Exercise 26, § 4

Let $G, G'$ be two groups, $f : G \to G'$ a mapping such that, for two arbitrary elements $x, y$ of $G$, $f(xy) = f(x)f(y)$ or $f(xy) = f(y)f(x)$. It is proposed to prove that $f$ is a homomorphism of $G$ into $G'$ or a homomorphism of $G$ into the opposite group ${G'}^0$ (in other words, either $f(xy) = f(x)f(y)$ for every ordered pair $(x, y)$ or $f(xy) = f(y)f(x)$ for every ordered pair $(x, y)$).

(a) Show that the set $N = f^{-1}(e')$ ($e'$ the identity element of $G'$) is a normal subgroup of $G$ and that $f$ factors into $G \xrightarrow{p} G/N \xrightarrow{g} G'$, where $g$ is injective. Attention may therefore be confined to the case where $f$ is injective, which will be assumed throughout what follows.

(b) Show that if $xy = yx$ then $f(x)f(y) = f(y)f(x)$ (consider $f(x^2y)$ and $f(x^2y^2)$, expressing them in several ways).

(c) Show that if $f(xy) = f(x)f(y)$ then $f(yx) = f(y)f(x)$. (Show with the aid of (b) that attention may be confined to the case where $xy \neq yx$.)

(d) Show that $f(xyx) = f(x)f(y)f(x)$. (Use (a) or (b) according to whether $xy = yx$ or $xy \neq yx$; in the second case, consider $f(x^2y)$ and $f(yx^2)$.)

(e) Let $A$ be the set of $x \in G$ such that $f(xy) = f(x)f(y)$ for all $y \in G$ and $B$ the set of $x \in G$ such that $f(xy) = f(y)f(x)$ for all $y \in G$. Show that $A \neq G$, $B \neq G$ and $A \cup B = G$ is an impossible situation. (By virtue of (b) there would then exist $x, y$ in $A$ such that $f(xy) = f(x)f(y) \neq f(y)f(x)$ and $u, v$ in $B$ such that $f(uv) = f(v)f(u) \neq f(u)f(v)$. Deduce from this a contradiction by considering successively the two possibilities $xu \in A, xu \in B$; in the first case consider $f(xuv)$ and in the second $f(yxu)$.)

The rest of the proof consists of proving that $A \cup B \neq G$ is impossible, arguing by reductio ad absurdum. If $A \cup B \neq G$, there exist $a, b, c$ in $G$ such that

(f) Show that $f(c)f(a)f(b) = f(b)f(a)f(c)$. (Consider two cases according to whether $bc \neq cb$ or $bc = cb$. In the first case, consider $f(bac)$; in the second, use (b) and (c) and consider $f(abc), f(bca)$ and $f(bac)$.) (g) Consider $f(abac)$ and obtain a contradiction.