MHB Groups as Groupoids with One Object

[Math Amateur]

I am reading Paolo Aluffi's book: "Algebra: Chapter 0"

I am currently focussed on Chapter 2: "Groups: First Encounter".

On page 41, Aluffi defines a group as follows:

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Definition: A group is a groupoid with a single object.

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Aluffi has already defined a groupoid on page 29 as follows:

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" ... ... there are categories in which every morphism is an isomorphism; such categories are called groupoids."

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I cannot see how the definition of a group as a groupoid with a single object results in what I have come to know as a group: that is a set G with a binary operation $$ \bullet \ : \ G \times G \to G $$ that is associative and is such that G possesses a unique identity and unique inverses for each element.

Can someone please help me to see how the definition via category theory is equivalent to the usual 'undergraduate' definition of a group. I would especially like to be able to give a rigorous and formal demonstration of how a category in which all morphisms are isomorphisms leads to the properties of a group.

Hope someone can help in this matter.
... ... Further to the above but possibly covered in the above is the following: ... ... On page 43 Aluffi gives the following proposition pertaining to groups:

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Proposition 1.7. The inverse is unique: if $$ h_1, h_2 $$ are both inverses of g in G, then $$ h_1 = h_2 $$

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Aluffi asks readers to construct a stand-alone proof of this based on the category theory definition of a group and the following proposition:

Proposition 4.2. The inverse of an isomorphism is unique.

I would really appreciate help in constructing such a proof.

PeterPeter

 

[Deveno]

Suppose we have a category $\mathcal{G}$ with a single object, let's just call it $S$, such that every arrow $g:S \to S$ is an isomorphism (in the categorical sense). We will further assume the collection of arrows is small enough to be a set.

Let us, then, examine the set $\text{Hom}_{\ \mathcal{G}}(S,S)$, that is to say, the arrows of $\mathcal{G}$.

For convenience, let's adopt the letter $G$ for $\text{Hom}_{\ \mathcal{G}}(S,S)$.

From the definition of a category, we see that arrows are composable, that is: given $g_1,g_2 \in G$ we can form the arrow:

$g_1\circ g_2 \in \text{Hom}_{\ \mathcal{G}}(S,S) = G$

It is evident, then, that we have an (associative) binary operation $\circ: G\times G \to G$.

Now in a category $\mathcal{C}$, there is a (unique!) identity arrow $1_A \in \text{Hom}_{\ \mathcal{C}}(A,A)$ for every object $A$.

For our one-object category $\mathcal{G}$, since we have only one object ($S$), we have a unique arrow:

$1_S:S \to S$ such that:

$g \circ 1_S = 1_S \circ g = g$.

This evidently serves as an identity for the binary operation $\circ$.

Furthermore, since every arrow $g$ is an isomorphism, we have an arrow $f$ such that:

$f\circ g = 1_S$
$g \circ f = 1_S$ (since the domain and co-domain of every $g$ is always the same).

Thus $f$ serves as an inverse to $g$ in $G$.

Therefore: $(\text{Hom}_{\ \mathcal{G}}(S,S),\circ)$ forms a group, in the usual sense.

Now, let's "go the other way": suppose we have a group $(G,\ast)$.

Now we define a category $\mathcal{C}$ with:

$\text{Obj}(\mathcal{C}) = \{G\}$.

and we define the arrows of $\mathcal{C}$ to be the functions:

$L_g:G \to G$ given by $L_g(a) = g\ast a$, for any given $g \in G$.

We need to show 3 things:

1. Arrows are composable
2. We have a unique identity arrow
3. All arrows are isomorphisms of $\mathcal{C}$.

(1) follows from $L_g \circ L_h = L_{g\ast h}$, note:

$(L_g \circ L_h)(a) = L_g(L_h(a)) = L_g(h\ast a) = g\ast(h\ast a) = (g \ast h)\ast a = L_{g\ast h}(a)$ for all $a \in G$. Look how associativity of the operation $\ast$ plays out, here.

(2) follows from the fact that $L_e = 1_G$ (where $e$ is the identity element of $G$).

(3) follows from the fact that:

$L_g \circ L_{g^{-1}} = L_{g \ast g^{-1}} = L_e = 1_G$,

$L_{g^{-1}} \circ L_g = L_{g^{-1}\ast g} = L_e = 1_G$ (this shows $L_g$ is an isomorphism in $\mathcal{C}$).

I leave it to you to prove that $\mathcal{G}$ and $\mathcal{C}$ as defined above, are isomorphic, AS CATEGORIES, when we take:

$G = (\text{Hom}_{\ \mathcal{G}}(S,S),\circ)$ to be our given group.

This means you must find a functor:

$F:\mathcal{G} \to \mathcal{C}$, defined on objects and arrows that preserves composition and identities, and a functor:

$H:\mathcal{C} \to \mathcal{G}$, such that $FH = 1_{\mathcal{C}}$ and $HF = 1_{\mathcal{G}}$, the identity functors on their respective categories.

*******************

To prove Proposition 1.7, consider the arrow (remember, we're in a category now) $h_1 \circ g \circ h_2$, using the associativity of $\circ$ (composition of arrows), and the defining properties of identity arrows.

In most "actual" groups, the group elements can be thought of as "things that do something", the identity element being "the thing that does nothing", and the inverse as the thing that "undoes" what the original element "did". The formalization of this is often called Cayley's Theorem:

A group $G$ is isomorphic to its (left) action on itself by (left) multiplication. This gives a "concrete" realization of an "abstract" thing: a cyclic group $C_n$ of order $n$ is the "same thing" (isomorphic to) as the permutation group on the set of elements of $G$ generated by an $n$-cycle.

Just so, the "one-object" category $\mathcal{G}$ is an abstract thing, which can be realized "concretely" by considering elements of $G$ as "multiplication maps" (arrows). The "groupness" of $G$ is then captured by the "isomorphism-ness" (if that's even a word) of the arrows.

More generally, and more usefully for category theory, we have that a one-object category is a MONOID, and that categories in general are "generalized monoids" (the corresponding concept for monoids, as groupoids are to groups), or "partial monoids" as they are often called. This is not, strictly speaking, totally accurate, because some categories are so big, they do not have a "set of objects", for example, for every set $X$, we have the free group $FX$, and thus at LEAST one group for every possible cardinality of a set. This is already "too many groups" to form a set, so (for example) the category $\mathbf{Grp}$ does not have an underlying "set of objects", and so does not form a partial monoid.

However, some categories are "small" (the collection of objects and arrows ARE small enough to be sets). Even when categories aren't "small", sometimes for any two objects $A,B$ the collection of arrows $A \to B$ is small enough to be a set, such categories are called "locally small". These often prove to be the best compromise.

One reason this is true, is that in discussing categories, the objects are pretty much irrelevant. Why? Because we can just replace them with the identity arrows. Nevertheless, this becomes "hard to relate to", we think of our world as not only "process", but "state" as well (compare: "the sky is darkening" versus the "sky is dark").

 

[Math Amateur]

Suppose we have a category $\mathcal{G}$ with a single object, let's just call it $S$, such that every arrow $g:S \to S$ is an isomorphism (in the categorical sense). We will further assume the collection of arrows is small enough to be a set.

Let us, then, examine the set $\text{Hom}_{\ \mathcal{G}}(S,S)$, that is to say, the arrows of $\mathcal{G}$.

For convenience, let's adopt the letter $G$ for $\text{Hom}_{\ \mathcal{G}}(S,S)$.

From the definition of a category, we see that arrows are composable, that is: given $g_1,g_2 \in G$ we can form the arrow:

$g_1\circ g_2 \in \text{Hom}_{\ \mathcal{G}}(S,S) = G$

It is evident, then, that we have an (associative) binary operation $\circ: G\times G \to G$.

Now in a category $\mathcal{C}$, there is a (unique!) identity arrow $1_A \in \text{Hom}_{\ \mathcal{C}}(A,A)$ for every object $A$.

For our one-object category $\mathcal{G}$, since we have only one object ($S$), we have a unique arrow:

$1_S:S \to S$ such that:

$g \circ 1_S = 1_S \circ g = g$.

This evidently serves as an identity for the binary operation $\circ$.

Furthermore, since every arrow $g$ is an isomorphism, we have an arrow $f$ such that:

$f\circ g = 1_S$
$g \circ f = 1_S$ (since the domain and co-domain of every $g$ is always the same).

Thus $f$ serves as an inverse to $g$ in $G$.

Therefore: $(\text{Hom}_{\ \mathcal{G}}(S,S),\circ)$ forms a group, in the usual sense.

Now, let's "go the other way": suppose we have a group $(G,\ast)$.

Now we define a category $\mathcal{C}$ with:

$\text{Obj}(\mathcal{C}) = \{G\}$.

and we define the arrows of $\mathcal{C}$ to be the functions:

$L_g:G \to G$ given by $L_g(a) = g\ast a$, for any given $g \in G$.

We need to show 3 things:

1. Arrows are composable
2. We have a unique identity arrow
3. All arrows are isomorphisms of $\mathcal{C}$.

(1) follows from $L_g \circ L_h = L_{g\ast h}$, note:

$(L_g \circ L_h)(a) = L_g(L_h(a)) = L_g(h\ast a) = g\ast(h\ast a) = (g \ast h)\ast a = L_{g\ast h}(a)$ for all $a \in G$. Look how associativity of the operation $\ast$ plays out, here.

(2) follows from the fact that $L_e = 1_G$ (where $e$ is the identity element of $G$).

(3) follows from the fact that:

$L_g \circ L_{g^{-1}} = L_{g \ast g^{-1}} = L_e = 1_G$,

$L_{g^{-1}} \circ L_g = L_{g^{-1}\ast g} = L_e = 1_G$ (this shows $L_g$ is an isomorphism in $\mathcal{C}$).

I leave it to you to prove that $\mathcal{G}$ and $\mathcal{C}$ as defined above, are isomorphic, AS CATEGORIES, when we take:

$G = (\text{Hom}_{\ \mathcal{G}}(S,S),\circ)$ to be our given group.

This means you must find a functor:

$F:\mathcal{G} \to \mathcal{C}$, defined on objects and arrows that preserves composition and identities, and a functor:

$H:\mathcal{C} \to \mathcal{G}$, such that $FH = 1_{\mathcal{C}}$ and $HF = 1_{\mathcal{G}}$, the identity functors on their respective categories.

*******************

To prove Proposition 1.7, consider the arrow (remember, we're in a category now) $h_1 \circ g \circ h_2$, using the associativity of $\circ$ (composition of arrows), and the defining properties of identity arrows.

In most "actual" groups, the group elements can be thought of as "things that do something", the identity element being "the thing that does nothing", and the inverse as the thing that "undoes" what the original element "did". The formalization of this is often called Cayley's Theorem:

A group $G$ is isomorphic to its (left) action on itself by (left) multiplication. This gives a "concrete" realization of an "abstract" thing: a cyclic group $C_n$ of order $n$ is the "same thing" (isomorphic to) as the permutation group on the set of elements of $G$ generated by an $n$-cycle.

Just so, the "one-object" category $\mathcal{G}$ is an abstract thing, which can be realized "concretely" by considering elements of $G$ as "multiplication maps" (arrows). The "groupness" of $G$ is then captured by the "isomorphism-ness" (if that's even a word) of the arrows.

More generally, and more usefully for category theory, we have that a one-object category is a MONOID, and that categories in general are "generalized monoids" (the corresponding concept for monoids, as groupoids are to groups), or "partial monoids" as they are often called. This is not, strictly speaking, totally accurate, because some categories are so big, they do not have a "set of objects", for example, for every set $X$, we have the free group $FX$, and thus at LEAST one group for every possible cardinality of a set. This is already "too many groups" to form a set, so (for example) the category $\mathbf{Grp}$ does not have an underlying "set of objects", and so does not form a partial monoid.

However, some categories are "small" (the collection of objects and arrows ARE small enough to be sets). Even when categories aren't "small", sometimes for any two objects $A,B$ the collection of arrows $A \to B$ is small enough to be a set, such categories are called "locally small". These often prove to be the best compromise.

One reason this is true, is that in discussing categories, the objects are pretty much irrelevant. Why? Because we can just replace them with the identity arrows. Nevertheless, this becomes "hard to relate to", we think of our world as not only "process", but "state" as well (compare: "the sky is darkening" versus the "sky is dark").

Thanks for this highly informative post, Deveno.

Most appreciative of your help.

Just working through the post carefully now ... reflecting on what you have written.

Thanks again,Peter