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A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

For a finite group with trivial centre of a group and a prime divisor $p$ of its order, some nonidentity conjugacy class has size coprime to $p$. Otherwise the class equation would give $|G|\equiv 1(\mathrm{mod}p)$. Primality is essential: in the symmetric group $S_3$, neither nonidentity class size is coprime to the composite divisor $6$.

The special linear group $S{L}_2(5)$ has $24$ nonzero first columns and five second columns giving determinant one, hence order $120$; its centre is $\{I,-I\}$. In the quotient by this centre, the nonidentity conjugacy classes have sizes $15,20,12,12$, corresponding to projective element orders $2,3,5,5$. These arise from trace zero, paired traces $\pm 1$, and paired unipotent traces $\pm 2$. No proper union of these classes with the identity has size dividing $60$, so the quotient is simple. The simple groups with order a power of two times fifteen result identifies it with $A_5$.

A conjugacy class is real when it is stable under inversion of conjugacy classes. Equivalently, one, and therefore every, member is conjugate to its inverse. Members need not themselves be involutions.

The rotations in three dimensions preserving a dodecahedron form a group of order $60$. There are $24$ nonidentity rotations about the six axes through opposite face centres, $20$ about the ten axes through opposite vertices, and $15$ half-turns about the fifteen axes through opposite edge midpoints. Its nonidentity conjugacy classes have sizes $12,12,20,15$: the centralizer of a rotation of order five or three consists of rotations about its oriented axis and has order five or three, respectively, while all edge half-turns are conjugate. A normal subgroup is a union of whole conjugacy classes containing the identity. No proper nontrivial sum of these class sizes plus one divides $60$, proving that it is a simple group. The centralizers of the fifteen involutions are five distinct Klein four-groups, each containing three involutions. Conjugation permutes these five subgroups nontrivially. The kernel is a normal subgroup, so simplicity makes this nontrivial action injective. Its sign map to a group of order two must be trivial, since simplicity would otherwise force an impossible injection of sixty elements into two. Thus its image lies in the alternating group $A_5$, and equality of orders makes the image all of $A_5$.

The orientation-preserving rotational symmetry group of a regular icosahedron has order sixty. Its conjugacy classes have sizes $1,15,20,12,12$: identity; half-turns about opposite-edge axes; rotations through $\pm 2\pi /3$ about opposite-face axes; and rotations through $\pm 2\pi /5$ and $\pm 4\pi /5$ about opposite-vertex axes, respectively. Conjugation carries an axis to its rotated axis and preserves the rotation angle. The transitive group action on edges, faces and vertices gives the indicated classes, with reversal of the axis identifying the two signs. A normal subgroup must contain whole conjugacy classes; none of the proper nonidentity sums of these class sizes is a divisor of sixty. Thus this group is a simple group.

A shortest conjugacy representative is a word having the fewest written letters among all words representing elements of a fixed conjugacy class. This minimum equals the least group word length of an element in the class; a longer spelling of that element is not itself a shortest representative. Such a word exists because lengths are nonnegative integers. It is freely and cyclically reduced: free cancellation shortens the same representative, while removing mutually inverse first and last letters shortens a conjugate. Every cyclic rotation has the same minimal length.

In the complex symmetric-group algebra, normalize a Young symmetrizer by $e_t=h_t/H_{\lambda }$. The row-column collision lemma proves $e_{t}A{e}_t=\mathbb{C}{e}_t$, so $e_t$ is a primitive idempotent and $A{e}_t$ is an irreducible left module. Different shapes are separated by the vanishing corner $h_{\lambda }A{h}_{\mu }=0$ for $\lambda >\mu$. Counting conjugacy classes then proves that these modules exhaust the simple modules.

For a finite subgroup $U$ of a group $T$ of Riemannian isometries of a closed manifold $N$, assume $U$ acts freely and define $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$. Invariance of the Riemannian heat kernel and change of variables show that $I_t$ is a class function on $T$. For $U\le T$, the heat kernel on a finite isometric quotient and integration over a fundamental domain give the displayed heat trace formula. Grouping by conjugacy classes proves equality for Gassmann equivalent subgroups. More generally the same argument allows fixed points of $T$ outside the freely acting subgroups; only the quotient under consideration must be smooth.

Suppose a finite group $G$ acts by Riemannian isometries on a compact Riemannian manifold $M$, and Gassmann equivalent subgroups $H_1,H_2$ act freely. The quotient manifolds $H_i\backslash M$ have equal Laplace-Beltrami operator spectra. For an eigenfunction space $E$, the quotient multiplicity is $|H_i{|}^{-1}{\sum }_{h\in H_i}\mathrm{tr}(h|E)$, and the equality follows by summing its character of a representation over conjugacy classes. Nonconjugacy inside $G$ alone does not guarantee that the quotients are nonisometric.

A permutation of order two is a product of $k$ disjoint transpositions. Its conjugacy class has the displayed size. For $k\ge 2$, equality with the transposition class requires $(n-2)!/(n-2k)!=2^{k-1}k!$. The left side grows strictly with $n$. At $n=2k$, its ratio to the right side is one only for $k=3$; at $k=2$ it jumps from below to above one between $n=4$ and $n=5$, and for $k\ge 4$ it is already greater than one. Thus degree six is the only possible exceptional degree for an group automorphism not preserving transpositions.

The proper rotational tetrahedral symmetry group is $A_4$, acting by even permutations of the vertices. Its conjugacy classes consist of the identity, the three half-turns about opposite-edge midpoint axes, four rotations through $+2\pi /3$ about outward-pointing vertex axes, and their four inverses. A 3-cycle has a three-element centralizer, so its class has size $12/3=4$. Conjugating it to its inverse would require an odd permutation on its three moved vertices, so the two orientations form different classes in $A_4$ even though they lie in one class in $S_4$.

In a finite Dehn presentation, every nonidentity finite-order element has a shortest conjugacy representative of length at most half the length of some relator. Indeed a positive power contains a Dehn segment. If the representative had length at least $k=\lfloor |r|/2\rfloor +1$, the first $k$ letters of that segment would fit into a cyclic rotation and could be replaced by fewer letters, contradicting minimality. Bounded relator lengths and a finite alphabet therefore give finitely many conjugacy classes of finite-order elements. With no relators the group is free and torsion-free, so only the identity class remains.

If an infinite group has a single nonidentity conjugacy class, it is a torsion-free group. Otherwise every nonidentity element has the same finite order $n$. Powers show $n$ is prime. For $n\ge 3$, choose $xg{x}^{-1}=g^2$; then $x^n=1$ gives $g=g^{2^n}$, impossible by Fermat's little theorem. For $n=2$, the group has exponent two and is abelian, so every conjugacy class is a singleton and the hypothesis would force the group to have only two elements. Infinitude excludes this final case.

For a finite field of odd characteristic and nonzero $a,b$, let $U_a=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An arbitrary determinant-one matrix $P=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ satisfies $P{U}_a=U_{b}P$ only if $t=0$ and $ar=bu$. The determinant condition $ru=1$ then gives $b/a=r^2$. Conversely $\mathrm{diag}(r,r^{-1})U_a\mathrm{diag}(r,r^{-1}{)}^{-1}=U_{a{r}^2}$. Thus nonzero upper unipotent matrices fall into two conjugacy classes under this square-class criterion. For example, $U_1$ and $U_3$ are conjugate over ${\mathbb{F}}_{11}$ because $5^2=3$, but not over ${\mathbb{F}}_5$. The criterion compares these particular upper unipotent representatives; it does not classify all determinant-one matrices.

A Von Neumann algebra whose center consists exactly of scalar multiples of the identity. A group von Neumann algebra is a factor exactly when every nonidentity conjugacy class is infinite. A central operator has a square-summable coefficient vector constant on those classes; an infinite class must therefore have coefficient zero. Conversely, the sum of left translations over a finite nonidentity class is a nonscalar central operator.