It’s knot learning time! Last time, we looked at the trefoil knot:

In that post, I showed you how you can write down a presentation for the fundamental group of a knot. So, you may want to review the procedure I gave in that post, because we’re going to do it again!

This time, I am going to compute the fundamental group of the following knot:

It is known as the *cinquefoil* knot, *cinque* indicating that it looks like a five-pointed star if drawn in the most symmetrical way possible. (To be clear, I know my drawing is *not* symmetrical).

I have already labelled the segments of the knot drawing with letters so you can follow along. Remember, the knot group is then generated by the letters $a,b,c,d,e$. To get the relations, you have to draw small oriented boxes around each crossing and write down the segments you intersect for each relator. The exponent for each segment will be $1$ if the segment is coming from the right and $-1$ if it is coming from the left. Believe me, it is explained in much more detail in the previous post on knots.

In this case, I am omitting the red-purple-blue crossing, and I get the following presentation:

$$\langle~a,b,c,d,e~|~ ada^{-1}c^{-1}, a^{-1}dbd^{-1}, b^{-1}d^{-1}be, ce^{-1}b^{-1}e~\rangle.$$ As it happens with knot groups, this presentation is rather inefficient. However, as it *also* happens with knot groups, it’s easy to reduce this presentation to a smaller one. I explained how to do that (with a knot group!) in the post on Tietze transformations. Basically, you take a relator likle $ada^{-1}c^{-1}$ which means $c = ada^{-1}$ and substitute it where you see a $c$, and then use the Tietze transformations to reduce. Since I already did a detailed example in the aforementioned post, I will just write down the end result, a one-relator group:

$$G = \langle~d,x~|~dx^2d = x^3~\rangle.$$ By doing this relatively painless procedure of reducing the knot group, it will make life a whole lot easier when you want to *prove* stuff about it. For example, intuitively, it seems the cinquefoil knot is not equivalent to the unknot. To show this rigorously, it suffices to show that this knot group $G$ is not ismorphic to $\Z$.

Also, it would be really nice if we could show that this group is not isomorphic to the knot group of the trefoil shown at the beginning of this post. For this, it suffices to exhibit a group $H$ and two elements $d,x\in H$ such that $dx^2d = x^3$ *and* such that $dx\not= xd$. Why this latter condition? Simply because this relation of course can hold in the integers (for example, set $x=2,d=1$). Whereas, if $dx\not= xd$, we will have shown that $G$ is *not* commutative and therefore can’t be the integers.

In fact, there is such a group: in the symmetric group $S^5$, we consider the following elements (by abuse of notation we also call them $d$ and $x$):

$$\begin{align*}d &= (1,5,4,3,2),\\ x &= (1,4,5,2,3).\end{align*}$$ I will leave it to the reader to show that they don’t commute, and

$$dx^2d = x^3 = (1,2,4,3,5).$$ Therefore, there exists a surjective homomorphism of $G$ onto a nonabelian subgroup of $S_5$ (can anyone tell me what this subgroup is?). We note that $S_5$ is actually the smallest symmetric group such that there exists two noncommuting elements $d,x$ for which the relation $dx^d = x^3$ holds. Because we found a surjective homomorphism from the trefoil knot group to $S_3$, this means that the trefoil and the cinquefoil are *not* equivalent as knots!