Confusing solution

Module 5 Day 7 Your Turn Part 3 MiniQuestion Solution
Let the GCD of a and b be g; then we can express a as g × A and b as g × B, where A and B don't have any factors in common. Then the LCM of a and b is g × A × B, and the LCM divided by the GCD would be A × B. Since this is equal to b, that means that g × B = A × B, which means that g = A. Then a = A^2, and b is a multiple of the square root of a.
Bold part is confusing 
@neatlobster (I'm gonna copy over the question first):
\(\text{Suppose that for integers } a \text{ and } b, \frac{\text{lcm}(a,b)}{\gcd(a,b)} =b \)
$$a=b^2 \\ \gcd(a,b)=1 \\ \text{One of the numbers is } b \text{ times the other} \\ a=1 \\ a \text{ is a square, and } b \text{ is a multiple of }\sqrt{a} $$
\(\text{Which of the following must be true?}\)Following the explanation, we have the following:
\( \gcd(a,b)=g \)
\(a = g \cdot A \)
\(b = g \cdot B \)
\(\text{lcm}(a,b)=g \cdot A \cdot B \)Remember, the problem gave us that \(\frac{\text{lcm}(a,b)}{\gcd(a,b)}=b \). Using our substitutions for the lcm & gcd, we have that
\(\frac{g \cdot A \cdot B}{g} = b \)
Now, here's the clever substitution: recall that \(b = g \cdot B\). So,
\(\frac{g \cdot A \cdot B}{g} = b \implies \frac{b \cdot A }{g} = b \implies \frac{A}{g} = 1 \implies A=g \)Since \(a = g \cdot A , a = A^2 \), so \(a\) is a perfect square.
And, since \(a=A^2 \implies \sqrt{a} = A = g\)
\(b = g \cdot B \implies b= \sqrt{a} \cdot B\) , so \(b\) is a multiple of \(\sqrt{a}\).
Hence, the answer is the fifth one.I know this is a lot, let me know if you have any more questions. I'd be glad to help!
P.S. Here's a hint: try using \(\gcd(a,b) \cdot \text{lcm}(a,b) = a \cdot b\). It's a very useful fact!

Thanks so much!

@neatlobster @quacker88 Thank you!!