2.5.6

a) The relation is defined as $$x \equiv y \iff \exists z \in G \text{ s.t. } x = zyz^{-1}.$$ We have reflexivity because $$x = exe = zxz^{-1} \implies x \equiv x.$$ We have transitivity because $$a \equiv b \text{ and } b \equiv c \implies \exists z_0, z_1 \in G \text{ s.t. } a = z_0bz_0^{-1}, b = z_1cz_1^{-1} \implies$$$$ a = z_0bz_0^{-1} = z_0z_1cz_1^{-1}z_0^{-1} = (z_0z_1)c(z_0z_1)^{-1}$$ and as $$z_0z_1 \in G$$ we finally have $$a \equiv c.$$ We have symmetry because $$x \equiv y \implies x = zyz^{-1} \implies z^{-1}xz = y \implies (z^{-1})x(z^{-1})^{-1} = y \implies y \equiv x.$$

b) Let G be a group, and let A(G) be the set of all elements a in G whose conjugacy class $$C_a$$ consists only of a. Then we claim A(G) = Z(G), where Z(G) is the center of G.

Recall that $$C_a = \{g \in G: \exists b \in G \; s.t. \; g = bab^{-1}\}$$. Let a be in A(G), then $$g \in C_a \implies g = a$$. Given any h in G, observe that $$hah^{-1}$$ is in $$C_a$$ by definition, so that $$hah^{-1} = a$$, thus $$ha = ah$$, hence a is in Z(G).

On the other hand, let a be in Z(G), so that a commutes with every element of G. Let g be in $$C_a$$. Then, $$g = bab^{-1}$$ for some b in G, but because a commutes with b, ab = ba and hence $$bab^{-1} = a $$, so that g = a, thus $$C_a = a$$ and so a is in A(G).

^_^