WEAK CAYLEY TABLE GROUPS

Stephen P. Humphries

Abstract. Let G, H be groups. We study bijections f : G → H which preserve conjugacy classes and satisfy: for all g, g′ ∈ G the elements f(gg′) and f(g)f(g′) are conjugate. These are related to 2-characters of groups, objects originally studied by Frobenius and more recently by K. W. Johnson et. al. If G = H, then any automorphism or anti-automorphism satisfies these conditions. We call these trivial. We show, amongst other things, that any such f : G → G for G being the symmetric group S_{n}, n > 0, dihedral group D_{n}, n ∈ ℕ ∪ {∞} or the free group F_{n}, n ≠ 3, is trivial.

§0 Introduction

There are many approaches to the problem of which properties of a group G characterise the group. For example, it is well-known that the characters of G do not characterise G; however if one extends the notion of character in a way originally considered by Frobenius (see [H1, H2] for a very readable account), to obtain k-characters χ^{(k)} : G^{k} → ℂ, then the 3-characters of a group do determine the (finite) group [HJ] (but the 2-characters do not [JS]). Here, for a character χ, the k-characters are defined inductively by χ^{(1)}(g) = χ(g) and for 1 < k ≤ deg(χ) we have

χ^{(k)}(g_{1}, g_{2}, …, g_{k}) = χ(g_{1})χ^{(k−1)}(g_{2}, …, g_{k})
− χ^{(k−1)}(g_{1}g_{2}, …, g_{k}) − χ^{(k−1)}(g_{2}, g_{1}g_{3}, …, g_{k})
− ⋯ − χ^{(k−1)}(g_{2}, …, g_{1}g_{k}).

For 1 < k ≤ deg(χ) it can be shown that the k-characters determine the (k − 1)-characters. Frobenius was led to this approach to his foundational work in representation theory by considerations of the group determinant, a subject recently resurrected in the work of Johnson and others; see for example [J,JS,JMS,FS].

Returning to our original question we note that the group determinant also determines the group [FS]. The definition is: if x_{g}, g ∈ G, are indeterminates corresponding to the elements of G and M is the matrix (x_{gh^{−1}}), for some ordering of the elements g_{1}, …, g_{n} of G, then the group determinant is det(M) and it has irreducible factors corresponding to the irreducible characters of G (a result of Frobenius [Fr,H1,H2]).

The 2-characters do not characterise the group [JS]; however, as noted above, the 2-characters do determine the ordinary characters. Knowledge of the 2-characters is equivalent to that in the weak Cayley table (WCT), which is the |G| × |G|-matrix with ij^{th} entry equal to the conjugacy class of g_{i}g_{j}. See [JS, p. 1022] or the introduction to [JMS] for an explanantion of this.

Typeset by AMS-TEX

Define a weak Cayley table morphism (WCTM) to be a function f : G → H satisfying the following two conditions:

(i) if g, g′ are conjugate, then so are f(g), f(g′); and (ii) f(gh) is conjugate to f(g)f(h) for all g, h ∈ G.

Here f is called a weak Cayley table isomorphism (WCTI) if f is a bijection. Define G and H to have the same WCT if there is a WCTI f : G → H. The WCT is in fact a strictly stronger invariant than the characters. For example, the dihedral group D_{4} and the quaternion group of order 8 have the same characters, but have different WCTs. We note that there are pairs of nonabelian groups of order p^{3} (p an odd prime) p^{4}, p^{7} and p^{11} which have the same WCT. See [JMS].

Let f : G → H be a WCTM which is onto. It easily follows that

ker(f) = {g ∈ G|f(g) = id_{H}}

is a normal subgroup of G. Let π : G → G/ker(f) be the quotient group homomorphism. Now, since f : G → H is onto, then f induces a WCTI ϕ : G/ker(f) → H defined in the natural way: ϕ(ker(f)g) = f(g). One checks that f = ϕπ. It follows that any WCTM which is onto is the composition of a group epimorphism and a WCTI. In this paper we will thus only consider WCTIs.

If G = H, then the set of all WCTIs f : G → G forms a group which we denote by W(G). Note that automorphisms and anti-automorphisms of G are WCTIs. We will say that W(G) is trivial if W(G) = W_{0}(G) where W_{0}(G) =< Aut(G), I > and I = I_{G} : G → G is defined by I(x) = x^{−1}. We will call I the inverse anti-automorphism. The map I_{G} lies in the centre of W(G) (see Lemma 1.2 (i)). Further, if f : G → H is a WCTI, then W(G) ≅ W(H).

We prove:

Theorem 1.7. Let G_{i}, H_{i}, i = 1, 2, be non-trivial groups with H_{1} indecomposable, not infinite cyclic and such that H_{1} is not isomorphic to a subgroup of G_{1} or G_{2}. Then there are no WCTIs f : G = G_{1} ∗ G_{2} → H = H_{1} ∗ H_{2}.

Corollary 1.8. Let G′, H′, i = 1, 2, be non-trivial, indecomposable groups such H′ is not infinite cyclic and not isomorphic to a subgroup of G′. Then there are no WCTIs f : G = G′ ∗ G′ → H = H′ ∗ H′. □

Theorem 1.9. Let G_{i}, H_{i}, i = 1, 2 be non-trivial, with H_{1}, H_{2} co-Hopfian and freely indecomposable. Suppose that G_{i} ≇ H_{j} for i, j = 1, 2. Then there are no WCTIs f : G = G_{1} ∗ G_{2} → H where H has a conjugacy closed subgroup isomorphic to H_{1} ∗ H_{2}.

Theorem 2.4. Let H_{1}, H_{2} be non-abelian infinite groups. Then W(H_{1} ∗ H_{2}) is trivial.

Theorem 2.11. Let F_{n} be the free group of rank n. Then the group W(F_{n}) is trivial for all n ≠ 3.

Proposition 2.12. Let f : F_{n} → G be a WCTI where G is generated by a set of n elements. Then G ≅ F_{n}.

We also prove a generalisation of Hölder’s theorem:

Theorem 3.1. For n > 0 the group W(S_{n}) is trivial.

Proposition 4.1. For n ∈ ℤ^{+} ∪ {∞} the group W(D_{n}) is trivial.

Theorem 5.5. Let H denote the integral Heisenberg group. Then the group W(H) has a subgroup meeting W_{0}(G) in only the identity which is isomorphic to ℤ^{∞}.

§1. Free Products

Let G be the free product G = G_{1} ∗ G_{2}. By [MKS, Theorem 4.1] any word w ∈ G can be written uniquely as w = a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r} where a_{i} ∈ G_{1} and b_{i} ∈ G_{2} and we can assume that all a_{i}, b_{i} are non-trivial elements except possibly for a_{1} and b_{r}. We will call any of the non-trivial a_{i}, b_{i} subwords of w. We will call a_{1} the first subword of w if a_{1} ≠ id otherwise b_{1} is. We denote this subword by w_{0}. The last subword is similarly defined and denoted by w_{∞}. We will call w an ij word if w_{0} ∈ G_{i} and w_{∞} ∈ G_{j}. If a ∈ G_{i}, then we will write ν(a) = i. We will let λ(w) denote its length as a product of words in G_{1} and G_{2}. Thus we can write w = w_{1} … w_{λ(w)} where w_{i} ∈ G_{j(i)} and j(i) ≠ j(i + 1). If u ∈ G, then we will let ũ denote any cyclically reduced word in the conjugacy class of u having minimal λ length. Let λ̃(u) = λ(ũ) for u ∈ G. The following lemma describes what happens to words under conjugation:

Lemma 1.1. Let u, v, w ∈ G.

(i) If λ(uvu^{−1}) < λ(v) + 2λ(u), then either ν(u_{∞}) = ν(v_{0}) or ν(v_{∞}) = ν(u_{∞}). (ii) λ(uvu^{−1}) ≥ λ(ṽ). (iii) If a ∈ G_{1} ∪ G_{2} and u ∈ G, then λ(ua) ≥ λ(u) − 1 and λ(au) ≥ λ(u) − 1. (iv) λ̃(uwu^{−1}a) ≥ λ(uwu^{−1}) − 1 for u ∈ G and a ∈ G_{1}, a ≠ id.

(v) Let a ∈ G_{1}, b ∈ G_{2} be non-trivial and let r ∈ G be such that

r ∉ (G_{1} ∪ G_{2})(ab)^{k}

for any integer k. Then for m ≠ 0 we have λ(r(ab)^{m}r^{−1}b^{−1}) ≥ 4 and λ̃(r(ab)^{m}r^{−1}b^{−1}) ≥ 3.

Proof. The proofs for (i)-(iv) are straightforward. (v) If ν(r_{∞}) = 1 and r_{∞} ≠ a^{−1}, then λ(rabr^{−1}) ≥ 5 and so by (iii) we have λ(rabr^{−1}b^{−1}) ≥ 4. If ν(r_{∞}) = 1 and r_{∞} = a^{−1}, then r = r_{1}a^{−1} and rabr^{−1}b^{−1} = r_{1}bar_{1}^{−1}b^{−1}. Now r ∉ (G_{1} ∪ G_{2})(ab)^{k} for any integer k implies that ν((r_{1})_{∞}) = 2 and (r_{1})_{∞} ≠ b^{−1}. This gives λ(r_{1}b) ≥ 2, λ(r_{1}ba ≥ 3, λ(r_{1}bar_{1}^{−1}) ≥ 5 and so by (iii) λ(r_{1}bar_{1}^{−1}b^{−1}) ≥ 4.

If ν(r_{∞}) = 2 and r_{∞} ≠ b, then λ(rabr^{−1}) ≥ 5 and so by (iii) λ(rabr^{−1}b^{−1}) ≥ 4. If ν(r_{∞}) = 2 and r_{∞} = b, then r = r_{1}b and so rabr^{−1}b^{−1} = r_{1}bar_{1}^{−1}b^{−1}. Now r ∉ (G_{1} ∪ G_{2})(ab)^{k} for any integer k implies that ν((r_{1})_{∞}) = 1 and (r_{1})_{∞} ≠ a. Thus we again have λ(r_{1}bar_{1}^{−1}b^{−1}) ≥ 4. We have now considered all cases. But λ̃(r(ab)^{m}r^{−1}b^{−1}) ≥ 3 now follows from what we have just proved and (iv). □

We will write ∼ or ∼_{G}to denote the equivalence relation of conjugacy in G.

We consider WCTIs f : G → H and let H_{1} ∗ H_{2} be a subgroup of H which is a non-trivial free product. We assume that H_{1} ∗ H_{2} is conjugacy closed in H, namely that if r, s ∈ H_{1} ∗ H_{2} are conjugate in H, then they are actually conjugate in H_{1} ∗ H_{2}. This is clearly the case if H = H_{1} ∗ H_{2} for example. Let us denote the elements of H_{1} by c_{j} and of H_{2} by d_{j}. We let a_{j} = f^{−1}(c_{j}), b_{j} = f^{−1}(d_{j}).

Lemma 1.2. Let f : G → H be a WCTI. Then

(i) f(id_{G}) = id_{H} and for all x ∈ G we have f(x^{−1}) = f(x)^{−1}; Assume that H_{1} ∗ H_{2} is conjugacy closed in H. Then we have (ii) for all i, j we have either f(a_{i}b_{j}) = f(a_{i})f(b_{j}) or f(a_{i}b_{j}) = f(b_{j})f(a_{i}).

Proof. (i) Let α = f^{−1}(id_{H}). Then f(α^{2}) ∼ f(α)^{2} = id_{H} implies that f(α^{2}) = id_{H}. But f is bijective and so α = α^{2}; this proves the first statement of (i). Next we have id_{h} = f(xx^{−1}) ∼ f(x)f(x^{−1}), which gives the second part of (i).

For (ii) we have f(a_{i}b_{j}) = rf(a_{i})f(b_{j})r^{−1} = rc_{i}d_{j}r^{−1} for some r ∈ H_{1} ∗ H_{2} since H_{1} ∗ H_{2} is conjugacy closed in H. Thus we have

c_{i} = f(a_{i}) = f(a_{i}b_{j}b_{j}^{−1}) = qrc_{i}d_{j}r^{−1}d_{j}^{−1}q^{−1},

where q ∈ H_{1} ∗ H_{2} since H_{1} ∗ H_{2} is conjugacy closed. This is now the situation of Lemma 1 (v) which then shows that we can assume that r ∈ H_{1} ∪H_{2}. If r ∈ H_{1} and r ≠ c_{i}^{−1}, then λ̃(rc_{i}d_{j}r^{−1}d_{j}^{−1}) ≥ 4, a contradiction. Thus if r ∈ H_{1}, then r = c_{i}^{−1} and the lemma follows in this case.

If r ∈ H_{2}, then by (i) we have f((a_{i}b_{j})^{−1}) = rd_{j}^{−1}c_{i}^{−1}r and so

d_{j}^{−1} = f(b_{j}^{−1}) = f((a_{i}b_{j})^{−1}a_{i}) = prd_{j}^{−1}c_{i}^{−1}r^{−1}c_{i}p^{−1},

where p ∈ H_{1} ∗H_{2} since H_{1} ∗H_{2} is conjugacy closed. If r ≠ d_{j}, id_{H}, then by Lemma 1.1 (iv) λ̃(rd_{j}^{−1}c_{i}^{−1}r^{−1}c_{i}) ≥ 4, a contradiction. Thus r = d_{j} or id_{H} and in either case we obtain the desired result. □

If f(a_{i}b_{j}) = f(a_{i})f(b_{j}), then we will say that the pair (a_{i}, b_{j}) preserves orientation. If f(a_{i}b_{j}) = f(b_{j})f(a_{i}), then we will say that the pair (a_{i}, b_{j}) reverses orientation.

Lemma 1.3. If a_{i}, b_{j} ≠ id_{G} and (a_{i}, b_{j}) preserves orientation, then so does any other pair (a_{k}, b_{m}).

Proof. Suppose that (a_{i}, b_{j}) preserves orientation and that (a_{k}, b_{j}) reverses orientation where a_{k} ≠ id. Then we have f(a_{i}b_{j}) = f(a_{i})f(b_{j}) and f(a_{k}b_{j}) = f(b_{j})f(a_{k}). Thus

f(a_{i}(a_{k})^{−1}) = f(a_{i}b_{j}(a_{k}b_{j})^{−1}) ∼ f(a_{i})f(b_{j})f(a_{k})^{−1}f(b_{j})^{−1}.

The left-hand-side of this equation is some c_{p}, which clearly cannot be conjugate to the right-hand-side since f(a_{i}), f(a_{k}) f(b_{j}) ≠ id . It follows that if some (a_{i}, b_{j}) preserves orientation, then for any k all (a_{k}, b_{j}) preserve orientation. A similar argument shows that if some (a_{i}, b_{j}) preserves orientation, then for any k all (a_{i}, b_{k}) preserve orientation. This proves the lemma. □

Composing a WCTI f with the inversion anti-homomorphism I_{H} or with an automorphism of H (or of G) gives another WCTI. Thus we see that the above result allows us to assume (after perhaps composing with I_{H}) that the WCTI f has the property that all (a_{i}, b_{j}) preserve orientation. This we will now assume.

Lemma 1.4. With the above assumptions we have

(i) f(b_{i}a_{j}) = f(b_{i})f(a_{j}) for all a_{i}, b_{j}; (ii) f(a_{i}a_{j}) = f(a_{i})f(a_{j}) for all a_{i}, a_{j}. (iii) f(b_{i}b_{j}) = f(b_{i})f(b_{j}) for all b_{i}, b_{j}.

Proof. (i) follows since by Lemma 1.2 (i) we have

f(b_{i}a_{j}) = f((b_{i}a_{j})^{−1})^{−1} = f(a_{j}^{−1}b_{i}^{−1})^{−1} = (f(a_{j}^{−1})f(b_{i}^{−1}))^{−1} = f(b_{i})f(a_{j}).

For (ii) we have

f(a_{i}a_{j})f(b_{k}) = f(a_{i}a_{j}b_{k}) = f(a_{i}(a_{j}b_{k})) = rf(a_{i})f(a_{j}b_{k})r^{−1}
= rf(a_{i})f(a_{j})f(b_{k})r^{−1},  (1.1)

where r ∈ H_{1} ∗ H_{2}. We now need Lemma 1.5. If c_{i}d_{j} ∼_{H_{1}∗H_{2}} c_{k}d_{m}, with c_{i}, d_{j}, c_{k}, d_{m} ≠ id, then c_{i} = c_{k} and d_{j} = d_{m}.

Proof. Suppose that c_{i}d_{j} = sc_{k}d_{m}s^{−1} where s ∈ H_{1} ∗ H_{2} has smallest λ length. It will suffice to show that s = id_{H}. So assume not. We consider the following cases.

If ν(s_{∞}) = 1 and s_{∞} ≠ c_{k}^{−1}, then λ(sc_{k}d_{m}s^{−1}) ≥ 3, a contradiction. If ν(s_{∞}) = 1 and s_{∞} = c_{k}^{−1}, then s = qc_{k}^{−1} and ν(q_{∞}) = 2. Here we can assume that q_{∞} ≠ d_{m}, since if this were the case, then we could reduce λ(s). Thus we have

c_{i}d_{j} = sc_{k}d_{m}s^{−1} = qd_{m}c_{k}q^{−1}.  (1.2)

From this it is clear that q_{∞}d_{m} is a subword of the right-hand-side, which shows that q_{∞}d_{m} = d_{j}. Also, since c_{k} is a subword of the right-hand-side, then we must have c_{k} = c_{i}. But it is also true that q_{∞}^{−1} = d_{m}d_{j}^{−1} is a subword of the right-hand-side of (1.2); but this now gives a contradiction, even if q_{∞} = id_{H} (which implies that q = id and so c_{i}d_{j} = d_{j}c_{i} by (1.2)).

If ν(s_{∞}) = 2 and s_{∞} ≠ d_{m}, then λ(sc_{k}d_{m}s^{−1}) ≥ 3, a contradiction. If ν(s_{∞}) = 2 and s_{∞} = d_{m}, then s = qd_{m} with ν(q_{∞}) = 1 and we again have (1.2). We similarly see that d_{m} is a subword of the right-hand-side and so d_{m} = d_{j}. Also we get c_{i} = c_{k}q_{∞}^{−1} and proceed to obtain a contradiction as in the above cases. □

Applying Lemma 1.5 to (1.1) concludes the proof of Lemma 1.4 (ii). The proof of Lemma 1.4 (iii) is similar. □

Thus we now know that f(ab) = f(a)f(b), f(ba) = f(b)f(a), f(aa′) = f(a)f(a′) and f(bb′) = f(b)f(b′) for all a, a′ ∈ f^{−1}(H_{1}) and b, b′ ∈ f^{−1}(H_{2}). The following is an easy consequence:

Corollary 1.6. For i = 1, 2 we let F_{i} = f^{−1}(H_{i}). Then the restriction f|_{F_{i}} : F_{i} → H_{i} is an isomorphism of groups. In particular, G contains subgroups isomorphic to H_{1} and H_{2}.

Proof. In view of Lemma 1.4 all we need to note is that since f(xx′) = f(x)f(x′) for all x, x′ ∈ F_{i} it follows that F_{i} is a subgroup of G. □

Recall that a group G is freely indecomposable if it cannot be written as a free product G = G_{1} ∗ G_{2} of non-trivial groups. Further, a group G is said to be co-Hopfian if every monomorphism from G to G is an automorphism. The above allows us to prove the following:

Theorem 1.7. Let G_{i}, H_{i}, i = 1, 2 be non-trivial groups with H_{1} freely indecomposable, not infinite cyclic and such that H_{1} is not isomorphic to a subgroup of G_{1} or G_{2}. Then there are no WCTIs f : G = G_{1} ∗ G_{2} → H = H_{1} ∗ H_{2}.

Proof. We will assume that f : G = G_{1} ∗ G_{2} → H = H_{1} ∗ H_{2} is a WCTI and obtain a contradiction. By Corollary 1.6 we see that F_{1} = f^{−1}(H_{1}) is a subgroup of G isomorphic to H_{1}. But by hypothesis H_{1} is freely indecomposable and not infinite cyclic and so the Kurosh subgroup theorem [MKS, Corollary 4.9.1] shows that F_{1} (hence H_{1}) is isomorphic (in fact conjugate) to a subgroup of either G_{1} or G_{2}. Thus no such WCTIs exists. □.

Corollary 1.8. Let G′, H′, i = 1, 2 be non-trivial, freely indecomposable groups such H′ is not infinite cyclic and not isomorphic to a subgroup of G′. Then there are no WCTIs f : G = G′ ∗ G′ → H = H′ ∗ H′. □.

Theorem 1.9. Let G_{i}, H_{i}, i = 1, 2 be non-trivial, with H_{1}, H_{2} co-Hopfian and freely indecomposable. Suppose that G_{i} ≇ H_{j} for i, j = 1, 2. Then there are no WCTIs f : G = G_{1} ∗ G_{2} → H where H has a conjugacy closed subgroup isomorphic to H_{1} ∗ H_{2}.

Proof. We first note that since all the groups are co-Hopfian, they are not infinite cyclic (since the infinite cyclic group is not co-Hopfian). As in the proof of the Theorem 1.7 we see that H_{i} is isomorphic to a subgroup of either G_{1} or G_{2}. Now (since f^{−1} : H → G is also a WCTI), for i = 1, 2 we similarly see that G_{i} is isomorphic to a subgroup of either H_{1} or H_{2}. By considering the small number of possibilities for the choice of “G_{1} or G_{2} ” and “H_{1} or H_{2} ” in the above we see that either (i) there are i, j = 1, 2 such that G_{i} is isomorphic to a subgroup of H_{j} and H_{j} is isomorphic to a subgroup of G_{i}; or (ii) (up to renumbering the H_{i}) we have monomorphisms

ϕ_{1} : G_{1} → H_{1}, ϕ_{2} : H_{1} → G_{2}, ϕ_{3} : G_{2} → H_{2}, ϕ_{4} : H_{2} → G_{1}.

For (i) let ϕ : G_{i} → H_{j} and ψ : H_{j} → G_{i} be the monomorphisms. Then ϕψ : H_{j} → H_{j} is a monomorphism. Since H_{j} is co-Hopfian it follows that ϕψ is an isomorphism; in particular ϕ is onto and so G_{i} ≅ H_{j}, a contradiction.

For (ii) we similarly note that ϕ_{1}ϕ_{4}ϕ_{3}ϕ_{2} : H_{1} → H_{1} is a monomorphism; whence ϕ_{1} is onto and H_{1} ≅ G_{1}, again giving a contradiction. □

§2 Free products continued

For a group G we let G′ denote the commutator subgroup of G. Now let γ_{1}(G) = G and for n > 1 we let γ_{n}(G) = [γ_{n−1}(G), G]. These are the groups of the lower central series of G.

Lemma 2.1. Let f : G → H be a WCTM. Then f(γ_{n}(G)) ⊂ γ_{n}(H). If f is a WCTI, then the restriction f|_{γ_{n}(G)} : γ_{n}(G) → γ_{n}(H) is a bijection.

Proof. The proof of the first part will be by induction on n ≥ 1, the case n = 1 being trivial. Let g ∈ γ_{n}(G), h ∈ G. Then f(hg^{−1}h^{−1}) = qf(g^{−1})q^{−1} for some q ∈ H and so

f(ghg^{−1}h^{−1}) ∼ f(g)f(hg^{−1}h^{−1}) ∼ f(g)qf(g^{−1})q^{−1} ∼ f(g)qf(g)^{−1}q^{−1} ∈ γ_{n+1}(H),

and so induction gives the first result.

Now if f is a WCTI, then f^{−1} is also a WCTI and so by what we have already proved we have f^{−1}(γ_{n}(H)) ⊂ γ_{n}(G). Thus f(γ_{n}(G)) = γ_{n}(H). This now easily gives the result. □

Proposition 2.2. Let f : G → H be a WCTM. Then for all n > 0 there is an induced map F^{n} : γ_{n}(G)/γ_{n+1}(G) → γ_{n}(H)/γ_{n+1}(H) which is a homomorphism of abelian groups and is an isomorphism if f is a WCTI.

Proof. Let F^{n} : γ_{n}(G)/γ_{n+1}(G) → γ_{n}(H)/γ_{n+1}(H) be defined by

F^{n}(γ_{n+1}(G)g) = γ_{n+1}(H)f(g)

for g ∈ γ_{n}(G). Then by Lemma 2.1 the function F^{n} is well-defined. We will first show that F^{n} is a homomorphism. Let g_{1}, g_{2} ∈ γ_{n}(G); then f(g_{1}g_{2}) = qf(g_{1})f(g_{2})q^{−1} for some q ∈ H and so

F^{n}(g_{1}g_{2}) = γ_{n+1}(H)f(g_{1}g_{2}) = γ_{n+1}(H)qf(g_{1})f(g_{2})q^{−1}

= γ_{n+1}(H)f(g_{1})f(g_{2}) = γ_{n+1}(H)f(g_{1})γ_{n+1}(H)f(g_{2}) = F^{n}(g_{1})F^{n}(g_{2}).

Thus F^{n} is a homomorphism.

Now assume that f is a WCTI. Lemma 2.1 shows that F^{n} : γ_{n}(G)/γ_{n+1}(G) → γ_{n}(H)/γ_{n+1}(H) is a bijection and a WCTM, and so F^{n} is an isomorphism of abelian groups. □

Corollary 2.3. (i) If we have a WCTI f : F_{n} → F_{m} between free groups F_{n}, F_{m} then n = m.

(ii) If G is a perfect group (G = G′) and F : G → H is a WCTI, then H is also a perfect group. □

The following is the main result about free products that we prove in this section:

Theorem 2.4. Let H_{1}, H_{2} be non-abelian infinite groups. Then W(H_{1} ∗ H_{2}) is trivial for all n ≥ 1.

Proof. In this proof we will consider the more general situation where we have a WCT f : G = G_{1} ∗ G_{2} → H_{1} ∗ H_{2}. We will be able to conclude that G ≅ H_{1} ∗ H_{2} and that f is an isomorphism or anti-isomorphism.

From Lemma 1.4 and the remarks preceeding it we obtain the following result:

Proposition 2.5. Let f : K_{1}∗K_{2} → J_{1}∗J_{2} be a WCTI where K_{1}, K_{2}, J_{1}, J_{2} are all non-trivial groups. Then there is ϵ ∈ {0, 1} such that I^{ϵ}f(x_{1}x_{2}) = I^{ϵ}f(x_{1})I^{ϵ}f(x_{2}) for all (x_{1}, x_{2}) ∈ K_{i} × K_{j}, i, j = 1, 2. □

As in §1 we will denote the elements of H_{1} by c_{i}, and of H_{2} by d_{j}; we also let a_{i} = f^{−1}(c_{i}), b_{j} = f^{−1}(d_{j}). We may clearly assume that ϵ = 0.

Lemma 2.6. Let f : G → H_{1} ∗ H_{2} be a WCTI, as above with ϵ = 0. Let α ∈ Aut(H_{1} ∗ H_{2}) be an automorphism which fixes H_{1} and let e_{j} = f^{−1}(α(d_{j})). Then c_{i}α(d_{j}) = f(a_{i}e_{j}) for all i, j.

Proof. Since α fixes the H_{1} factor we can also decompose the free product H_{1} ∗ H_{2} as α(H_{1} ∗H_{2}) = H_{1} ∗α(H_{2}). Thus Proposition 2.5 shows that there is ϵ′ = 0, 1 such that I^{ϵ′}f(x_{i}y_{j}) = I^{ϵ′}f(x_{i})I^{ϵ′}f(y_{j}) for all (x_{i}, y_{j}) ∈ f^{−1}(H_{1})×f^{−1}(H_{1}), f^{−1}(H_{1})× f^{−1}(α(H_{2})), f^{−1}(α(H_{2}))×f^{−1}(H_{1}), f^{−1}(α(H_{2}))×f^{−1}(α(H_{2})). But we have f(a_{1}a_{2}) = f(a_{1})f(a_{2}), and if ϵ′ = 1, then for all a_{1}, a_{2} ∈ f^{−1}H_{1} with c_{i} = f(a_{i}) we have

c_{2}^{−1}c_{1}^{−1} = f(a_{2})^{−1}f(a_{1})^{−1} = f(a_{1}a_{2})^{−1}
= If(a_{1}a_{2}) = If(a_{1})If(a_{2})
= f(a_{1})^{−1}f(a_{2})^{−1} = c_{1}^{−1}c_{2}^{−1},

which shows that H_{1} is an abelian group. This contradiction shows that ϵ′ = 0. The result follows. □

Now for c ∈ H_{1} we let α_{c} ∈ Aut(H_{1} ∗ H_{2}) be the automorphism which is the identity on H_{1} and which on H_{2} is conjugation by c. Thus by Lemma 2.6 we have f^{−1}(c_{i}α_{c}(d_{j})) = a_{i}e_{j} for all i, j. Thus

b_{j}a_{i}^{−1} = f^{−1}(d_{j}c_{i}^{−1}) = f^{−1}(c_{i}^{−1}c_{i}d_{j}c_{i}^{−1}) = f^{−1}(c_{i}^{−1}α_{c}(d_{j})) = a_{i}^{−1}e_{j},

which shows that e_{j} = a_{i}b_{j}a_{i}^{−1} for all i, j. From this we see that f^{−1}(c_{k}c_{i}d_{j}c_{i}^{−1}) = a_{k}a_{i}b_{j}a_{i}^{−1} for all i, j, k, which in turn shows that f^{−1}(c_{k}d_{j}c_{i}^{−1}) = a_{k}b_{j}a_{i}^{−1} for all i, j, k. We similarly have f^{−1}(d_{k}c_{j}d_{i}^{−1}) = b_{k}a_{j}b_{i}^{−1} for all i, j, k.

This is the start for a proof by induction using the following inductive statement: Φ(r), r ≥ 1: for all non-trivial c_{1}, …, c_{r}, c_{r+1} ∈ H_{1} , d_{1}, …, d_{r}, d_{r+1} ∈ H_{2} we have (i) f(a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r}) = c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}; (ii) f(a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r}a_{r+1}) = c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}c_{r+1}; (iii) f(b_{1}a_{1}b_{2}a_{2} … b_{r}a_{r}) = d_{1}c_{1}d_{2}c_{2} … d_{r}c_{r}; (iv) f(b_{1}a_{1}b_{2}a_{2} … b_{r}a_{r}b_{r+1}) = d_{1}c_{1}d_{2}c_{2} … d_{r}c_{r}d_{r+1}.

What we have proved above is really the case r = 1. In what follows we will prove that (i) and (ii) for Φ(r) follow inductively from Φ(r − 1), a symmetric argument working for conditions (iii) and (iv).

Suppose that Φ(r − 1) is true and that f(a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r}) ≠ c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}. Then by Φ(r − 1) we see that

f(a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r}) = f((a_{1}b_{1})(a_{2}b_{2} … ac_{r}b_{r})) = qc_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}q^{−1}

for some q ∈ H_{1} ∗ H_{2} \ {id}. We may assume that no right subword of q is equal to (c_{1}d_{1} … c_{i}d_{i})^{−1} or (c_{1}d_{1} … c_{j}d_{j}c_{j+1})^{−1} with i > r/2 or j ≥ r/2 (else we could replace q by some conjugate (c_{1}d_{1} … c_{r}d_{r})^{u}q(c_{1}d_{1} … c_{r}d_{r})^{−u} which has smaller λ-length).

If q is an i1 word, then we can write q as q = q_{1}q_{2} where

λ(q_{2}c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}) = 2r − λ(q_{2})

and q_{2} is chosen so that λ(q_{2}) is maximal with this property. If q is an i2 word, then we interchange H_{1} and H_{2} and replace c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r} by (c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r})^{−1} etc. in what follows.

We consider: case (i) λ(q_{2}) is even, in which case we have q_{2} = d_{k}^{−1}c_{k}^{−1} … d_{1}^{−1}c_{1}^{−1}; or case (ii) λ(q_{2}) is odd, in which case we have q_{2} = c_{k+1}^{−1}d_{k}^{−1}c_{k}^{−1} … d_{2}^{−1}c_{2}^{−1}d_{1}^{−1}c_{1}^{−1}.

For (i) we have

qc_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}q^{−1} = q_{1}c_{k+1}d_{k+1} … c_{r}d_{r}c_{1}d_{1} … c_{k}d_{k}q_{1}^{−1}.

We consider two subcases: (a) q_{1} ≠ id; and (b) q_{1} = id. If we have (a), then qc_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}q^{−1} is an ii-word for some i = 1, 2.

If i = 1, then for any b ∈ f^{−1}(H_{2}) and d = f(b) we have

f(a_{1}b_{1}a_{2}b_{2} … a_{r−1}b_{r−1}b) = f((a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r})(b_{r}^{−1}a_{r}^{−1}b))
∼ q_{1}c_{k+1}d_{k+1} … c_{r}d_{r}c_{1}d_{1} … c_{k}d_{k}q_{1}^{−1}d_{r}^{−1}c_{r}^{−1}d.

But λ̃(f(a_{1}b_{1}a_{2}b_{2} … a_{r−1}b_{r−1}b)) ≤ 2(r − 1) whereas

λ̃(q_{1}c_{k+1}d_{k+1} … c_{r}d_{r}c_{1}d_{1}c_{2}d_{2} … c_{k}d_{k}q_{1}^{−1}d_{r}^{−1}c_{r}^{−1}d) ≥ 2r + 2.

This is a contradiction.

If i = 2, then

d_{1}c_{2}d_{2} … c_{r}d_{r} = f(b_{1}a_{2}b_{2} … a_{r}b_{r})
= f(a_{1}^{−1}.a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r})
∼ c_{1}^{−1}q_{1}^{−1}c_{k+1}d_{k+1} … c_{r}d_{r}c_{1}d_{1} … c_{k}d_{k}q_{1}^{−1}.

Comparing λ̃ values of the first and last terms in this equation also gives a contradiction and this now proves case (a).

If we have (b), then qc_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}q^{−1} = c_{k+1}d_{k+1} … c_{r}d_{r}c_{1}d_{1}c_{2}d_{2} … c_{k}d_{k}. We will need the following:

Lemma 2.7. Suppose that d_{1}, c_{2}, d_{2}, …, c_{r}, d_{r}, d′_{1}, c′_{2}, d′_{2}, …, c′_{s}, d′_{s} are all non-trivial (with c_{i}, c′_{i} ∈ H_{1} and d_{i}, d′_{i} ∈ H_{2}) and that

c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r} ∼ c_{1}d′_{1}c′_{2}d′_{2} … c′_{s}d′_{s}

for all c_{1} ∈ H_{1}. Then r = s and d_{1} = d′_{1}, c_{2} = c′_{2}, …, c_{r} = c′_{r}, d_{r} = d′_{r}.

Proof. Two such cyclically reduced words in H_{1}∗H_{2} are conjugate if and only if they are cyclic permutes of each other. Thus in our situation we certainly have r = s. Now since H_{1} is infinite there is c_{1} ∈ H_{1} which is distinct from c_{2}, …, c_{r}. Since c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r} ∼ c_{1}d′_{1}c′_{2}d′_{2} … c′_{s}d′_{s} we must now actually have c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r} = c_{1}d′_{1}c′_{2}d′_{2} … c′_{s}d′_{s} from which we deduce that d_{1} = d′_{1}, c_{2} = c′_{2}, …, d_{r} = d′_{r}. □

Now for a ∈ f^{−1}(H_{1}) and c = f(a) we have

cc_{2}d_{2} … c_{r}d_{r} = f(aa_{2}b_{2} … a_{r}b_{r})
= f(ab_{1}^{−1}a_{1}^{−1}a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r})
∼ cd_{1}^{−1}c_{1}^{−1}c_{k+1}d_{k+1}c_{k+2}d_{k+2} … c_{r}d_{r}c_{1}d_{1} … c_{k}d_{k}.

Lemma 2.7 now gives c_{1} = c_{k+1}, d_{1} = d_{k+1} otherwise they do not have the same λ̃-length) and c_{2} = c_{k+2}, d_{2} = d_{k+2}, …, c_{r} = c_{k}, d_{r} = d_{k}, which in turn shows that

f(a_{1}b_{1} … a_{r}b_{r}) = c_{k+1}d_{k+1}c_{k+2}d_{k+2} … c_{r}d_{r} … c_{k}d_{k}
= c_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}.

This concludes case (i).

For case (ii) we have q_{2} = c_{k+1}^{−1}d_{k}^{−1}c_{k}^{−1} … d_{2}^{−1}c_{2}^{−1}d_{1}^{−1}c_{1}^{−1}. Thus

qc_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}q^{−1} = q_{1}d_{k+1} … c_{r}d_{r}c_{1}d_{1}c_{2}d_{2} … c_{k}d_{k}c_{k+1}q_{1}^{−1}.

We again consider the two subcases: (a) q_{1} ≠ id; and (b) q_{1} = id. If (a), then q_{1}d_{k+1} … c_{r}d_{r}c_{1}d_{1}c_{2}d_{2} … c_{k}d_{k}c_{k+1}q_{1}^{−1} is an ii-word for some i = 1, 2.

If i = 1, then for any b ∈ f^{−1}(H_{2}) with d = f(b) we have

f(a_{1}b_{1}a_{2}b_{2} … a_{r−1}b_{r−1}b) = f((a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r})(b_{r}^{−1}a_{r}^{−1}b))

∼ q_{1}d_{k+1} … c_{r}d_{r}c_{1}d_{1}c_{2}d_{2} … c_{k}d_{k}c_{k+1}q_{1}^{−1}d_{r}^{−1}c_{r}^{−1}d.

But λ̃(f(a_{1}b_{1}a_{2}b_{2} … a_{r−1}b_{r−1}b)) = 2(r − 1) whereas

λ̃(q_{1}d_{k+1} … c_{r}d_{r}c_{1}d_{1}c_{2}d_{2} … c_{k}d_{k}c_{k+1}q_{1}^{−1}d_{r}^{−1}c_{r}^{−1}d) ≥ 2r + 2.

This is a contradiction. If i = 2, then we consider

d_{1}c_{2}d_{2} … c_{r}d_{r} = f(b_{1}a_{2}b_{2} … a_{r}b_{r}) ∼ c_{1}^{−1}q_{1}d_{k+1} … c_{r}d_{r}c_{1}d_{1} … c_{k}d_{k}c_{k+1}q_{1}^{−1}

to again obtain a contradiction. This proves case (a).

If we have (b), then

qc_{1}d_{1}c_{2}d_{2} … c_{r}d_{r}q^{−1} = d_{k+1}c_{k+2} … c_{r}d_{r}c_{1}d_{1}c_{2}d_{2} … c_{k}d_{k}c_{k+1}.

Let b ∈ f^{−1}(H_{2}) and d = f(d). Then we have

dc_{2}d_{2} … c_{r}d_{r} = f(ba_{2}b_{2} … a_{r}b_{r}) = f(bb_{1}^{−1}a_{1}^{−1}.a_{1}b_{1}a_{2}b_{2} … a_{r}b_{r})
∼ dd_{1}^{−1}c_{1}^{−1}d_{k+1}c_{k+2} … c_{r}d_{r}c_{1}d_{1} … c_{k}d_{k}c_{k+1}.

This equation gives a contradiction upon comparing the λ̃ values of both sides. This proves (ii) and so completes the proof of the induction statement Φ(r) for all r ≥ 1.

Elements of H_{1} ∗ H_{2} have the form c_{1}d_{1} … c_{r}d_{r}, c_{1}d_{1} … c_{r}d_{r}c_{r+1}, d_{1}c_{1} … d_{r}c_{r} or d_{1}c_{1} … d_{r}c_{r}d_{r+1}. This and the fact that f is a bijection, together with the proof of Φ(r) for all r ≥ 1 shows that f is an isomorphism and that the group G is isomorphic to H_{1} ∗ H_{2}. This concludes the proof of Theorem 2.4. □

Remark 2.8. We remark that in the proof of Theorem 2.4 we never made use of the fact that WCTIs permute conjugacy classes, although we did use the fact that f^{−1} is also a WCTI. This thus gives a stronger result.

Corollary 2.9. If n > 3, then W(F_{n}) is trivial. □

We now consider W(F_{2}). Consider the standard presentation for ℤ^{2} = F_{2}/N where F_{2} =< x, y > and N is the normal closure of xyx^{−1}y^{−1}. Note that by Lemma 2.1 we see that f(N) = N. In fact in general we have the following result:

Lemma 2.10. Let the group G have the presentation < x_{1}, …, x_{n}|r_{1}, …, r_{t} >. Let f ∈ W(< x_{1}, …, x_{n}| >). Then G also has the presentation

< x_{1}, …, x_{n}|f(r_{1}), …, f(r_{t}) > .

Proof. Let N(w_{1}, …, w_{t}) denote the normal closure of w_{1}, …, w_{t} in the free group < x_{1}, …, x_{n}| >. It will suffice to show that if w ∈ N(r_{1}, …, r_{t}), then f(w) ∈ N(f(r_{1}), …, f(r_{t})). This we accomplish by induction on u = u(w) where w can be written as a product of u conjugates of the given relators (or their inverse), i.e. we have w = c_{1}c_{2} … c_{u} where each c_{i} is a conjugate of some r_{j}^{±1}. Note that since c_{i} is a conjugate of a relator r_{j}^{±1}, then f(c_{i}) is a conjugate of f(r_{j}^{±1}). This is the start of the induction (u = 1). Now suppose that w = c_{1}c_{2} … c_{u} and that the result is true for all words w′ with u′ = u(w′) < u. Then

f(w) = f(c_{1}c_{2} … c_{u}) = f((c_{1})(c_{2} … c_{u})) ∼ f(c_{1})f(c_{2} … c_{u})

as required. The Lemma follows. □

Thus we must have ℤ^{2} =< x, y|xyx^{−1}y^{−1} >=< x, y|f(xyx^{−1}y^{−1}) > and in particular the elements xyx^{−1}y^{−1} and f(xyx^{−1}y^{−1}) have the same normal closures in the free group < x, y >. A result of Magnus ([M] or [B, p.50]), in the context of one-relator groups, then tells us that (xyx^{−1}y^{−1})^{δ} and f(xyx^{−1}y^{−1}) have to be conjugates for some choice of δ = ±1.

Now let u = f(x), v = f(y). Then the above shows that

(xyx^{−1}y^{−1})^{δ} ∼ f(xyx^{−1}y^{−1}) ∼ uvu^{−1}v^{−1}.

Now we have the following equivalent statements in the situation where F_{2} =< x, y >:

(1) {x, y} and {a, b} are Neilsen-equivalent;

(2) {a, b} is a set of free generators for < x, y >;

(3) xyx^{−1}y^{−1} and (aba^{−1}b^{−1})^{γ} are conjugate for some γ = ±1. For a proof of these equivalences and the definition of Nielsen equivalence see [MKS p. 165]. Thus we conclude that u, v are the images of x, y under some automorphism of F_{2}. We therefore may now assume that u = x, v = y.

The next goal will be to show that

f(x^{i}y^{j}x^{k}) = x^{i}y^{j}x^{k} and f(y^{i}x^{j}y^{k}) = y^{i}x^{j}y^{k} for all i, j, k ∈ ℤ.

This will consist of some more case checking.

First note that that f(x^{2}) = qx^{2}q^{−1} and so we have x = f(x) ∼ x^{−1}qx^{2}q^{−1} and so x^{−1}qx^{2}q^{−1} = pxp^{−1}. We may assume that either q = id or that q ≠ x^{m}. In the latter case one obtains a contradiction by considering the various possibilities for q and p. This starts a proof by induction which enables one to prove f(x^{m}) = x^{m} for all m ∈ ℤ. One similarly shows that one may assume f(y^{m}) = y^{m} for all m ∈ ℤ.

Now consider f(xy) = qxyq^{−1}. Since x = f(x) = f(xyy^{−1}) ∼ qxyq^{−1}y^{−1} and y = f(y) ∼ x^{−1}qxyq^{−1} one can show that we must have either f(xy) = xy or f(xy) = yx. In the latter case let α be the automorphism such that α(x) = x^{−1}, α(y) = y^{−1}. Then composing f with Iα shows that we may assume that f(x) = x, f(y) = y and f(xy) = xy.

The above shows the following: for all α ∈ Aut(F_{2}) there is β = β(α) ∈ Aut(F_{2}) such that

f(α(x^{m})) = β(α(x^{m})), f(α(y^{m})) = β(α(y^{m}))

for all m ∈ ℤ and f(α(xy)) = β(α(xy)). We have chosen β(id) = id.

Now f(xyx^{−1}) = qyq^{−1} and since f(x) and f(xyx^{−1}) must generate F_{2} we must have q = x^{m}. But x^{−1} = f(x^{−1}) ∼ y^{−1}x^{−1}x^{m}yx^{−m} which shows that either m = 0 or m = 1. But f is injective and f(y) = y, so we must have m = 1 and so f(xyx^{−1}) = xyx^{−1}. Similarly we show that f(x^{−1}yx) = x^{−1}yx. By the above paragraph we see that we have now shown that f(xy^{m}x^{−1}) = xy^{m}x^{−1} and f(x^{−1}y^{m}x) = x^{−1}y^{m}x for all m ∈ ℤ. We similarly have f(yx^{m}y^{−1}) = yx^{m}y^{−1} and f(y^{−1}x^{m}y) = y^{−1}x^{m}y for all m ∈ ℤ.

Now as in the above we must have f(x^{2}y) = x^{a}yx^{b} and since f(x^{2}y) ∼ x^{2}y we see that a + b = 2. Now since x^{2}(yxy^{−1}) ∼ f(x^{2}yxy^{−1}) = f((x^{2}y)xy^{−1}) ∼ x^{a}yx^{b}xy^{−1} we see that either a = 2 or a = −1. Similarly we have x ∼ f(x^{2}yy^{−1}x^{−1}) ∼ x^{a}yx^{b}y^{−1}x^{−1} which shows that either b = 0 or a = 1. The only situation satisfying both of these requirements is a = 2, b = 0 i.e.f(x^{2}y) = x^{2}y. Similarly we have f(x^{−2}y) = x^{−2}y, f(yx^{2}) = yx^{2}, f(yx^{−2}) = yx^{−2}, f(y^{2}x) = y^{2}x etc.

Now f(x^{2}yx^{−2}) = x^{a}yx^{−a}; so x^{2} = f(x^{2}) = f(x^{2}yx^{−2}(x^{2}y^{−1})) ∼ x^{a}yx^{−a}(x^{2}y^{−1}), which shows that a = 0 or a = 2. But f is a bijection, so a ≠ 0. Thus f(x^{2}yx^{−2}) = x^{2}yx^{−2}.

The above starts and indicates how a proof by induction goes for proving that f(x^{a}yx^{−a}) = x^{a}yx^{−a}, f(x^{a}y) = x^{a}y, f(xy^{a}) = xy^{a}, … for a ∈ ℤ.

Now we prove that f(x^{a}yx^{b}) = x^{a}yx^{b} for a, b ∈ ℤ. This is already proved if ab = 0. So assume ab ≠ 0. Note that we must have f(x^{a}yx^{b}) = x^{c}yx^{d} where c + d = a + b. Then

x^{a} = f(x^{a}) = f(x^{a}yx^{b}(x^{−b}y^{−1})) ∼ x^{c}yx^{d}x^{−b}y^{−1},

which shows that either c = 0 or d = b. But c ≠ 0 since f is a bijection and ab ≠ 0; thus d = b and so c = a.

Define the following inductive statement analogous to that called Φ(r) in the proof of Theorem 2.4: Φ′(r), r ≥ 1: for all non-trivial i_{1}, …, i_{r}, i_{r+1}, j_{1}, …, j_{r}, j_{r+1} ∈ ℤ we have

(i) f(x^{i_{1}}y^{j_{1}}x^{i_{2}}y^{j_{2}} … x^{i_{r}}y^{j_{r}}) = x^{i_{1}}y^{j_{1}}x^{i_{2}}y^{j_{2}} … x^{i_{r}}y^{j_{r}}; (ii) f(x^{i_{1}}y^{j_{1}}x^{i_{2}}y^{j_{2}} … x^{i_{r}}y^{j_{r}}x^{i_{r+1}}) = x^{i_{1}}y^{j_{1}}x^{i_{2}}y^{j_{2}} … x^{i_{r}}y^{j_{r}}x^{i_{r+1}}; (iii) f(y^{i_{1}}x^{j_{1}}y^{i_{2}}x^{j_{2}} … y^{i_{r}}x^{j_{r}}) = y^{i_{1}}x^{j_{1}}y^{i_{2}}x^{j_{2}} … y^{i_{r}}x^{j_{r}}; (iv) f(y^{i_{1}}x^{j_{1}}y^{i_{2}}x^{j_{2}} … y^{i_{r}}x^{j_{r}}y^{i_{r+1}}) = y^{i_{1}}x^{j_{1}}y^{i_{2}}x^{j_{2}} … y^{i_{r}}x^{j_{r}}y^{i_{r+1}}. In the above we have proved the case r = 1. The proof of this inductive statement is analogous to that of the corresponding statement, called Φ(r), used in the proof of Theorem 2.4. The reader is left to check that the proof works in this situation too. Thus we have proved that (after composing f with an automorphism or an anti-automorphism) we have f(w) = w for all w ∈ F_{2}.

Theorem 2.11. The group W(F_{n}), n ≠ 3, is trivial. □

We note the following consequence of a standard result in combinatorial group theory.

Proposition 2.12. Let f : F_{n} → G be a WCTI where G is generated by a set of n elements. Then G ≅ F_{n}.

Proof. By Lemma 2.1 we have γ_{m}(G)/γ_{m+1}(G) ≅ γ_{m}(F_{n})/γ_{m+1}(F_{n}) for all m ≥ 1. The result now follows from [MKS §5.7, Problem 2, p. 346]. □

§3. Symmetric groups

In this section we will prove

Theorem 3.1. For n ≥ 1 the group W(S_{n}) is trivial.

Proof. The idea of the proof is outlined in the following result:

Lemma 3.2. Suppose that G is a group containing a non-trivial conjugacy class C such that:-

(i) G =< C >; if f ∈ W(G), then there is f′ ∈ W_{0}(G) such that the following three statements are true: (ii) ff′(C) = C and ff′|_{C} = id_{C}; (iii) ff′(C^{2}) = C^{2} and ff′|_{C^{2}} = id|_{C^{2}} where C^{2} is the set of elements of G of length 2 relative to the generating set C; (iv) ff′(C^{3}) = C^{3} and ff′|_{C^{3}} = id|_{C^{3}} where C^{3} is the set of elements of G of length 3 relative to the generating set C;

(v) for all x, y ∈ G with x ∼ y, x ≠ y there is c ∈ C ∪ C^{2} such that xc ≁ yc. Then W(G) is trivial.

Proof. We replace ff′ by f in the proof below. Let λ = λ_{G,C} be the length function for G relative to the generating set C. Let x ∈ G be a shortest word such that y = f(x) ≠ x. Then λ(x) > 2 by (ii) and (iii), and we can write x = x′c where c ∈ C and λ(x′) < λ(x). Then we have f(x′) = x′ and so y = f(x′c) ∼ f(x′)f(c) = x′c = x. Thus by hypothesis (v) there is d ∈ C ∪ C^{2} such that yd ≁ xd. However using hypothesis (iv) we also have

yd ∼ f(xd) = f(x′cd) ∼ f(x′)f(cd) = x′cd = xd.

This contradiction gives the result. □

We will apply this result in the situation where G = S_{n} and C is the set of all transpositions (ij) in S_{n}. We check that we satisfy (i)-(v). A standard result shows that (i) is true.

Proposition 3.3. Let C be the set of all 2-cycles and let f : S_{n} → S_{n}, n > 2, be a WCTI. Then f(C) = C and, up to composing with an element of W_{0}(S_{n}), we have f|C = id_{C}.

Proof. We first prove that f(C) = C. If x ∈ S_{n} has order 2 and y = f(x), then id = f(xx) ∼ y^{2} shows that y also has order 2. Thus if C′ is a conjugacy class of elements of order 2, then so is f(C′). Now in the proof of Hölder’s Theorem (that every automorphism of S_{n}, n ≠ 2, 6 is inner) found in for example [KM, p. 43] it is shown that the other conjugacy classes of elements of order 2 in S_{n} have sizes different from the size of C, except in the case n = 6. Thus if n ≠ 6, then f(C) = C. If n = 6, then by [M] there is an (outer) automorphism of S_{6} which interchanges the two conjugacy classes having the same size as the size of C. Thus composing f with this automorphism if necessary we may assume that f(C) = C. Thus in all cases we may assume that f(C) = C.

The proof of the rest of this result will be by induction. We have to prove the result for S_{3} < S_{n} first. Now by conjugating we may assume that f((12)) = (12). Let f((13)) = (rs). Now

f((132)) = f((12)(13)) ∼ f((12))f((13)) = (12)(rs).

By the above remarks f((132)) does not have order 2. But if {r, s} ∩ {1, 2} = ∅, then f((12)(13)) has order 2, a contradiction. Thus {r, s} ∩ {1, 2} ≠ ∅; however we do not have {r, s} = {1, 2} since f is a bijection. Assume r = 1, 2 and s > 2. By conjugating by (3s) and by (12) (if necessary) we may assume that f((13)) = (13) (while maintaining f((12)) = (12)). Now let f((23)) = (uv). Then the above argument shows that {u, v} ∩ {1, 2} ≠ ∅ and {u, v} ∩ {1, 3} ≠ ∅.

We now show that f((12)(32)) = (123) or (132). First note that f((132)) = f((12)(13)) ∼ (12)(13) and so f((132)) = (ijk). But then f((132)(13)) ∼ (13)(ijk) which shows that 1, 3 ∈ {i, j, k}. Similarly 1, 2 ∈ {i, j, k} and so (ijk) = (123) or

Now if 1 ∈ {u, v}, then f((23)) = (1r), r ≠ 1, 2, 3 and so f((132)(23)) = f((13)) = (13) = (132)(1r) or (123)(1r) which shows that r = 1, 2, 3, each such choice giving a contradiction since f is a bijection. Thus f((23)) = (uv) = (23).

The rest of the proof is by induction : assume that f|S_{m} ∩ C = id_{S_{m}∩C} for m ≥ 3. We will show that f can be changed by an inner automorphism of S_{n} which is the identity on S_{m} such that we have f|_{(S_{m+1}∩C)} = id_{(S_{m+1}∩C)}. If i ≤ m and f((i, m + 1)) = (rs), then for 1 ≤ j ≠ i ≤ m we have f((ij)(i, m + 1)) ∼ (ij)(rs). But this gives {i, j} ∩ {r, s} ≠ ∅ for all such j. The bijectivity of f shows that we can assume that r = i and by conjugating (if necessary) we can obtain s = m + 1. Thus we can now assume f((i, m + 1)) = (i, m + 1). If we have i > 1 and f((i, m + 1)) = (is), then f((1, m + 1)(i, i + 1)) ∼ (1, m + 1)(i, s) ∼ (123) shows that we must have s = m + 1. Thus f((i, m + 1)) = (i, m + 1) for all i ≤ m. This shows that f|_{(S_{m+1}∩C)} = id_{(S_{m+1}∩C)} as required. The proof follows. □

Lemma 3.4. In the situation of the previous result we also have f|_{C^{2}} = id|_{C^{2}}.

Proof. We have already seen that we can assume that f((123)) = (123). Now for r ≠ 1, 2, 3 we have f((12)(2r)) ∼ (12)(2r) = (12r) and so f((12)(2r)) = (ijk). But we have f((2r)) = f((12)(12)(2r)) ∼ (12)(ijk) which shows that 1, 2 ∈ {i, j, k}; similarly we see that 2, r ∈ {i, j, k}. Thus (ijk) = (12r) or (21r). But if (ijk) = (21r), then

f((1r)(23)) = f((12)(2r)(123)) ∼ (21r)(123) = (23r),

a contradiction since (1r)(23) has order 2. Thus f((12)(2r)) = (12)(2r). We can similarly see that f((12)(1r)) = (12)(1r).

Now suppose that r, s ≠ 1, 2. Then f((12)(rs)) ∼ (12)(rs) and so f((12)(rs)) = (ij)(km) where i, j, k, m are distinct and we may assume that i < j, k, m. Now we have f((12)(12)(rs)) ∼ (12)(ij)(km) which is only possible if (ij) = (12). Similarly f((12)(rs)(rs)) ∼ (12)(km)(rs) shows that (km) = (rs). This concludes the proof of the Lemma. □

Lemma 3.5. In the situation of the previous result we also have f|_{C^{3}} = id|_{C^{3}}.

Proof. Let c ∈ C, d ∈ C^{2}. Note that f(dc) = f(cd^{−1})^{−1}; this allows us to only have to consider elements of C^{3} having the form cd where c ∈ C, d ∈ C^{2}. Then f(c) = c, f(d) = d and so f(cd) ∼ f(c)f(d) = cd. We may also assume that cd ∉ C. Further, we may, without loss, assume that d = (1, 2, 3) or d = (1, 2)(3, 4).

First assume that d = (1, 2, 3). Then, again without loss, we assume that c ∈ {(12), (13), (23), (14), (24), (34), (45)}. If c ∈ {(12), (13), (23)}, then cd ∈ C, contrary to the above assumptions. If c = (1, 4), then f(cd) ∼ cd = (1, 2, 3, 4) and so we have f(cd) = (i, j, r, s). This gives

(1, 2, 3) = d = f(d) = f(c.cd) ∼ f(c)f(cd) = (14)(ijrs).

which implies that 1, 4 are adjacent in (i, j, r, s). By considering (134) = f((134)) = f((23)(cd)) ∼ (23)(ijrs) we see that 2, 3 are adjacent in (i, j, r, s). Similarly we see that 1, 2 and 3, 4 are also adjacent in (ijrs). Thus (ijrs) = (1234)^{±1}. Since (345)(1234) = (124)(35) and (345)(4321) = (1532) are not conjugate we see that we must in fact have (ijrs) = (1234). This does the subcase c = (14) and the cases c = (24), (34) are similar. Now assume that c = (45). Then f(cd) = (ijk)(rs) where i, j, k, r, s are all distinct. Since (123) = d = f(d) = f(c.cd) ∼ c(ijk)(rs) we see that (rs) = c = (45). By considering f(e.cd) where e ∈ {(12), (23), (13)} we easily see that (ijk) = (123)^{±1}. But if (ijk) = (132), then we have

(45) = f((45)) = f(d^{−1}.cd) ∼ (132)(132)(45) = (123)(45)

which is a contradiction. Thus in all cases where d = (123) we have f(cd) = cd.

Now assume that d = (12)(34). Then without loss we can assume that c ∈ {(12), (34), (13), (14), (23), (24), (15), (25), (35), (45), (56)}. For c ∈ {(12), (34)} we have cd ∈ C and so f(cd) = cd. Now suppose that c = (13). Then f(cd) ∼ cd = (1234) and so f(cd) = (ijrs) where i, j, r, s are distinct. As in the above paragraph it is easy to show that f(cd) = (1234)^{±1} and if we have f(cd) = (1234)^{−1}, then f((345)(cd)) ≁ (345)f(cd). This does the subcase c = (13) and the subcases c ∈ {(14), (23), (24)} are similar. Now assume that c = (15). Then cd = (125)(34) and so f(cd) = (ijk)(rs) where i, j, k, r, s are distinct. As in the above one shows that (ijk)(rs) = (123)^{±1}(34) and then that the case (ijk)(rs) = (123)^{−1}(34) is not possible. This does this case and all of the cases c = (25), (35), (45) are similar. Lastly, if c = (56), then f(cd) = (ij)(rs)(uv) and one shows that each pair 1, 2; 3, 4; 5, 6 have to be adjacent in (ij)(rs)(uv). Thus f(cd) = cd in all cases. □

The last two results show that we now satisfy conditions (iii) and (iv) of Lemma 3.2.

Proposition 3.6. Let x, y ∈ S_{n} where x ≠ y and x ∼ y. Then there is e = (ij) ∈ C ∪ C^{2} such that xe ≁ ye.

Proof. Suppose first that x has a cycle containing the numbers i ≠ j and that in y the numbers i, j occur in different cycles. Then one checks that x(ij) ≁ y(ij). Thus we may assume that all the numbers in a cycle of x occur in some cycle of y.

Now if 1 ≤ i < j − 1 < j ≤ n, then

(123 … n)(ij) = (i + 1 i + 2 … j)(j + 1 … n12 … i).

This shows that if i, j occur in a cycle c of x and are not adjacent in c, then they are in the same cycle in y and are separated by the same number of elements of the cycle. In particular, if they are adjacent in x, then they are adjacent in y. Thus if the cycle c occurs in x, then in y we must have either the cycle c or the cycle c^{−1}. Assume the latter to be true for some cycle c. Then c has length greater than 2. Now if i, j, k are adjacent in c, then one checks that x(ijk) ≁ y(ijk) and so we are done. □

Proposition 3.6 shows that we now satisfy (v) of Lemma 3.2 and so concludes the proof of Theorem 3.1. □

§4. Dihedral groups

Let D_{n}, n ∈ ℤ^{+} ∪ {∞}, denote the dihedral group of order 2n with presentation < a, x|a^{2}, x^{n}, axa = x^{−1} >.

Proposition 4.1. Every WCTI in W(D_{n}) is trivial.

Proof. Note that an element of D_{n} has the form id, x^{k} or ax^{k}. Note that all n elements of order 2 are of the form ax^{k}, 0 ≤ k < n and two such elements ax^{k}, ax^{m} are conjugate if and only if k ≡ m mod (2, n). All other non-trivial conjugacy classes have the form {x^{k}, x^{−k}}, 0 < k < n/2. We also note that Aut(D_{n}) acts transitively on the set of reflections in D_{n} and that the only other elements of order 2 in D_{n} are central (and so are their own conjugacy class). Thus if f ∈ W(G), then we may assume (by composing f with an automorphism if necessary) that f(a) = a.

Let f ∈ W(G) with f(a) = a. Then by the above we must have f(x^{k}) = x^{α(k)} where α : ℤ/(nℤ) → ℤ/(nℤ) is a bijection and f(ax^{k}) = ax^{β(k)} where β : ℤ/(nℤ) → ℤ/(nℤ) is a bijection. Thus

x^{α(k+m)} = f(x^{k+m}) = f(x^{k}x^{m}) ∼ x^{α(k)}x^{α(m)} = x^{α(k)+α(m)};
x^{α(k)} = f(x^{k}) = f(a.ax^{k}) ∼ a.ax^{β(k)} = x^{β(k)};
x^{α(m−k)} = f(x^{m−k}) = f(ax^{k}ax^{m}) ∼ ax^{β(k)}ax^{β(m)} = x^{β(m)−β(k)}

These equations give the following

α(k + m) = ±(α(k) + α(m));  (4.1)
α(k) = ±β(k);  (4.3)
α(m − k) = ±(β(m) + β(−k))  (4.3)

for all k, m ∈ ℤ/(nℤ). Further α, β satisfy α(0) = 0, β(0) = 0 and α(−k) = −α(k), β(−k) = −β(k) for all k ∈ ℤ/(nℤ).

Now let α(1) = r; then by (4.1) and the fact that f is a bijection we must have (r, n) = 1. Then α(−1) = −r and so α(2) = ±2r. Since α(3) = ±(±2r + r) and α is a bijection we see that f(2) = 2r. Then α(−2) = −2r and so α(3) = ±3r. Since α(4) = ±(±3r + r) and α is a bijection we see that α(3) = 3r. Continuing we see that α(k) = kr for all k and so α is an automorphism.

By (4.1), (4.2) and (4.3) we see that β(k + m) = ±(β(k) + β(m)) for all k, m. Thus a similar arguement to that in the above paragraph shows that β is also an automorphism with β(1) = ±r. If β(1) = −r, then we compose with I_{G} to get β(1) = r. One now checks that f is an automorphism: since elements of D_{n} are of the form x^{k} or ax^{k} we check:

f(x^{k}.x^{m}) = x^{α(k+m)} = x^{α(k)}x^{α(m)} = f(x^{k})f(x^{m});
f(ax^{k}.x^{m}) = ax^{α(k+m)} = ax^{α(k)}x^{α(m)} = f(ax^{k})f(x^{m});
f(x^{k}.ax^{m}) = ax^{α(−k+m)} = x^{α(k)}.ax^{α(m)} = f(x^{k})f(ax^{m});
f(ax^{k}.ax^{m}) = x^{α(−k+m)} = ax^{α(k)}.ax^{α(m)} = f(ax^{k})f(ax^{m}).

This proves the result. □

§5 Examples of groups with non-trivial WCTIs

We first indicate the results of various computer calculations of the groups W(G) and then give a family of groups all of which have non-trivial WCTs.

Example 5.1. Computer calculations show that the groups Q and T_{12} each have trivial WCTs. Computer calculations show that the group W(A_{4}) is a group of order 288, and so has non-trivial WCTs since Aut(A_{4}) ≅ S_{4}. Thus we now have Aut(A_{4}) ≅ Aut(S_{4}), but W(A_{4}) ≇ W(S_{4}). If G_{21} is the non-abelian group of order 21, then G_{21} has non-trivial WCTIs and the group W(G_{21}) has a subgroup of order Example 5.2. If G_{20} is the non-abelian, non-dihedral group of order 20, then G_{20} has non-trivial WCTIs, |Aut(G_{20})| = 20 and the group W(G_{20}) has a subgroup of order 230400.

Let G_{p} denote the subgroup of SL(3, p) of all upper triangular matrices with 1’s on the diagonal. Let E_{ij} be the matrix differing from the identity only in the ij entry, which is a 1. Let a = E_{12}, b = E_{23}. Then G_{p} =< a, b > and we let c = b^{−1}a^{−1}ba. Then we have [a, c] = [b, c] = 1 and an element g of G_{p} can be represented as g = a^{r}b^{s}c^{t} for r, s, t ∈ ℤ/(pℤ). Then we have

a^{r}b^{s}a^{−r} = b^{s}c^{−rs}, b^{s}a^{r}b^{−s} = a^{r}c^{rs}, a^{r}b^{s}c^{t}.a^{u}b^{v}c^{w} = a^{r+u}b^{s+v}c^{t+w+us}.

Further, (a^{r}b^{s}c^{t})^{−1} = a^{−r}b^{−s}c^{−t+rs} and the conjugacy class containg the element g = a^{r}b^{s}c^{t} where (r, s) ≠ (0, 0) consists of all elements of the form a^{r}b^{s}c^{t′}, t′ ∈ ℤ/(pℤ). Let π_{p} : G_{p} → ℤ_{p}^{2} = G_{p}/Z(G_{p}) be the quotient map.

We now look for β : (ℤ/(pℤ))^{3} → ℤ/(pℤ) such that the function

f : G_{p} → G_{p}, f(a^{r}b^{s}c^{t}) = a^{r}b^{s}c^{t+β(r,s,t)}

is a WCTI. Here we also require that for fixed r, s the function t ↦ t + β(r, s, t) is a bijection. The above characterisation of conjugacy classes (and the fact that p is a prime) shows that such a function f satisfies the requirement that f(gg′) ∼ f(g)f(g′) as long as π_{p}(gg′) ≠ (0, 0). If π_{p}(gg′) = (0, 0), then we check that the condition that β needs to satisfy is

β(0, 0, t + w − rs) = β(r, s, t) + β(−r, −s, w)  (5.1)

for all r, s, t, w ∈ ℤ/(pℤ). The following result gives one function β that satisfies these requirements:

Lemma 5.3. Let u, v ∈ ℤ/(pℤ) \ {0}. Define β = β_{u,v} by

β(r, s, t) = \{ \begin{matrix}ur if r ≠ 0; \\ vs if r = 0 \end{matrix}

Then f(a^{r}b^{s}c^{t}) = a^{r}b^{s}c^{t+β(r,s,t)} defines a WCTI.

Proof. We just need to check that (5.1) holds and this is obvious. □

Corollary 5.4. Each group G_{p}, where p is a prime, has a non-trivial WCTI.

Proof. One checks that for non-zero u, v the function β_{u,v} defined in Lemma 5.3 gives a WCTI f = f_{u,v} which is not an automorphism or an anti-automorphism. Thus such an f is non-trivial. □

Let H denote the integral Heisenberg group i.e. the group of all integral uppertriangular 3 × 3-matrices with 1’s on the diagonal. Let a = E_{12}, b = E_{23} be the generators as in the above with c = E_{13} generating the centre.

Theorem 5.5. The group W(H), has a subgroup of non-trivial WCTIs which is isomorphic to ℤ^{∞}.

Proof. To construct such WCTIs we follow the idea used in the proof of Corollary 5.4. So we find a function β(r, s, t) : ℤ^{3} → ℤ such that (i) for fixed r, s the function t ↦ t + β(r, s, t) is a bijection. (ii) for all r, s, t, u, v, w ∈ ℤ we have

β(r, s, t) + β(u, v, w) − β(r + u, s + v, t + w + rs) ∈< r + u, s + v >

where < r + u, s + v > is the subgroup of ℤ generated by r + u and s + v. This will guarantee that f = f(β) : H → H, defined by f(a^{r}b^{s}c^{t}) = a^{r}b^{s}c^{t+β(r,s,t)} is a non-trivial WCTI. Here we are using the fact that in H the elements a^{r}b^{s}c^{t} and a^{u}b^{v}c^{w} are conjugate if and only if r = u, s = v, and t − w ∈< r, s >.

Since H/Z(H) ≅ ℤ^{2} is abelian one can show that any WCTI is a product of some element of GL(2, ℤ) and some element of the above type.

For n ∈ ℤ, n > 0, define the function

β_{n}(r, s, t) = r^{2n+1} + s^{2n+1}.

Then one checks that β_{n} satisfies (i) and (ii) above. The corresponding function f_{n} = f(β_{n}) is thus a WCTI for any n > 0 and one easily checks that it is not a homomorphism or an anti-homomorphism since n > 0. It is also easy to check that for any n > 0 the subgroup < f_{1}, f_{2}, …, f_{n} > is isomorphic to ℤ^{n}, from which it follows that < f_{1}, f_{2}, ⋯ >≅ ℤ^{∞}. □

§6 Simple groups

Proposition 6.1. If G is a finite simple group and f : G → H is a WCTI, then H ≅ G.

Proof. This follows from the fact [JS, JMS] that the WCT of G determines the character table of G, together with the result of Chen [C] which says that if G is a simple group and H has the same character table as G, then H ≅ G. This latter result uses the classification of finite simple groups. □

For a group G let T = T (G) be the subgroup of G generated by id_{G} and all elements of the form f(ab)f(b)^{−1}f(a)^{−1} for all f ∈ W(G) \ W_{0}(G) and all a, b ∈ G.

Lemma 6.2. The subgroup T is characteristic in G and T ⊂ [G, G].

Proof. Let f ∈ W(G) \ W_{0}(G) and a, b ∈ G; let ι ∈ Aut(G). Then ιf ∈ W(G) \ W_{0}(G) and we have

ι(f(ab)f(b)^{−1}f(a)^{−1}) = ιf(ab)ιf(b)^{−1}ιf(a)^{−1} ∈ T,

which shows that T is characteristic in G. Since f(ab) is a conjugate of f(a)f(b) we have T ⊂ [G, G]. □

We note that W(G) is trivial if and only if T (G) is the trivial group. We use this to prove:

Corollary 6.3. If G is a simple group, then either W(G) is trivial or T (G) = G.

Proof. Since G is simple and T is characteristic (and so normal) in G we have either T (G) = {id} or T (G) = G. □

Acknowledgement. The author is indebted to Ken Johnson for introducing him to the subject of this paper and for conversations relating to it.

References

[B] Baumslag G., A survey of groups with a simgle defining relation, Proceeding of Groups - St. Andrews 1985 (E.F. Robertson, C.M. Campbell, eds.), London Math. Soc. Lecture Notes Series, vol. 121., 1986, pp. 30-58.

[C] Chen G. Y., A new characterization of finite simple groups, Chinese Science Bulletin 40 (1995), 446-450.

[FS] Formanek E., Sibley D., The group determinant determines the group, Proc. Amer. Math. Soc. 112 (1992), 649-656.

[Fr] Frobenius, G., Über die Primfactoren der Gruppendeterminante, Sber. Preuss. Akad. Wiss. Berlin (1896), 1343-1382.

[H1] Hawkins T., New light on Frobenius’ creation of the theory of group characters, Arch. Hist. Exact Sci. 12 (1974), 217-243.

[H2] Hawkins T., The origins of the theory of group characters, Arch. Hist. Exact Sci. 7 (1971), 142-170.

[HJ] Hoenke H.-J., Johnson K.W., The 3-characters are sufficient for the group determinant, Contemporary Mathematics, vol. 184, 1995, pp. 193-206.

[J] Johnson K.W., On the group determinant, Math. Proc. Camb. Phil. Soc 109 (1991), 299-311.

[JMS] Johnson K.W., Mattarei S., Sehgal S., Weak Cayley tables, Preprint (1997).

[JS] Johnson K.W., Sehgal S., The 2-character table does not determine the group, Proc. Amer. Math. Soc. 119 (1993), 1021-1027.

[KM] Kargapolov M.I., Merzljakov Ju.I., Fundamentals of the Theory of Groups, Springer-Verlag, 1979.

[M] Magnus W., Untersuchungen uber einige undendliche diskontinuierliche Gruppen, Math. Ann. 105 (1931), 52-74.

[MKS] Magnus W., Karrass A., Solitar D., Combinatorial Group Theory, Dover, 1976.