Detect a knot from its fundamental group I'm studying the braid closure and I ended up with the knot $K= (\sigma_1\,\sigma_2\,\sigma_1\,\sigma_2\,\sigma_1)_*$, here the notation is according Murasugi but it does not really matter. By using a Snappy software, I know that the fundamental group of $K$ has the following presentation
$$
\pi_1(S^3 \setminus K) = \langle\, a,b \mid ab^2=b^2a \,\rangle,
$$
thus $K$ is not trivial. Snappy doesn't recognize $K$, which means that $K$ is a composite know (i.e. is not prime). My questions are:
Which knot is K? Can I find its prime factors (i.e K1#K2 = K)?

and
For a generic K, is there a way to detect K's name from its fundamental group?

 A: Note: In the braid group on three strands, $\sigma_1 \, \sigma_2 \, \sigma_1 = \sigma_2 \, \sigma_1 \, \sigma_2$,
so your braid is equivalent to
$$
\sigma_1 \, (\sigma_1 \, \sigma_2 \, \sigma_1) \, \sigma_1 
= \sigma_1^2 \, \sigma_2 \, \sigma_1^2.
$$
By Markov's theorem, you can cyclically permute the the braid word without changing the link closure, so
$$
\bigl( \sigma_1^2 \, \sigma_2 \, \sigma_1^2 \bigr)_* 
= \bigl( \sigma_1^2 \, \sigma_1^2 \, \sigma_2 \bigr)_* 
= \bigl( \sigma_1^4 \, \sigma_2 \bigr)_*.
$$
Using the other Markov move (stabilization), the closure of $\sigma_1^4 \, \sigma_2$ in $B_3$ is isotopic to the closure of $\sigma_1^4$ in $B_2 \cong \mathbb{Z}$ (just twists of two strands).
This particular $4$-twist link is sometimes called Solomon's knot despite not being a knot by mathematical definitions. (It's a two component link.)
A: Given the braid presentation, I drew the knot in Knotfolio:

It was recognized as the prime link L4a1, which the other answer points out is known as "Solomon's knot" (though in knot theory only one-component links are called "knots"). In particular, as an oriented link it is L4a1{1} and not L4a1{0}.
Supposing it's prime, this is the only 2-component link with up to 5 crossings with the same Jones polynomial and Alexander polynomial, so Knotfolio has correctly identified it. The diagram, through Seifert's algorithm, gives a Seifert surface of genus 1 ("can. genus" in Knotfolio) so the link is indeed prime.
