Show that for natural numbers $a$ and $b$:
$$ \phi([a, b]) \times \phi(\gcd(a, b)) = \phi(a) \times \phi(b)$$
where $[a, b] = LCM (a, b)$.
From Exercise (5.1).25, we have
$$ \phi(a b) = \frac{\phi(a) \times \phi(b) \times \gcd(a,b)}{\phi(\gcd(a,b))} $$
This gives us
$$ \phi(a) \times \phi(b) = \frac{\phi(a b) \times \phi(\gcd(a,b))}{\gcd(a,b)} $$
Proposition (2.22) tells us that $ab = [a,b] \times \gcd(a,b)$ for natural numbers $a,b$, and so
$$ \phi(a) \times \phi(b) = \frac{\phi([a,b] \times \gcd(a,b)) \times \phi(\gcd(a,b))}{\gcd(a,b)} $$
Applying the result of Exercise (5.1).25 again
$$ \phi(a) \times \phi(b) = \frac{\phi([a,b]) \times \phi(\gcd(a,b)) \times \gcd([a,b],\gcd(a,b))}{\phi( \gcd([a,b],\gcd(a,b)))} \times \frac{\phi(\gcd(a,b))}{\gcd(a,b)} $$
Now, $\gcd([a,b], \gcd(a,b))=\gcd(a,b)$, and so we can simplify
$$ \phi(a) \times \phi(b) = \frac{\phi([a,b]) \times \phi(\gcd(a,b)) \times \bcancel{\gcd(a,b)}}{\cancel{\phi( \gcd(a,b))}} \times \frac{\cancel{\phi(\gcd(a,b))}}{\bcancel{\gcd(a,b)}} $$
Which leaves us with the desired result
$$ \phi([a, b]) \times \phi(\gcd(a, b)) = \phi(a) \times \phi(b)$$