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The free group is often defined by its universal property. A group $F$ is said to be free on a subset $S$ with inclusion map $\iota : S \rightarrow F$ if for every group $G$ and set map $\phi:S \rightarrow G$ there exists a unique homomorphism $\overline{\phi}:F \rightarrow G$ such that $\overline{\phi} \circ \iota (s) = \phi (s)$ $ \forall s \in S$.

It is said that the existence of $\overline{\phi}$ is what determines there are no relations. My question is (without defining the (free) group of reduced words) can you show there are no relations just by picking groups G to map into? That is, can you show no reduced words on $S^{\pm1}$ (excluding the empty word) are equal to the identity element in the free group?

Example attempt:
Suppose the reduced word $w$ is not the empty word, so it contains some letter $a$. Suppose adding all the powers of $a$ in the word $w$ gives the integer $k$. Define the set map $\phi : S \rightarrow \mathbb{Z}/( \lvert k \rvert +1)\mathbb{Z} $ by

$\phi (s) = \left\{ \begin{array}{@{}l@{\thinspace}l} 0 & \text{ if } s \neq a\\ 1 & \text{ if } s = a\\ \end{array} \right.$

then the homomorphism from $F$ to $\mathbb{Z}/( \lvert k \rvert +1)\mathbb{Z}$ extending the set map $\phi$ sends $w$ to $k$. As homomorphisms preserve group identities, and k is not the identity in $\mathbb{Z}/( \lvert k \rvert +1)\mathbb{Z}$, then $w$ is not the identity in $F$.

This was my first idea, but fails because it misses the case where $k=0$.

cede
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    I really like this question. It is asking, in other words, if there exist any laws which must be satisfied by all groups, but are not direct consequences of the word-reduction rules. As you note, constructing the free group as a group of reduced words shows that the answer is "no". But I wonder if we can instead use some kind of probabilistic argument about finite symmetric groups. Somehow show that, for $n \gg 0$, the probability that $(w(\sigma_1, \dots, \sigma_k))(x) = x$ is strictly less than $1$ as $\sigma_1, \dots, \sigma_k$ range over $S_n$ and $x$ ranges over ${1, \dots, n}$. – diracdeltafunk Oct 24 '23 at 03:35
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    @diracdeltafunk: this is true because the free groups are residually finite (which does not follow from the universal property, and the analogous property for other algebraic structures is sometimes false, iirc). But I don't know how easy it will be to prove. – Qiaochu Yuan Oct 24 '23 at 06:32

2 Answers2

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Let $u$ be a reduced word, for instance $u = ab\bar a$. Now, spell your word in the Cayley graph of the free group, for instance $$ 1 \xrightarrow{a} 2 \xrightarrow{b} 3 \xleftarrow{a} 4 $$ and complete $a$ and $b$ into permutations, for instance $a = (1,2)(3,4)$ and $b = (1)(2,3)(4)$. In the permutation group generated by $a$ and $b$, $u$ will not be the identity, since it maps $1$ to $4$.

This is basically the idea behind the Stallings automaton.

EDIT. Another example. Let $u = a^2b^{-3}cb^2$. Spelling $u$ yields the following diagram: $$ 1 \xrightarrow{a} 2 \xrightarrow{a} 3 \xleftarrow{b} 4 \xleftarrow{b} 5 \xleftarrow{b} 6 \xrightarrow{c} 7 \xrightarrow{b} 8 \xrightarrow{b} 9 $$ Now complete $a$, $b$ and $c$ into permutations, for instance $a = (1, 2, 3)$, $b = (6, 5, 4, 3)(7,8,9)$ and $c = (6,7)$.

J.-E. Pin
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    That's very cool! Can you maybe add a more complicated example, say $u = a^2 b^{-3} c b^2$ in the free group on ${a,b,c}$, just so that we understand this procedure? – Martin Brandenburg Oct 24 '23 at 06:24
  • How do you know that $1 \ne 4$ in the Cayley graph of the free group? – ronno Oct 24 '23 at 11:42
  • Unless you mean the graph on reduced words with edges given by concatenating generators, in which case this is the same argument as in @Martin's answer – ronno Oct 24 '23 at 12:01
  • I think the point of this answer is that even though the proof is motivated by the Cayley graph, it does not really rely on it. The permutation groups are what we only need. – Martin Brandenburg Oct 24 '23 at 12:50
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More generally, the universal property can be used to describe the elements in a coproduct of groups. Notice that the free group is a coproduct of groups $\mathbb{Z}$. See me answer at

Specifically, let $W$ be the set of reduced words over $S$, and define a map $S \to \mathrm{Sym}(W)$ which sends $s \in S$ to the map $w \to \text{reduction of } sw$ (you can write this down more explicitly, see my answer there; you also need to prove that it is bijective). The universal property allows us to extend this to a homomorphism $F(S) \to \mathrm{Sym}(W)$. Then one checks that $F(S) \to \mathrm{Sym}(W) \xrightarrow{\text{ev}_1} W$ is left inverse to $W \to F(S)$, so that $W \to F(S)$ is injective as claimed.

I have investigated similar questions of this type (derive the element structure from the universal property) in the past. Here are some examples. Maybe this will be of interest for you.

Martin Brandenburg
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  • That map is a bijection (with inverse $w \mapsto \text{reduction of } s^{-1}w$) because reduction is a well-defined operation (on $s^{-1}s w$). – ronno Oct 24 '23 at 11:59