Name for the group generated by a groupoid

Question

Is there a name for a group generated by a groupoid?

I'm defining a groupoid as a category $\mathcal{C}$ where every arrow has a unique inverse that is both a left inverse and a right inverse.

I'm interested in ways of taking a groupoid and scrunching it down into a group. I'm secondarily curious about how much of the groupoid structure this construction preserves.

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Suppose I have a groupoid $\mathcal{G}$ and I'm trying to turn it into a group $G$. Let's call the group operation $+$ for convenience regardless of whether it is commutative or not.

Let $X$ be the set of finite formal sums of $\mathcal{G}$ arrows.

Let's define the negation of a formal sum $a_1 + \cdots + a_n$ as $(-a_n) + \cdots + (-a_1)$.

Here are some rewrites on elements of $X$.

Suppose we have a finite sum $\cdots + a + b + \cdots $.

If $a + b$ is defined in $\mathcal{G}$, rewrite this to $\cdots + w + \cdots$ where $w$ is $a+b$ in $\mathcal{G}$.

Suppose any identity element of $\mathcal{G}$ appears anywhere in a sum, $\cdots + 0 + a + \cdots$. Let's remove it giving $\cdots + a + \cdots$.

Now let's define an equivalence relation.

Suppose $a \in X$ and $b \in X$. Then, $a \equiv b$ holds if and only if $a - b$ reduces to the empty formal sum.

Let's check that $\equiv$ is an equivalence relation.

• $a \equiv a$ holds because in $a + (-a)$ we can keep converting the innermost terms to identities and then dropping them.
• $a \equiv b$ holds if and only if $b \equiv a$ holds because any series of reduction steps that leads to the empty formal sum on the LHS can be applied in reverse to the RHS and vice versa.
• $a \equiv b$ and $b \equiv c$ implies $a \equiv c$. We can conclude $a - b \equiv 0$ and $b - c \equiv 0$ and thus $a -c \equiv 0$.

$\equiv$ is also a congruence.

I think this construction is equivalent to taking all the arrows of $\mathcal{G}$ as the set of generators and then imposing as relations:

1. $\text{id}_X = 0$ for all objects $X$ in $\mathcal{G}$.
2. All true equations about arrows of $\mathcal{G}$.

Answer 1

$ \newcommand\G{\mathcal G} \newcommand\F{\mathbb F} $Preliminaries

1. If $X$ is any set, the pair groupoid $P_X$ consists of the set $X\times X$, with operation defined by $$ (x,y)(z,w)=\left\{\matrix{ (x,w), & \text {if } y=z,\cr \text{undefined}, & \text {otherwise}.\cr }\right. $$

2. If $G$ is any group one may define a groupoid structure on the cartesian product $G\times P_X$ by introducing the multiplication $$ (g,p)(h,q)=\left\{\matrix{ (gh,pq), & \text {if $pq$ is defined} ,\cr \text{undefined}, & \text {otherwise}.\cr }\right. $$

3. It is not hard to show that any groupoid is a disjoint union of a family of groupoids of the form $G\times P_X$ (the multiplication being undefined when the factors come from two different subgroupoids).

OK, so now back to the question.

Given a groupoid $\G$ let us denote by $U(\G)$ the group obtained by scrunching down $\G$, namely the group generated by a set $\{[\gamma ]: \gamma \in \G\}$ subject to the relations $[\alpha ][\beta ]=[\alpha \beta ]$, whenever $\alpha \beta $ is defined in $\G$.

If $\G$ is the disjoint union of subgroupoids $\G_1$ and $\G_2$, it is fairly obvious that $$ U(\G) = U(\G_1) * U(\G_2), $$ (free product) so it is enough to describe $U(\G)$ for connected, or transitive, groupoids, all of which necessarily have the form $G\times P_X$, as above.

Given such a groupoid, I claim that $U(G\times P_X)$ is isomorphic to the free product $G*\F$, where $\F$ is the free group generated by a set with one element less than $X$.

In order to prove it, let us first observe that $\F$ is isomorphic to the subgroup of $\F(X)$ (free group on $X$) generated by the subset $\{xy^{-1}: x, y\in X\}$, so this is the model for $\F$ we shall adopt.

Fixing some $x_0$ in $X$, observe that the map $$ \varphi :\big [g,(x,y)\big ] \in G\times P_X \mapsto xx_0^{-1}\cdot g\cdot x_0y^{-1}\in G*\F, $$ is compatible with the relations defining $U(G\times P_X)$, so it induces a group homomorphism $$ \Phi :U(G\times P_X)\to G*\F, $$ which I claim is an isomorphism.

In order to exhibit an inverse for $\Phi $, let us first consider the group homomorphism $$ \Delta :\F(X) \to U(G\times P_X), $$ specified on the generating set $X$ by $$ \Delta (x) = [1,(x, x_0)]. $$ It might be worth observing that $ \Delta (xy^{-1}) = [1,(x, y)], $ for every $x$ and $y$ in $X$. We then consider the unique group homomorphism $$ \Psi :G*\F \to U(G\times P_X), $$ such that $$ \Psi (g)= [g,(x_0,x_0)],\quad \forall g\in G, $$ and $$ \Psi (w)=\Delta (w),\quad\forall w\in \F. $$ It is now easy to see that $\Phi $ and $\Psi $ are each other's inverse.