Is the free group on an empty set defined?

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.

Solution 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.

Solution 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.