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By Lagrange's theorem, a stabilizer subgroup of $G$ has order $p^j$ for some $0\le j\le n$. The orbit-stabilizer theorem therefore givesThus an orbit of a group action either has one element or has size divisible by $p$, even if the acted-on set is infinite.

Apply this to the conjugation action. Its singleton orbits of a group action are the elements of the centre of a group, while every other orbit has size divisible by $p$. Partitioning the finite group into these orbits gives the class equation, and henceSince the identity belongs to $Z(G)$, its order is positive and divisible by $p$. ConsequentlyNow suppose that $|Z(G)|=p^{n-1}$. It is then proper, so choose $x\notin Z(G)$. The centralizeris a subgroup: commuting with $x$ is preserved by multiplication and inverses. It contains $Z(G)$ and also $x$, so $|C_G(x)|>|Z(G)|=p^{n-1}$. By Lagrange's theorem, its order is a power of $p$ dividing $p^n$, forcing $C_G(x)=G$. But that would make $x$ central, a contradiction. Therefore

A simple group has no normal subgroup except the identity subgroup and the whole group. To prove the Simplicity of the alternating group A5, we compute its conjugacy classes. Since the sign homomorphism is onto, $|A_5|=5!/2=60$. Its possible nonidentity cycle types are a three-cycle, a product of two disjoint transpositions, and a five-cycle.

There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles. The centralizer in $S_5$ of $(123)$ consists of its three powers and an independent interchange of the two unused letters. Only the three powers are even. Its centralizer in $A_5$ therefore has order three, giving one conjugacy class of size $60/3=20$ by the orbit-stabilizer theorem.

There are $5\cdot 3=15$ products of two disjoint transpositions: choose the fixed letter and then a pairing of the other four letters. For $(12)(34)$, a commuting permutation fixes $5$ and may interchange the two pairs and swap within them, giving a centralizer of order eight in $S_5$. It contains an odd element, $(12)$, so exactly half its elements are even. The centralizer in $A_5$ has order four, and these fifteen elements form one conjugacy class of size $60/4=15$.

Finally, there are $4!=24$ five-cycles. A permutation commuting with a five-cycle is determined by the image of one letter and must be a power of that cycle. All five powers are even, so the centralizer in $A_5$ has order five. Each such conjugacy class has size $60/5=12$, giving two classes of size twelve.

A normal subgroup must be a union of conjugacy classes and must include the identity. Its order therefore has the formThe resulting distinct possibilities areOf these, only $1$ and $60$ divide $60$. Lagrange's theorem rules out every other possible order. Hence

A permutation cycle $(a_1{a}_2\cdots a_p)$ sends $a_1$ to $a_2$, each successive listed point to the next, and $a_p$ to $a_1$, fixing every unlisted point. Its length is $p$. For any permutation $h$,Thus conjugate group elements in the symmetric group have the same cycle length. Conversely, two cycles of the same length are conjugate: send their entries, in cyclic order, to one another and extend this bijection arbitrarily to the unused points. Length-one cycles represent the identity and cause no exception to this conclusion.

An odd-length permutation cycle is an even permutation, since its sign is $(-1{)}^{p-1}$. Let $c,d$ be such cycles on $p+2$ points and choose $h\in S_{p+2}$ with $hc{h}^{-1}=d$. If $h$ is odd, let $t$ be the transposition of the two points unused by the listed cycle $c$. Then $tc=ct$, and $ht$ is even withHence all the $p$-cycles are conjugate within $A_{p+2}$. The two spare letters supply the parity correction; they are the reason the same argument need not work with only one spare letter.

In $A_4$ the answer is no. For example, $(123)$ and $(132)$ are conjugated by the odd transposition $(23)$ in $S_4$. The centralizer of $(123)$ in $S_4$ is exactly its order-three cyclic subgroup: a commuting permutation fixes the unique fixed point $4$ and acts as a power of the cycle on the other three points. All these centralizing elements are even. Every other conjugating element differs from $(23)$ by a centralizing element, so it is also odd. There is no conjugator in $A_4$. Equivalently, the eight three-cycles split into two conjugacy classes of size $12/3=4$, illustrating the alternating conjugacy class splitting criterion.

The conjugacy class of $x$ is its orbit under the conjugation action, and its stabilizer is the centralizer $C_G(x)$. The orbit-stabilizer theorem givesBy Lagrange's theorem, the order of the centralizer divides $p^n=|G|$, so the index is a power of $p$. A conjugacy class has size one exactly when its element belongs to the center of a group.

Separate these singleton classes in the class equation:where the sum takes one representative of each nonsingleton conjugacy class. Every summand is divisible by $p$. Thus $p\mid |Z(G)|$, and the identity ensures $|Z(G)|>0$. Consequently $|Z(G)|\ge p$. This is the nontrivial center of a finite p-group property.

A soluble group, also called a solvable group, has a terminating derived series:For a subgroup $S\le G$, induction gives $S^{(j)}\le G^{(j)}$, so subgroups of soluble groups are soluble. For a surjective group homomorphism $q:G\to Q$, $q(G^{(j)})=Q^{(j)}$, so quotients of soluble groups are soluble. Finally, in a group extension $1\to K\to E\to Q\to 1$, suppose $K^{(r)}=1$ and $Q^{(s)}=1$. Then $E^{(s)}\le K$ and $E^{(s+r)}=1$. Soluble groups are closed under subgroups, quotients and group extensions.

A virtually soluble group contains a soluble group as a finite-index subgroup. The finite-index facts proved in parts (i)–(iii) imply closure under subgroups and quotients: intersect a finite-index soluble subgroup with the chosen subgroup, or take its image under the quotient map.

For group extensions, no finite-generation hypothesis may be inserted. We first establish the finite-index characteristic soluble subgroup lemma. If $K$ is a virtually soluble group, the kernel of its action on the cosets of a finite-index soluble subgroup is a soluble normal subgroup $K_0$ of finite index. Among soluble normal subgroups containing $K_0$, choose $R$ with maximal $|R/K_0|$, possible because $K/K_0$ is finite. If $S$ is any soluble normal subgroup of $K$, then $RS$ is soluble: it is an extension of $R$ by $S/(R\cap S)$. Maximality forces $S\le R$. Thus $R$ is the unique largest soluble normal subgroup of $K$, making it a characteristic subgroup, and it has finite index.

Now suppose $1\to K\to G\stackrel{\pi }{\to }Q\to 1$ has both $K$ and $Q$ virtually soluble. Replace $G$ by the preimage $G_0$ of a finite-index soluble subgroup of $Q$. The subgroup $R$ just constructed is characteristic in $K$ and therefore normal in $G_0$. In $E=G_0/R$, the subgroup $A=K/R$ is a finite normal subgroup, and $E/A$ is a soluble group. The centralizer $C_E(A)$ has finite index in $E$, since conjugation gives a map $E\to \mathrm{Aut}(A)$ with finite image. Its intersection with $A$ is the center of a group $Z(A)$, an abelian group, while its quotient by $Z(A)$ embeds in the soluble group $E/A$. Thus $C_E(A)$ is a soluble group. Its preimage in $G_0$ is an extension by $R$, so it too is soluble and has finite index in $G$. Virtually soluble groups are closed under group extensions.

For the final example take the restricted direct sum of groupswhere $A_5$ is the nonabelian simple group of even permutations on five letters. Every finite collection of elements lies in a product of finitely many finite factors, so $D$ is a locally finite group and hence a torsion group. It cannot contain a nonabelian free group, which is a torsion-free group.

To show that $D$ is not a virtually soluble group, let $H$ be any finite-index subgroup and let $N\le H$ be the kernel of the finite coset action. Each coordinate $A_5$ maps either injectively or trivially into the finite quotient $D/N$, by Simplicity of the alternating group A5. The nontrivial images of distinct factors commute, and each has trivial centre, so any $k$ of them generate a direct product of groups of order $60^k$. Only finitely many such images can occur in a finite quotient. Therefore $N$, and hence $H$, contains a whole coordinate copy of $A_5$, which is not soluble: its nontrivial commutator subgroup is normal and therefore equals $A_5$. No finite-index subgroup of $D$ is soluble.

The identity belongs to the center of a group. If $z,w$ commute with every $g$, then $zw$ does too; rearranging $gz=zg$ also shows that $z^{-1}$ commutes with $g$. Hence $Z(G)$ is a subgroup. Moreover $gz{g}^{-1}=z$ for $z\in Z(G)$, so it is a normal subgroup.

Suppose $|G|=p^n>1$. The conjugation action partitions $G$ into conjugacy classes. The orbit-stabilizer theorem gives class size $[G:C_G(x)]$, where $C_G(x)$ is the centralizer of $x$. By Lagrange's theorem, this size is a power of $p$; it is one exactly when $x$ is central. Thus the class equation has the formIt follows that $p\mid |Z(G)|$. Since the identity is central, $|Z(G)|\ge 1$, and thereforeThis is the nontrivial-centre property of a finite p-group.

The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two double cosetsAn $H$-bimodule-equivariant set map is determined by $u=\Phi (1)$ and $v=\Phi (t)$. Equivariance requires $u\in Z(H)$ andbecause $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi (h)=hu$ and $\Phi (ht{h}^{\prime })=hv{h}^{\prime }$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.

For $n\ge 5$, the center of $H$ and this centralizer are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\dots ,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\ge 3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(12)\}$. The choiceis well defined and satisfies all the printed equivariance conditions, but sends $(34)$ to $(12)$ rather than to the identity. Uniqueness holds for $n\ge 5$; it is false at $n=4$ as printed. Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.

In the matrix expression, $t^a={\lambda }^a/2$ denotes a fundamental generator, whereas the component adjoint covariant derivative uses adjoint generators. With $\mathrm{Tr}(t^d{t}^h)={\delta }^{dh}/2$,Using antisymmetry to write $f^{acd}=f^{dac}$ and $f^{bed}=f^{dbe}$ gives exactly the component mass term. This is the gauge-boson mass matrix from an adjoint scalar.

For ${\Phi }_0=v\mathrm{diag}(1,1,-2)$, the Gell-Mann matrices $t^1,t^2,t^3,t^8$ commute with ${\Phi }_0$. Their gauge bosons are therefore massless. The generators $t^4,t^5,t^6,t^7$ connect the first two eigenspaces to the third; the eigenvalue difference is $3v$. Directly taking the matrix trace giveswith all remaining entries zero. ThusThe factor $1/2$ is the canonical real-vector mass term normalization. The unbroken Lie algebra is $\mathfrak{su}(2)\oplus \mathfrak{u}(1)$, giving the usual description SU(2) × U(1). More precisely, the centralizer of the vacuum expectation value inside $SU(3)$ isThis global quotient refines the adjoint-Higgs breaking of SU(3) to SU(2) and U(1) without changing the local gauge boson spectrum.

Conjugating a cycle merely relabels its entries:Thus conjugate permutations have the same cycle type. Conversely, if $\sigma$ and $\tau$ have the same cycle type, match the entries of each cycle of $\sigma$ bijectively and in cyclic order with those of a cycle of $\tau$ of the same length. Extending these matches to a permutation $g$ gives $g\sigma g^{-1}=\tau$.

Let $C_{S_n}(\sigma )$ be the centralizer of $\sigma$. Its $S_n$-conjugacy class remains one $A_n$-conjugacy class exactly when $C_{S_n}(\sigma )$ contains an odd permutation. Indeed, in that case $C_{A_n}(\sigma )$ has index two in $C_{S_n}(\sigma )$, so the orbit-stabilizer theorem givesIf the centralizer contains only even permutations, the $A_n$ class has half the size and the $S_n$ class splits into two $A_n$ classes.

The even cycle types in $S_5$ areThe classes of the identity, a three-cycle, and a double transposition do not split: their centralizers contain an odd permutation. The centralizer of a five-cycle is its cyclic subgroup of order five, which lies in $A_5$, so that class splits in two. Therefore

For $t\in \mathfrak{t}$, the root-space decomposition gives its centralizerEvery root space is one-dimensional, so this centralizer has the minimum possible dimension $\ell =\dim \mathfrak{t}$ exactly when no summand on the right occurs. By the Regular element criterion in a Cartan subalgebra,

Let $D$ act on the group algebra $kG$ by conjugation. Its fixed-point algebra isand the centralizer $C_G(D)$ consists of the elements of $G$ commuting with every element of $D$. The Brauer morphism isThus $\beta$ deletes the coefficients of basis elements outside $C_G(D)$. The nonfixed $D$-orbits have cardinality divisible by $p$, so the usual orbit argument shows that this projection is a unital ring homomorphism. Hence

The decomposition matrix separates into two connected components. The first contains ${\varphi }_1,{\varphi }_2$ and ${\chi }_1,{\chi }_{3A},{\chi }_{3B},{\chi }_4$; it is the principal block $B_0$. The second contains only ${\varphi }_3$ and ${\chi }_5$; since ${\varphi }_3(1)=5$ contains the full 5-part of $|A_5|=60$, this is a defect-zero representation and its block $B_1$ has defect group $1$.

For $D=1$, the normalizer is $N_G(1)=G$, so the Brauer correspondence is the identity and $B_1$ corresponds to itself.

The defect group of the principal block is a Sylow 5-subgroup $P\cong C_5$. There are six Sylow 5-subgroups in $A_5$, so the orbit-stabilizer theorem givesThe centralizer of a 5-cycle in $A_5$ is $P$, and an involution in the normalizer acts on $P$ by inversion. HenceIn characteristic five the simple $kN$-modules are inflated from $N/P\cong C_2$: their Brauer characters are ${\psi }_{+}=(1,1)$ and ${\psi }_{-}=(1,-1)$ on the identity and involution classes. If $1,\epsilon ,{\rho }_1,{\rho }_2$ are the two one-dimensional and two two-dimensional ordinary characters of $D_{10}$, their reductions areThe resulting decomposition matrix is connected, so these characters form the unique 5-block $b_0$ of $N$, with defect group $P$. By the Brauer first main theorem,This is the complete 5-modular blocks of A5 correspondence.

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 p-subgroup is p-radical if it equals the p-core of its normalizer. Every defect group of a block is p-radical by defect groups are centralizer-conjugate Sylow intersections. Being an intersection of arbitrary Sylow subgroups alone is weaker than the centralizer-conjugate statement used in that argument.