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Conjugation by a permutation merely permutes the transpositions. Their sum $C$ is therefore a conjugacy class sum and belongs to the center of an associative algebra $\mathbb{C}{S}_n$. Since the complex Specht module is an irreducible representation, Schur lemma says that $C$ acts on it as a scalar, say $c(\lambda )$.

Let $P_{\bullet }\to \mathbb{Z}$ be a projective resolution of the trivial $\mathbb{Z}G$-module. The projective-resolution definition of group cohomology isThis is independent, up to a natural isomorphism, of the chosen projective resolution.

The degreewise natural isomorphismscommute with the coboundary maps. Taking cohomology proves that group cohomology commutes with finite direct sums:

Now restrict $P_{\bullet }$ from $G$ to a subgroup $K$. The group ring $\mathbb{Z}G$ is free as a $\mathbb{Z}K$-module, so restriction carries free modules to free modules and projective modules to projective modules. Thus the restricted complex is a projective resolution of the trivial $\mathbb{Z}K$-module. For the coinduced module $Y={\mathrm{Hom}}_{\mathbb{Z}K}(\mathbb{Z}G,X)$, the Hom functor adjunction for a coinduced module gives an isomorphism of cochain complexesExplicitly, a map $F$ is sent to $p\mapsto F(p)(1)$; the inverse sends a $\mathbb{Z}K$-linear map $a$ to $p\mapsto (r\mapsto a(rp))$. Taking cohomology proves Shapiro's lemma:

For the conjugation module of a group ring $N=\mathbb{Z}{G}_{\mathrm{conj}}$, the basis $G$ is the disjoint union of its conjugacy classes. Hence $N$ is the direct sum of the integral permutation modules on those classes. The class of a representative $g_i$ is the transitive $G$-set $G/C_G(g_i)$, where $C_G(g_i)$ is its centralizer. Since $G$ is finite, this permutation module is both induced and coinduced from the trivial $C_G(g_i)$-module $\mathbb{Z}$. Applying group cohomology commutes with finite direct sums and Shapiro's lemma yields the group cohomology of a conjugation module:

The symmetric group $S_3$ has three conjugacy classes, represented by the identity, a transposition, and a three-cycle. Their centralizers are respectivelyFor any finite group $L$ acting trivially on $\mathbb{Z}$,because a group homomorphism sends an element of finite order to an element of finite order, while the additive group of the integers contains no nonzero torsion elements. ThereforeThe periodic resolution of a finite cyclic group alternates the maps $t-1$ and $1+t+\cdots +t^{m-1}$. After applying ${\mathrm{Hom}}_{\mathbb{Z}{C}_m}(-,\mathbb{Z})$ with the trivial action, these become alternately zero and multiplication by $m$, provingCombining this calculation with the supplied $H^2(S_3,\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$ gives

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 $6$.

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.

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.

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.