Why is the Whitehead double of a knot always prime?

Question

I was looking for a proof that there are infinitely many prime knots and one said "take your favorite (prime) knot and consider all its Whitehead double", implying that all Whitehead doubles of a (prime?) knot are prime. But why is that so?

Answer 1

It's fairly easy to see that if a knot has knot genus $1$, then it is prime. Recall that if $K$ and $K'$ are knots, $K\# K'$ is their connect sum, and $g(-)%$ is the genus of a knot, then $$g(K\# K')=g(K)+ g(K').$$That is, the genus of a knot is additive under connect sum. It follows that if $g(K)=1$, then $K$ is prime.

Now, a twisted Whitehead double of any (non-trivial) knot has genus $1$ and so it is prime$^*$. As there are infinitely many twisted Whitehead doubles of any knot, and they can be distinguished by their Alexander polynomials, it follows that there are infinitely many prime knots.

$*$ There's a great explanation of why such a knot has genus $1$ given in this answer.