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Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

We use the core-quotient bijection for partitions: an ordered $p$-tuple $({\nu }_0,\dots ,{\nu }_{p-1})$ and a fixed $p$-core determine a unique partition of an integer, withThe irreducible characters in both $B$ and $C$ therefore correspond to the same ordered tuples of total size $w$. Replacing the empty core of a partition by $\gamma$ while keeping the quotient of a partition fixed gives a bijection ${\chi }^{\lambda }\mapsto {\chi }^{\mu }$ from $B$ to $C$. In particular $|B|=|C|$.

For such a pair, all levels of the core tower of a partition above level zero agree. At level zero, their sizes are $0$ and $m$, respectively. Put $D={\sum }_{r\ge 1}|T^C(\lambda {)}_r|={\sum }_{r\ge 1}|T^C(\mu {)}_r|$. When $0\le m<p$, the base-p digit sum satisfiesThe P-adic valuation of a symmetric-group character degree from the core tower then givesThus the same bijection preserves degree coprimality to $p$ and restricts to a bijection between the indicated sets of irreducible characters of degree coprime to p. ConsequentlyFor $m=0$, take $\gamma$ to be the empty partition; the construction becomes the identity.

By character orthogonality, the squared multiplicities are the character inner product norm of the restriction of a character:Since $\chi$ is an irreducible character, its norm on $G$ is one, and henceThe omitted summands are nonnegative, so the character restriction norm bound is an equality precisely when

Work with the selected transitive orbit $X\cong G/H$. By Maschke's theorem, decompose its permutation character aswhere the ${\chi }_j$ are distinct nontrivial irreducible characters. Character orthogonality givesConsequently the augmentation summand $V$ is irreducible exactly when this inner product equals $2$.

The double cosets $H\backslash G/H$ are the $H$-orbits on $G/H$. Because $G$ acts transitively on $X$, these in turn correspond to the $G$-orbits on $X\times X$: move the first coordinate to the base point $H$, after which its stabilizer is $H$. One orbit is the diagonal. There are exactly two orbits precisely when $H$ is transitive on $X\setminus \{H\}$, equivalently when the action is two-transitive. Part (ii) therefore provesThis is the irreducible augmentation criterion for a transitive group action.

Finally, transitivity implies that the fixed-vector space $\mathbb{C}[X{]}^G$ consists only of constant vectors and has dimension one. If the nonzero representation $V$ were a trivial representation, it would contribute another fixed vector, a contradiction. Thus $V$ is not the trivial representation.

Burnside lemma says that the number of orbits of a finite group $G$ on a finite set $X$ isIndeed, double-count the set $\{(g,x):gx=x\}$. Counting first by $g$ gives the numerator. Counting first by $x$ gives ${\sum }_x|G_x|$; each orbit $O$ contributes $|O||G_x|=|G|$ by the orbit-stabilizer theorem, proving the formula.

Let $\pi (g)=|X^g|$ be the character of the permutation representation. Thenis, by Burnside's lemma, the number of orbits on $X\times X$. A two-transitive group action has exactly two such orbits: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity also gives $⟨\pi ,1_G⟩=1$. If $\pi =1_G+\sum m_i{\chi }_i$ is its irreducible character decomposition, then $2=1+\sum m_i^2$, so exactly one nontrivial irreducible character occurs, with multiplicity one. Thus the permutation character has precisely two distinct irreducible summands.

Write the symmetry group of a square as $D_4=⟨r,s\mid r^4=s^2=e,srs=r^{-1}⟩$, with $r$ a counterclockwise quarter-turn and $s$ reflection across the horizontal axis. The conjugacy classes of the symmetry group of a square areHere $r^2$ is the half-turn and $r^3$ the clockwise quarter-turn. The two members of the fourth class reflect across the horizontal and vertical axes. Those of the fifth reflect across $y=x$ and $y=-x$. The identity and half-turn are central elements, giving singleton classes. The relations $srs=r^3$, $rs{r}^{-1}=r^{2}s$, and $r(rs)r^{-1}=r^{3}s$ show that each displayed pair consists of conjugate elements. More generally, conjugating $r^{k}s$ by $r$ changes $k$ to $k+2$, and conjugating by $s$ changes $k$ to $-k$; parity is preserved, so the two reflection pairs cannot merge.

In a one-dimensional representation, the relation $srs=r^{-1}$ forces the image of $r$ to equal its inverse. Thus $r$ and $s$ can be assigned independent signs $\alpha ,\beta \in \{1,-1\}$. These four choices give four irreducible characters. The ordinary two-dimensional action on the plane gives one more character: a rotation through $\theta$ has trace $2\cos \theta$, and a reflection has trace zero. Therefore the character table of the symmetry group of a square isFor instance the last row has squared character norm $(4+4)/8=1$, so the planar representation is irreducible; its inner product with each one-dimensional row is zero. All five rows obey character orthogonality. Their squared dimensions sum to $1+1+1+1+4=8$, so they exhaust the irreducible representations of this order-eight group.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

Write the virtual character as $\psi ={\sum }_{\chi }{a}_{\chi }\chi$ with integer coefficients and irreducible characters $\chi$. Character orthogonality gives $⟨\psi ,\psi ⟩={\sum }_{\chi }{a}_{\chi }^2$. Norm one forces exactly one coefficient to be $1$ or $-1$, and positive degree selects the positive sign.

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.

Every finite monomial group is a solvable group. Thus expressing every irreducible character as induction from a linear character imposes a group-theoretic solvability condition.