# Definition Given a [[group]] $G$, a [[Group Element]] $b$ conjugate to $a$ is defined as: $ b = g a g^{-1} $ for some arbitrary $g \in G$. The two elements are said to be conjugate to each other. This is an [[Equivalence Relation]] whose [[Equivalence Class|equivalence classes]] are called conjugacy classes. Every element of the group belongs to precisely one conjugacy class, and the classes $Cl(a)$ and $Cl(b)$ are equal if and only if $a$ and $b$ are conjugate, and disjoint otherwise. The conjugacy class is defined is $ Cl(a) = \left\{ g a g^{-1} | g \in G \right\} $ In physics, each conjugacy physicall corresponds to a different kind of symmetry (e.g. rotation by $\pi$ about equivalent two-fold axes, rotation by $2\pi/3$ about equivalent three-fold axes, etc.). The identity element is always in its own conjugacy class and it is the only one that forms a group. # Abelian Groups The elements of an [[Abelian group]] are all self-conjugate since $g a g^{-1} = a$ and we have $Cl(a) = {a}$. An Abelian group has as many classes as elements. # Theorem 1 If $b$ is conjugate to $a$ and $c$ is conjugate to $b$ then $c$ is conjugate to $a$ ### Proof $ \begin{align} b &= g a g^{-1}, \\ c &= h b h^{-1} = hga g^{-1} h^{-1} = hg a (hg)^{-1} \end{align} $ # Theorem 2 ^7674f5 If $a$ and $b$ are conjugate, then so are $a^k$ and $b^k$. ### Proof $ b^k = (g a g^{-1})(g a g^{-1})(g a g^{-1}) \ldots (g a g^{-1}) = g a^k g^{-1} $ # Theorem 3 All elements of the same conjugacy class have the same [[Group Element Order|order]]. ### Proof Assume $a$ has order $n$. We know from the [[Conjugacy Class#^7674f5|previous theorem]] that $a^n$ and $b^n$ are conjugate if $a$ and $b$ are conjugate. Then: $ b^n = g a^n g^{-1} = g e g^{-1} = e $ and $a$ and $b$ have the same order. # Theorem 4 The [[Character of a Representation|character]] for each [[Group Element|element]] in a conjugacy class is the same. ### Proof Let $a$ and $b$ be elements in the same conjugacy class, i.e., $ a = g^{-1} b g $ where $g \in G$. Choose a representation where each element can be represented by a [[unitary operator|unitary matrix]] $D$, which is always possible for most groups of interest according to the theorem on [[Theorem on the Unitarity of Representations]]. $ D(a) = D(g^{-1})D(b)D(g) = D^{-1}(g)D(b)D(g) $ Thus: $ \chi(a) = \text{trace}[D(a)] = \text{trace}[D^{-1}(g)D(b)D(g)] = \text{trace}[D(b)] = \chi(b) \qquad \square $ and the characters of $a$ and $b$, two elements of the same conjugacy class, are the same. ### Consequences This theorem is quite important. As stated earlier, elements in the same conjugacy class represent the same type of symmetry, and now we know that elements with the same character also represent the same type of symmetry. This was called the "great beauty of character" by van Vleck. This can allow us to test, e.g., whether a pair of two-fold symmetry axes are equivalent or not.