Exercise 33, § 7

Let $E = G *_A G'$ be the sum of two groups amalgamated by a common subgroup $A$. Let $P$ (resp. $P'$) be a subset of $G$ (resp. of $G'$) satisfying condition (A) of I, p. 80. Let $n$ be an integer $\geq 1$. We denote by $L_n$ the set of the $x \in E$ whose reduced decomposition is of length $\leq 2n - 1$. We denote by $M_n$ the set of the $x \in E$ of the form $ap_1 p'_1 p_2 p'_2 \ldots p_n p'_n$, with $a \in A, p_i \in P - \{e\}, p'_i \in P' - \{e\}$; analogously, we denote by $M'_n$ the set of elements of the form $ap'_1 p_1 p'_2 p_2 \ldots p'_n p_n$, where $a, p_i, p'_i$ satisfy the same conditions. (a) Now prove that $L_n, M_n$ and $M'_n$ are disjoint and that their union is the set of the elements of $E$ whose reduced decomposition is of length $\leq 2n$. (b) Construct a bijection $\varepsilon$ of $M_n$ onto $M'_n$ such that $\varepsilon(ax) = a \varepsilon(x)$ if $a \in A, x \in M_n$. (c) Let $X_n = L_n \cup M_n$ and $X'_n = L_n \cup M'_n$; now prove that one has $G.X_n = X_n$ and $G'.X'_n = X'_n$. (d) We extend $\varepsilon$ to a bijection of $X_n$ onto $X'_n$ by putting $\varepsilon(x) = x$ if $x \in L_n$. Now prove that one can make $E$ operate on $X_n$ in such a way that $g(x) = gx$ if $g \in G, x \in X_n$ and $g'(x) = \varepsilon^{-1}(g'\varepsilon(x))$ if $g' \in G', x \in X_n$. Now prove that, if $g \in X_n$, one has $g(e) = g$. (e) Let $\tau_n : E \to \mathfrak{S}_{X_n}$ be the homomorphism defined by the action of $E$ on $X_n$ described above, and let $R_n$ be the kernel of $\tau_n$. Now prove that $R_n \cap X_n = \{e\}$; deduce that the intersection of the $R_n$ is reduced to $e$. (f) When $G$ and $G'$ are finite, now prove that $E$ is residually finite (cf. I, p. 129, exerc. 5). (Observe that the $R_n$ are then subgroups of finite index.)