.e., $e \gt 1$. This means $e = fg$ for some $f \mid x$ and $g \mid y$, where $1$ or both of $f$ and $g$ are $\gt 1$. This gives
$$z = \left(\frac{x}{f}\right)\left(\frac{y}{g}\right) \tag{4}\label{eq4A}$$
WLOG, assume $f \gt 1$. This gives
$$\begin{equation}\begin{aligned}
(ab)^{zg} & \equiv ((ab)^z)^{g} \pmod{m} \\
a^{zg}b^{zg} & \equiv 1 \pmod{m} \\
\left(a^{\left(\frac{x}{f}\right)y}\right)\left(b^{\left(\frac{x}{f}\right)y}\right) & \equiv 1 \pmod{m} \\
\left(a^{\left(\frac{x}{f}\right)y}\right)\left(\left(b^{y}\right)^{\left(\frac{x}{f}\right)}\right) & \equiv 1 \pmod{m} \\
a^{\left(\frac{x}{f}\right)y} & \equiv 1 \pmod{m}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This shows $x \mid \left(\frac{x}{f}\right)y$ so, with $\gcd(x,y) = 1$, this means
$$x \mid \frac{x}{f} \tag{6}\label{eq6A}$$
However, this shows $x \le \frac{x}{f}$, which is impossible with $f \gt 1$. This proves the assumption $z \lt xy$ is false, which means $z = xy$. Thus, $z$ is congruent to $xy$ in any modulo, including $\phi(m)$, so \eqref{eq3A} is true.
