Why is the commutator subgroup of the group associated to a finite quandle finitely generated? A quandle $Q$ is a set with one binary operation $(x, y) \mapsto  x ∗ y$ which satisfies the following three axioms:
i) $\forall x \in Q: x ∗ x=x$
ii) the map $S_{x}: y \mapsto y ∗ x$ is a bijection on $Q$ for all $x \in Q$
iii)$(x∗ y)∗ z=(x∗ z)∗ (y∗ z)$ for all $x,y,z \in Q$
Given a quandle $(Q,∗)$, denote by $G_{Q}$ the group generated by all the elements of $Q$ and the set of relations
$x ∗ y = y^{−1}xy$  for all $x, y \in Q$.
I came across a theorem that said: let $(Q, ∗)$ be a finite quandle , then the commutator subgroup $G^{'}_{Q }$ of $G_{Q }$ is finitely generated. But why is that? I mean $G^{'}_{Q }= \langle g^{-1}h^{-1}gh|g,h \in G_Q \rangle$, but you still can generate infinitely many elements of $G_Q$ by using finitely many elements of $Q$. Can someone help me out?
 A: A quick Google search gave me this paper: https://arxiv.org/pdf/1705.10607.pdf
Proposition 3.3 (2).
The title is "Automorphism groups of quandles and related groups" and it's by Bardakov, Nasybullov, and Singh.
Notice that in this case the commutator subgroup is generated by $ghg^{-1}h^{-1}$ for $g,h\in Q$. In general the commutator subgroup is the same as the normal subgroup generated by these elements, but the commutators of generators here are invariant under conjugation. Note that if $gag^{-1}a^{-1}$ and $hah^{-1}a^{-1}$ can be expressed this way, then
$$(gh)a(gh)^{-1}a^{-1}=g(hah^{-1}a^{-1})g^{-1}(gag^{-1}a^{-1})$$
Normally $g(hah^{-1}a^{-1})g^{-1}$ would not remain a product of commutators of generators, but in the enveloping group of a quandle it does. This makes the proof work.
A: It's pretty simple observation, really, and is true for more general class of groups. 
Let $G$ be a group normally generated by single element $g_0$ and denote its conjugacy class as $X$. Then $G'$ is generated by elements $gg_0^{-1}$, where $g$ lies in $X$. 
Proof: observe that $ H := \langle fh^{-1} \, | \, f, h \in  X \rangle$ is normal and factor $G/H$ is cyclic, therefore $G' < H$, but every element of the form $fh^{-1}$ is a commutator of two elements in $X$ because $X$ is a complete conjugacy class, so $H = G'$. Substituting $fg^{-1} = fg_0^{-1}(hg_0^{-1})^{-1}$ we obtain that $\mathrm {rk} \, G' \leq |X| - 1$.
In fact, our quandle is not always transitive, but anyway factor $G_Q/\langle fh^{-1} \,|\, f, h \in Q\rangle$ will be free abelian on orbits of $Q$, and for similar reasons kernel lies in commutator.
