RCG3:

The morphisms of a category \\(\mathscr{C}\\) presented by graph (\\( \mathscr{G} = (E,\{\bullet\})\\)) with only one node, and equations (\\(Eq \subset E^*\x E^*\\) where \\(Eq = [\forall x y, x\circ y = y \circ x] \cup Eq' where (\forall(x = y)\in Eq', y = id,\,\land\,\forall e\in E, \exists x, e \in x \land (x = id) \in Eq'\\) are an albien group under composition.

Proof: We know that all one object categories are monoids (and vice versa). So we just have to show that each element has an unique inverse.
By the definition of presented in: \\[\forall m \in \mathscr{Hom}(\bullet,\bullet), \exists w \in E^* of length n, m = w_0\circ w_1 \circ \dots \circ w_{n-1}\\] in fact there are many such w because of the equations in Eq.

Proceed by Induction:
\\[w=\epsilon,\\]
in which case m = id = id^{-1}\\]
\\[\exists m' m'^{-1}, \m = w_0\circ m'\\ \land m'\circ m'_{-1}=id)

If w_0 has an inverse then: \\[w_0 m'\circ [m'^{-1}w_0^{-1}] = m\circ m'^{-1}w_0^{-1} =id = [m'^{-1}w_0^{-1}] \circ m\\]

So we just need to find an inverse to the singleton word [w_0]. By assumption, \\[\exists x, w_0 \in x \land (x,id) \in Eq'\\] by the \\(\forall x y, x\circ y = y\circ x\\) equation we can pull an w_0 to the front (using \\(\backslash\\) as list difference, _not_ set difference): \\[w_0 (x \backslash w_0) = id\\] so \\(w_0^{-1} = x \backslash w_0\\).

And we have what we need.

I am pretty sure its not that hard to extend this too the non ablian case. But cases like "fgfgh=id" give me headaches.

I also have no idea how to tell if a group is an symetry group.