Subgroups of Free Groups and Schreier Rewriting — Examples

The two right cosets are $H$ and $Ha$, so $\mathcal{T}=\{1,a\}$ is a Schreier system. Its nontrivial Schreier generators are $s(1,b)=b$, $s(a,a)=a^2$, and $s(a,b)=ab{a}^{-1}$, while $s(1,a)=1$.

By [L1], the three nontrivial generators from step 1.1 form a free basis of $H$. Since $[F(a,b):H]=2$, [L2] also gives $\mathrm{rank}(H)=1+2(2-1)=3$, agreeing with the computed basis.

The right cosets of $H$ are $H{a}^n$ for $n\in \mathbb{Z}$, so $\mathcal{T}=\{a^n:n\in \mathbb{Z}\}$ is a Schreier system by [L1].

For every $n\in \mathbb{Z}$, one has $s(a^n,a)=1$ and $s(a^n,b)=a^{n}b{a}^{-n}$. Thus the nontrivial Schreier generators are exactly the displayed conjugates.

By [L2], those generators form a free basis of $H$.

The quotient records only the parities of the exponent sums of $a$ and $b$, so the four right cosets are exactly $H$, $Ha$, $Hb$, and $Hab$. The labeled Schreier graph therefore has $a$-edges $H\leftrightarrow Ha$ and $Hb\leftrightarrow Hab$, and $b$-edges $H\leftrightarrow Hb$ and $Ha\leftrightarrow Hab$.

The three displayed edges form a rooted spanning tree, so [L2] gives the corresponding Schreier system $\{1,a,b,ab\}$. The five positive edges not in that tree yield the nontrivial generators $a^2$, $b^2$, $a{b}^2{a}^{-1}$, $aba{b}^{-1}$, and $ba{b}^{-1}{a}^{-1}$.

By [L3], those five elements form a free basis of $H$.

The quotient by $H$ has cosets $H$ and $Hb$, so $\mathcal{T}=\{1,b\}$ is a right transversal. The nontrivial Schreier generators are $x=s(1,a)=a$, $z=s(b,a)=ba{b}^{-1}$, and $y=s(b,b)=b^2$.

The defining relator is $r=ba{b}^{-1}a$. Reidemeister-Schreier rewrites the two conjugates determined by $\mathcal{T}$ as $\tau (r)=zx$ and $\tau (br{b}^{-1})=yx{y}^{-1}z$. Therefore [L1] yields the presentation $⟨x,y,z\mid zx,yx{y}^{-1}z⟩$.

The first relator gives $z=x^{-1}$. Substituting this into the second relator yields $yx{y}^{-1}{x}^{-1}$. Hence the presentation from step 2.1 simplifies to $⟨x,y\mid yx{y}^{-1}{x}^{-1}⟩$, so the presented group is the free abelian group on two generators, that is, ${\mathbb{Z}}^2$. Thus $H$ is a surface subgroup of torus type.

The right cosets of $H$ are $H{a}^n$ for $n\in \mathbb{Z}$, so $\mathcal{T}=\{a^n:n\in \mathbb{Z}\}$ is a Schreier system. Its nontrivial Schreier generators are the infinitely many elements $a^{n}b{a}^{-n}$ for $n\in \mathbb{Z}$.

By [L1], the family in step 1.1 is a free basis of $H$. Since that basis is infinite, $H$ has infinite rank.

The subgroup $K$ consists of words with even exponent sum in $a$, so it has index $2$ in $F(a,b)$. The basis computation $K=⟨b,a^2,ab{a}^{-1}⟩$ shows that $H$ is generated by two members of that free basis.

Hence $K$ splits as the free product of $H$ with the cyclic subgroup generated by $ab{a}^{-1}$, so $H$ is a free factor of $K$. Also $a\mathrm{\notin }K$ because its exponent sum in $a$ is odd. This is a concrete separating finite-index overgroup of the kind promised abstractly by [L1].

Let $H\le F(a,b)$ be the subgroup of words with even exponent sum in $a$. Its two right cosets are $H$ and $Ha$. The set $\mathcal{T}=\{1,ab\}$ is a transversal, but it is not a Schreier system because the initial segment $a$ of $ab$ is not in $\mathcal{T}$.

Using [L1], the nontrivial raw transversal elements are $a{b}^{-1}{a}^{-1}$, $b$, $aba$, and $ab{a}^{-1}$. Indeed, the last one is $r_{\mathcal{T}}(ab,b)=a{b}^2(ab{)}^{-1}=ab{a}^{-1}$. This list is redundant because $a{b}^{-1}{a}^{-1}=(ab{a}^{-1}{)}^{-1}$.

By [L3], the reduced Schreier basis is guaranteed only for Schreier systems. Step 2.1 shows that the arbitrary transversal $\{1,ab\}$ instead gives a redundant list, so it does not yield the reduced Schreier basis. The Schreier initial-segment condition is load-bearing.