The Reidemeister-Schreier presentation theorem

By [L3], the set $S$ of nontrivial Schreier generators is a free basis of $H^{\mathrm{\prime }}$. Therefore the inclusion $S\hookrightarrow H^{\mathrm{\prime }}$ extends to an isomorphism $\rho :F(S)\to H^{\mathrm{\prime }}$.

By [L1], the ambient quotient is $G=F(X)\mathrm{/}N$ with $N=⟨\text{\hspace{0.17em}⁣}⟨R⟩\text{\hspace{0.17em}⁣}{⟩}_{F(X)}$. The restriction of the quotient map to $H^{\mathrm{\prime }}$ has image $H$ and kernel $H^{\mathrm{\prime }}\cap N$, so [L5] gives $H\cong H^{\mathrm{\prime }}\mathrm{/}(H^{\mathrm{\prime }}\cap N)$. By [L2], every element of $H^{\mathrm{\prime }}\cap N$ is a finite product of conjugates $u{r}^{\pm 1}{u}^{-1}$ with $u\in F(X)$ and $r\in R$. Writing $u=ht$ with $h\in H^{\mathrm{\prime }}$ and $t\in \mathcal{T}$ turns each such conjugate into $h(t{r}^{\pm 1}{t}^{-1})h^{-1}$, so $H^{\mathrm{\prime }}\cap N$ is contained in the normal closure in $H^{\mathrm{\prime }}$ of the elements $tr{t}^{-1}$. Conversely, each $tr{t}^{-1}$ lies in $H^{\mathrm{\prime }}\cap N$, and $H^{\mathrm{\prime }}\cap N$ is normal in $H^{\mathrm{\prime }}$, so that normal closure is exactly $H^{\mathrm{\prime }}\cap N$.

Let $q:H^{\mathrm{\prime }}\to H^{\mathrm{\prime }}\mathrm{/}(H^{\mathrm{\prime }}\cap N)$ be the quotient map, and put $\bar{\rho }=q\circ \rho :F(S)\to H^{\mathrm{\prime }}\mathrm{/}(H^{\mathrm{\prime }}\cap N)\cong H$. Fix $t\in \mathcal{T}$ and $r\in R$, and write $tr{t}^{-1}=a_1\cdots a_n$. Let $t_j$ and ${\sigma }_j$ be the successive representatives and rewriting factors from [L4]. If $a_j=x\in X$, then [L6] gives ${\sigma }_j=s(t_{j-1},x)=t_{j-1}x{t}_j^{-1}$, so $t_{j-1}{a}_j={\sigma }_j{t}_j$. If $a_j=x^{-1}$, then $t_{j-1}$ is the chosen representative of the coset $H^{\mathrm{\prime }}{t}_{j}x$, so [L6] gives ${\sigma }_j=s(t_j,x{)}^{-1}=t_{j-1}{x}^{-1}{t}_j^{-1}$ and again $t_{j-1}{a}_j={\sigma }_j{t}_j$. Multiplying these identities yields $tr{t}^{-1}={\sigma }_1\cdots {\sigma }_n{t}_n=\rho (\tau (tr{t}^{-1}))$, because $tr{t}^{-1}\in H^{\mathrm{\prime }}$ forces $t_n=1$ by [L4]. Since $tr{t}^{-1}\in H^{\mathrm{\prime }}\cap N$, every rewritten relator $\tau (tr{t}^{-1})$ lies in $\ker \bar{\rho }$.

Conversely, if $w\in \ker \bar{\rho }$, then $\rho (w)\in H^{\mathrm{\prime }}\cap N$. By step 1.2, $\rho (w)$ is a finite product of conjugates in $H^{\mathrm{\prime }}$ of the elements $tr{t}^{-1}$ and their inverses. Replacing each $tr{t}^{-1}$ by the equal element $\rho (\tau (tr{t}^{-1}))$ from step 2.1 and applying the isomorphism ${\rho }^{-1}$ shows that $w$ lies in the normal closure of the words $\tau (tr{t}^{-1})$ in $F(S)$. Therefore $\ker \bar{\rho }=⟨\text{\hspace{0.17em}⁣}⟨\tau (tr{t}^{-1}):t\in \mathcal{T},r\in R⟩\text{\hspace{0.17em}⁣}{⟩}_{F(S)}$.

The map $\bar{\rho }$ is surjective onto $H^{\mathrm{\prime }}\mathrm{/}(H^{\mathrm{\prime }}\cap N)\cong H$, so [L5] gives $F(S)\mathrm{/}\ker \bar{\rho }\cong H$. Substituting the kernel description from step 3.1 yields $H\cong ⟨S\mid \tau (tr{t}^{-1})\text{ for }t\in \mathcal{T},r\in R⟩$, which is the Reidemeister-Schreier presentation.