Free group ($F_n$)

The free group $F(S)$ on a set $S$ consists of reduced words in letters $s^{\pm 1}$, with multiplication followed by cancellation. Every map from $S$ to a group extends uniquely to a homomorphism from $F(S)$.

An endomorphism of $F_n$ has image generated by at most $n$ elements. If this image has finite index $d$, the Nielsen–Schreier formula gives rank $1+d(n-1)$. Its abelianization requires at least that many generators. For $n\ge 2$, comparison with $n$ forces $d=1$.

// Target: geometry-and-topology.bigb

In $F(a,b)$, any set of distinct conjugates $b^{-i}a{b}^i$ freely generates its subgroup. Combine adjacent powers of the same conjugate in a reduced word. After expansion, distinct successive indices leave nonzero intervening powers of $b$, and the intervening nonzero powers of $a$ cannot cancel in the free product $⟨a⟩\ast ⟨b⟩$. The same argument works with $a,b$ interchanged and for finite or countably infinite sets of indices.

If $a,b$ freely generate, then $a^{i}b{a}^{-i}$ for any distinct integer indices freely generate their subgroup. Group a reduced word in these new generators into blocks with nonzero exponents and different successive indices. Expansion produces alternating nonzero powers of $b$ and interior powers $a^{i_{j+1}-i_j}$, so it cannot freely reduce to the empty word. Consequently every finite-rank free group embeds in a rank-two free group; free rank is not monotone under subgroup inclusion.

Suppose a group homomorphism $\pi :G\to Q$ is injective on a subgroup freely generated by $x_1,\dots ,x_n$. If $y_i$ have $\pi (y_i)=\pi (x_i)$, then the $y_i$ freely generate their subgroup: a relation among them projects to a relation among the free basis images, so the corresponding abstract reduced word must be empty. In particular, this applies to $y_i=x_{i}g$ when the normal closure of $g$ intersects $⟨x_1,\dots ,x_n⟩$ trivially.

A subset $X$ is a free basis of a group $G$ when the natural map from the free group on $X$ to $G$ is an isomorphism. Equivalently, $X$ generates $G$ and no nonempty freely reduced word in $X\cup X^{-1}$ equals the identity. A generating set is not automatically a free basis.

A word in generators and their formal inverses is freely reduced if no consecutive letters are mutual inverses. Deleting such inverse pairs gives the unique reduced representative in a free group. In a general group presentation, free reduction preserves the represented group element but may not produce a shortest representative.

Free reduction removes adjacent inverse-letter pairs. A stack algorithm scans the letters, canceling the top letter when the next is its inverse. The resulting reduced word is unique: deleting an inverse pair before scanning does not change the final stack, so every cancellation sequence gives the same result. In any group presentation these deletions preserve the represented element, and in a free group the reduced word is the identity exactly when it is empty.

A freely reduced word is cyclically reduced if its first and last letters are not inverse, with the empty word also allowed. Every cyclic permutation of a nonempty cyclically reduced word is freely reduced, represents a conjugate, and has the same length. Positive powers are freely reduced because no cancellation occurs at copy boundaries.

A reduced word contains no adjacent generator and inverse-generator pair. Cancel such pairs to obtain the unique reduced representative in a free group. One proof constructs the free group itself from reduced words, with multiplication given by concatenation followed by cancellation; its action by these operations on words distinguishes distinct reduced representatives.

For every function $f:S\to G$ from a set into a group, there is a unique group homomorphism $\bar{f}:F(S)\to G$ whose restriction to $S$ is $f$.

Suppose nontrivial groups $G_1,G_2$ act on a set and there are disjoint nonempty subsets $X_1,X_2$ such that every nonidentity element of $G_1$ maps $X_2$ into $X_1$, while every nonidentity element of $G_2$ maps $X_1$ into $X_2$. Assume at least one factor has at least three elements. Then the subgroup generated by $G_1$ and $G_2$ is their free product. The cardinality hypothesis excludes the counterexample in which $G_1=G_2=C_2$ acts by swapping $X_1$ and $X_2$. Versions allowing both factors to have order two need an additional separation hypothesis.

Every finitely generated free group is a residually finite group. For $F_2$, the ping-pong lemma embeds it into ${\mathrm{SL}}_2(\mathbb{Z})$, and reduction modulo a prime in an integral matrix group separates each nonidentity element in a finite quotient.