• Let \(G\) be a group and \(\sigma : G \rightarrow G\) be an involution. If \(f : G \rightarrow \mathbb{C}\) is a solution of equation (3.1) (i.e. \(f(xy)f(x\sigma(y)) = f(x)^2 - f(y)^2\)), then \[K_f(G) = N_f(G) = M_f(G)\]

• First let's show that \(K_f(G) = N_f(G)\). Let \(u \in K_f(G)\). Then \(f(u) = 0\) by definition. If we replace \(x\) by \(xu\) and \(y\) by \(x^{-1}\) in (3.1), we have for all \(x \in G\) we find equation (3.28):

\[f(xu)^2 -f(x)^2 \\ =f(xu)^2 - f(x^{-1})^2 \\ =f(xux^{-1})f(xu(\sigma(x^{-1})) = 0\]

The first equality here is by Lemma 3.2 equation (3.5). The final equality here is by the normality of \(K_f(G)\). Thus \(u \in N_f(G)\), so \(K_f(G) \subseteq N_f(G)\).

Next let \(u \in N_f(G)\). Then by definition of \(N_f(G)\), we find that \[f(xu)^2 = f(x)^2\] for all \(x \in G\). If we let \(x = u^{-1}\), then we find \(f(e)^2 = f(u^{-1})^2\). We know from Lemma 3.2 that \(f(e) = 0\), so \(f(u^{-1})^2 = 0\). But again by Lemma 3.2, we know that \(f(u^{-1})^2 = f(u)^2\), so we find that \(f(u)^2 = 0\) which implies that \(f(u) = 0\). Therefore we can conclude that \(u \in K_f(G)\). This proves that \(N_f(G) \subseteq K_f(G)\) so we have proven that \(K_f(G) = N_f(G)\).

• Now we want to show that \(K_f(G) = M_f(G)\). Let \(u \in K_f(G)\). This implies that \(f(u) = 0\). We know, \(K_f(G) = N_f(G)\), so \(f(xu)^2 = f(x)^2\) for all \(x \in G\). Now,

\[f(xu^2)f(xu\sigma(u))\\ = f(xu)^2 - f(u)^2 \\ = f(xu)^2 \\ =f(x)^2 \\ = f(xu\sigma(u))^2\]

The first equality here follows directly from (3.1). The second is because \(f(u) = 0\). The third follows because \(N_f(G) = K_f(G)\). The last follows from equation (3.7) from Lemma 3.2.

Thus we find equation (3.29) stating \[f(xu^2)f(xu\sigma(u)) = f(xu\sigma(u))^2\] for all \(x \in G\). Suppose \(f(xu\sigma(u)) \neq 0\). Then the last equality yields \(f(xu^2) = f(xu\sigma(u))\). Suppose \(f(xu\sigma(u)) =0\). Then \(xu\sigma(u) \in K_f(G)\). By Lemma 3.2, \(u\sigma(u) \in K_f(G)\), so \(x \in K_f(G)\). Since \(u \in K_f(G)\), \(u^2 \in K_f(G)\). So, \(xu^2 \in K_f(G)\), which implies \(f(xu^2) = 0\) and \(f(xu^2) = f(xu\sigma(u))\). So \(K_f(G) \subseteq M_f(G)\).

Let \(u \in M_f(G)\). Then \(f(xu^2) = f(xu\sigma(u))\) for all \(x \in G\). Letting \(x = u^{-1}\), we have \(f(u) = f(\sigma(u))\). By Lemma 3.2, \(f(\sigma(u)) = -f(u)\) and thus we have

\[f(u) = f(\sigma(u)) = -f(u)\]

which yields \(f(u) =0\). So \(u \in K_f(G)\) and \(M_f(G) \subseteq K_f(G)\). Hence \(M_f(G) = K_f(G)\). So we're done.

• QED

Let \(G\) be a 2-divisible group and \(\mathbb{C}\) the field of complex numbers. Let \([G,G]\) be the 2-divisible commutator subgroup of \(G\). Let \(\sigma : G \rightarrow G\) be an involution. If \(f : G \rightarrow \mathbb{C}\) is a solution of (3.1) satisfying \(f(x) = f(xy\sigma(y))\) for all \(x,y \in G\), then \(f\) is a function on the quotient group \(G/[G,G]\).

Replacing \(x\) by \(xu\) and \(y\) by \(x^{-1}\) in (3.1) for \(u \in [G,G]\), we obtain

\[f(xux^{-1})f(xu\sigma(x^{-1})) = f(xu)^2 - f(x^{-1})^2.\]

Since \(u \in [G,G]\) and the latter is a normal subgroup of \(G\), this means that \(xux^{-1} \in [G,G]\). Therfore \(f(xux^{-1}) =0\) and hence we have

\[f(xu)^2 = f(x^{-1})^2\]

for all \(x \in G\) and \(u \in [G,G]\).

By property (3.5), the above equality can be reduced to

\[f(xu)^2 = f(x)^2\]

for all \(x \in G\) and \(u \in [G,G]\). Using Lemma 3.5, we find that

\[f(xu^2) = f(xu\sigma(u)).\]

Since \(f(x) = f(xy\sigma(y))\) for all \(y \in G\), we get that \(f(xu^2) = f(x)\). Since \([G,G]\) is 2 -divisible the previous equality can be written as \(f(xu) = f(x)\). This means that \(f\) is a finction of the quotient group \(G/[G,G]\).

• QED