Is the free group on an empty set defined?

Question

I'm guessing that the free group on an empty set is either the trivial group or isn't defined. Some clarification would be appreciated.

Answer 1

It is defined. Free groups are defined for any sets just by the following universal property:

Let $S$ be a set. A group $F(S)$ with a map $\iota\colon S\rightarrow F(S)$ is called free over $S$, if for any group $H$ and any map $g\colon S\rightarrow H$ there is exactly one group homomorphism $h\colon F(S)\rightarrow H$, such that $g=h\circ\iota$.

(With this universal property, the free group over a fixed set is defined up to isomorphism and one possible choice in this isomorphism class is just the construction with "words" in $S$, which you propably know.)

Let's try this for the empty set $\emptyset$. The guess was, that it could be the trivial group $\{0\}$ with the only possible map $\iota\colon \emptyset\rightarrow \{0\}$, namely the empty map. So let $H$ be an arbitary group and $g\colon \emptyset\rightarrow H$ a map. Since there is no such map except the empty map, h must be the empty map. Now we are searching for a group homomorphism $h\colon \{0\}\rightarrow H$, such that $g=h\circ\iota$. Since there is exactly one group homomorphism from $\{0\}$ to another group, namely the one, which send $0$ to the neutral element of $H$ and $g=h\circ\iota$ holds for this map (both sides are just the empty map), we get the result: The free group over the empty set is just the trivial group.

If you know some basic category theory, you could do this way more elegant and things become more natural.

Answer 2

Short version: It's the trivial group. The only element is the empty word.

Long version: To elaborate, we write the set of words in $A$ as $\mathfrak{W}_A$ (considered as the monoid generated by $A\cup A^{-1}$ under concatenation) and the free group on $A$ as $\mathfrak{F}_A$. $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{ccc} \mathfrak{W}_\emptyset & \ra{\delta} & \mathfrak{F}_\emptyset \\ & \searrow & \da{\mu}\\ & & 1 & \end{array}$$ While $\delta:\mathfrak{W}_A\rightarrow \mathfrak{F}_A$ is not usually injective (since $\mathfrak{F}_A$ is the set of equivalence classes of cyclically reduced words), in this case, the only word with letters in $\emptyset$ is the empty word, so actually $\delta$ is an isomorphism (of groups!), as is $\mu:\mathfrak{F}_\emptyset\rightarrow 1$ for the same reasons.

Answer 3

Left adjoints preserve colimits, in particular the empty colimits, i.e. initial objects. In particular, the free group functor takes the initial set, i.e. $\emptyset$, to the initial group, i.e. the trivial group. Another example: The polynomial ring without variables is the base ring.

A must read: sexy vacuity

Answer 4

My guess is that would be a group with one element. There should be results about what direct products of groups have to do with sets of generators. What happens when you take the direct free product of a no generators and hence no relations, with a group with some other specified set of generators and relations?

PS: For now, I've struck out "direct" but left it visible, while substituting "free". But could this work just as well either way?

PPS: 15 minutes ago I was going to post that by hindsight I should definitely have said "free product". I've spent the past 15 minutes attempting to log in, without success. Now I'm out of time for a couple of hours, so maybe in two or three hours I'll try again.