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The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]

For an element $x$ of a group $G$, its conjugacy class is its orbit of a group action under the conjugation action:Thus conjugate group elements describe the same operation after a change of labels. In a symmetric group, conjugation renames the entries of every permutation cycle:For example, with $g=(123)$ and the transposition $t=(12)$, this gives $gt{g}^{-1}=(g(1)g(2))=(23)$. Products use function composition of permutations, with the rightmost factor applied first. The complete six-element example is conjugacy classes of the symmetric group on three letters.

The conjugacy classes form a partition of a set of $G$, since conjugacy is an equivalence relation. In a symmetric group, two permutations lie in the same conjugacy class precisely when they have the same cycle type: renaming preserves the lengths of permutation cycles, and matching equal-length permutation cycles constructs a renaming in the other direction. This compresses many labelled permutations into a few structural types. See McKernan, Conjugation in Sn, lemma 6.2 and proposition 6.6.

The stabilizer subgroup of $x$ under the conjugation action is its centralizer $C_G(x)$. Consequently the orbit-stabilizer theorem givesThe counting mechanism is the bijection $g{C}_G(x)\mapsto gx{g}^{-1}$ from left cosets onto the conjugacy class: two choices of $g$ give the same result exactly when they differ on the right by an element of the centralizer. A large centralizer means many choices of labels leave the operation unchanged, hence a small conjugacy class. A conjugacy class has size one precisely when its element belongs to the center of a group. The class equation counts these orbits of a group action together.

The symmetric group $S_3$ has exactly six permutations, written using permutation cycles:Here $e$ is the identity element; $(12)$, $(13)$ and $(23)$ are transpositions; and $(123)$ and $(132)$ are permutation cycles of length three. With $t=(12)$ and $r=(123)$, rightmost-first function composition gives $r^2=(132)$ and $r^{-1}=r^2$. The conjugacy classes and representative centralizers are

An explicit check of the conjugation $gt{g}^{-1}$ with $g=r=(123)$ sends $1\mapsto 3\mapsto 3\mapsto 1$, $2\mapsto 1\mapsto 2\mapsto 3$, and $3\mapsto 2\mapsto 1\mapsto 2$, where each chain applies $g^{-1}$, then $t$, then $g$. Thus $1$ is fixed and $2,3$ are exchanged: $gt{g}^{-1}=(23)$. Relabelling also produces $(13)$, so the three transpositions share one conjugacy class. A transposition reverses the length-three permutation cycle, since $tr{t}^{-1}=(213)=(132)=r^{-1}$, putting both length-three permutation cycles in one conjugacy class.

Checking each of the six possible permutations against $gx=xg$ gives the displayed centralizers. Only $e$ has a one-element conjugacy class, so the center of a group here is $Z(S_3)=\{e\}$. The class equation is $6=1+3+2$. The second diagram in centralizer and normalizer uses the same $r$ and $t$ to show how preserving a cyclic subgroup differs from fixing its group generator.

For subgroups $A,B$ write $[A,B]$ for the subgroup generated by their group commutators. A nilpotent group has a central series: a finite chain $1=G_0\le G_1\le \cdots \le G_c=G$ with $[G_i,G]\le G_{i-1}$. Equivalently its upper central series, defined by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, reaches $G$. Indeed every central series term $G_i$ lies in $Z_i$ by induction, and the upper central series itself is central. A third equivalent definition is that the lower central series ${\gamma }_1=G$, ${\gamma }_{i+1}=[{\gamma }_i,G]$ terminates at ${\gamma }_{c+1}=1$: reversing this series gives a central series, and induction along any central series gives ${\gamma }_{c+1}=1$.

The least such $c$ is the nilpotency class. Nontrivial abelian groups have class one. Upper-unitriangular $3\times 3$ matrices have class two: their group commutators lie in the central top-right-entry subgroup, and further group commutators vanish. The same example works over a finite field or over the integers, so nilpotence is not a finiteness property. Subgroups and quotients inherit nilpotence by intersecting or mapping a central series; finite direct products of groups have class at most the largest factor class. Every nilpotent group is soluble, since its derived series is contained successively in the lower central series. The converse fails: $S_3$ is soluble but its center is trivial, so it cannot be a nontrivial nilpotent group. Extensions of nilpotent groups need not be nilpotent, as the extension $C_3◃S_3$ with quotient $C_2$ shows.

To prove the finite classification, first note that every nontrivial finite p-group has nontrivial center. In its class equation, every noncentral conjugacy class has size divisible by $p$, so the center's order is divisible by $p$. Induct on the group order: the smaller p-group $P/Z(P)$ is nilpotent, and lifting its central series and adjoining $Z(P)$ proves $P$ nilpotent. A finite direct product of groups of prime-power-order groups is consequently nilpotent.

Conversely suppose $G$ is finite and nilpotent. We need the normalizer condition for nilpotent groups. For any proper subgroup $H<G$, choose the first upper-central term $Z_i$ not contained in $H$. Then $Z_{i-1}\le H$. Every $z\in Z_i$ commutes with $H$ modulo $Z_{i-1}$, so it normalizes $H$. Choosing $z\notin H$ proves $H<N_G(H)$.

Let $P$ be a Sylow subgroup and put $M=N_G(P)$. Since $P$ is normal in $M$, it is the unique Sylow p-subgroup of $M$. Any element normalizing $M$ must therefore normalize $P$, so $N_G(M)=M$. If $M$ were proper, the normalizer condition would contradict this equality. Hence $N_G(P)=G$: every Sylow subgroup is normal. For Sylow subgroups $P,Q$ belonging to different primes, $[P,Q]\le P\cap Q=1$, so they commute. Their product has order the product of their orders, namely $|G|$, and has trivial intersections between each factor and the product of the others. Multiplication therefore givesThis proves finite nilpotent group decomposition in both directions without assuming the classification in the normalizer argument. It also shows that a finite group is nilpotent exactly when all its Sylow subgroups are normal. The finite hypothesis is essential to the prime-power direct-product statement; an infinite cyclic group is nilpotent but is not such a finite product.

A subnormal series is a finite chain $1=G_r⊴G_{r-1}⊴\cdots ⊴G_0=G$, in which each term is normal in the preceding term. Its factors are the quotient groups $G_i/G_{i+1}$. A normal series of a group imposes the stronger condition that every term be normal in $G$. These conditions should not be conflated: inserting a subgroup normal in an intermediate term need not preserve normality in the whole group. A refinement inserts additional terms, and two series are equivalent when their nontrivial factors can be paired by group isomorphisms.

The basic refinement mechanism is the Zassenhaus butterfly lemma. For $A^{\prime }⊴A$ and $B^{\prime }⊴B$, let $C=A\cap B$ and $D=(A^{\prime }\cap B)(A\cap B^{\prime })$. Both factors of $D$ are normal in $C$, so $D⊴C$. The map from $C$ onto $A^{\prime }C/A^{\prime }(A\cap B^{\prime })$ has kernel $D$: if an element of $C$ equals $a^{\prime }{b}^{\prime }$ with $a^{\prime }\in A^{\prime }$ and $b^{\prime }\in A\cap B^{\prime }$, then $a^{\prime }=c{b}^{\prime -1}\in A^{\prime }\cap B$. Conversely each product in $D$ is in the kernel. The first isomorphism theorem therefore givesThe denominators are normal in their numerators: conjugation by $C$ preserves them, while conjugation by $A^{\prime }$ or $B^{\prime }$ changes their elements only by a factor already in that normal subgroup.

Given two subnormal series $(G_i)$ and $(H_j)$, refine the first by $G_{i+1}(G_i\cap H_j)$, and the second by $H_{j+1}(H_j\cap G_i)$. The endpoints are $G_i,G_{i+1}$ and $H_j,H_{j+1}$ respectively, and the Zassenhaus butterfly lemma pairs the successive factors indexed by $(i,j)$ and $(j,i)$. This proves the Schreier refinement theorem.

A composition series is a strict subnormal series all of whose factors are simple groups. Every finite group has one: choose a maximal proper normal subgroup, repeat inside it, and use decreasing orders to terminate. A simple factor cannot admit a nontrivial refinement. Applying the preceding refinement theorem to two composition series proves the Jordan–Hölder theorem: their lengths and multisets of simple factors agree, up to group isomorphism. The factors are the Jordan–Hölder factors, and their number is the composition length. If $N⊴G$, lifting a composition series of $G/N$ and adjoining one for $N$ gives

These factors do not determine the whole group, because they do not determine how its group extensions are assembled. For example, $C_4$ and the Klein four-group both have two factors $C_2$, but one contains an element of order four and the other does not. The symmetric group $S_3$ has a composition series $1<C_3<S_3$, with factors $C_3,C_2$, whereas the simple group $A_5$ has the single factor $A_5$.

A chief series is a maximal strict normal series of a group; each chief factor is a minimal nontrivial normal subgroup in the appropriate quotient of $G$. Finite chief factors are characteristically simple. To justify their structure, let $N$ be any nontrivial finite characteristically simple group and choose a minimal nontrivial normal subgroup $M$ of $N$. Its distinct images under automorphisms of $N$ are minimal normal subgroups, so distinct ones intersect trivially and commute. Choose a maximal family giving an internal direct product of groups. If any other image were not contained in that product, its intersection with the product would be trivial by minimality, and it could be added, a contradiction. Hence this product is characteristic and equals $N$. Each factor $M$ is simple: a proper nontrivial normal subgroup of $M$ would be normal in the whole direct product $N$, contrary to the choice of $M$. Thus $N$ is a direct product of groups of isomorphic finite simple groups. This applies to chief factors by conjugation from $G$.

Chief and composition factors need not coincide. In $A_4$, the chain $1<V_4<A_4$ is a chief series, since conjugation cycles the three order-two subgroups of $V_4$. A composition series instead inserts any one of those subgroups, giving factors $C_2,C_2,C_3$. Its inserted order-two subgroup is normal in $V_4$ but not in $A_4$. Chief series retain the ambient group's conjugation information that composition series discard.

The derived series gives a canonical criterion for a group to be soluble: put $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic, and each factor is Abelian. A finite group is soluble exactly when this series reaches the identity, equivalently when all its composition factors are cyclic of prime order. For one direction, the derived series supplies Abelian factors, whose simple refinements are cyclic of prime order. Conversely an extension of two soluble groups is soluble: if $(G/N{)}^{(r)}=1$, then $G^{(r)}\subseteq N$; if $N^{(s)}=1$, then $G^{(r+s)}=1$. Induction up a composition series with cyclic prime factors proves the other direction. In a finite soluble group, each minimal normal subgroup is elementary Abelian: its characteristic commutator subgroup must be trivial, its primary components force a single prime, and its characteristic subgroup of $p$th powers must be trivial. Thus every chief factor is a vector space over some ${\mathbb{F}}_p$. In a general finite group the Jordan–Hölder factors can instead be nonabelian simple groups.

The central series demand more than Abelian factors: each factor must be central in the corresponding quotient of the whole group. Define the upper central series by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, and the lower central series by ${\gamma }_1=G$ and ${\gamma }_{i+1}=[{\gamma }_i,G]$. A group is nilpotent exactly when the upper series reaches $G$, equivalently when the lower series reaches the identity. Indeed, for any ascending central series $1=C_0\le \cdots \le C_c=G$, induction gives $C_i\subseteq Z_i$ and ${\gamma }_{c+1-i}\subseteq C_i$; reversing the lower series gives an ascending central series when it terminates. The least such $c$ is the nilpotency class. A central series has Abelian factors, so every nilpotent group is soluble; the converse fails, as $S_3$ is soluble but has trivial group center.

For a finite p-group, the class equation shows that the group center is nontrivial, since every noncentral conjugacy class has size divisible by $p$. Applying the same argument to successive central quotients makes the upper central series reach the group. More generally a finite group is nilpotent exactly when it is the direct product of groups of its Sylow subgroups. Here is the nontrivial direction. A central series gives the normalizer condition for nilpotent groups: for a proper subgroup $H$, choose the first $Z_i$ not contained in $H$. Then $[Z_i,H]\subseteq Z_{i-1}\subseteq H$, so $Z_i\subseteq N_G(H)$ and $N_G(H)>H$. If a Sylow subgroup $P$ had proper normalizer $N$, then $N_G(N)>N$; but $P$ is the unique Sylow subgroup of $N$, hence characteristic there, and $N_G(N)\subseteq N_G(P)=N$, a contradiction. Thus all Sylow subgroups are normal, different ones commute because their intersection is trivial, and they form an internal direct product. Conversely each finite Sylow subgroup is nilpotent, and products of their central series give a central series for their direct product.

Composition series record simple building blocks, chief series record ambient normal structure, derived series detect solubility, and central series detect nilpotence. Their associated factors and extension data explain why the same group can have very different useful series.

For the transfer homomorphism, choose representatives $T$ of the right cosets $H\backslash G$. For $t\in T$ and $g\in G$, write $tg=h(t,g)t_g$, where $h(t,g)\in H$ and $t_g\in T$. The permutation $t\mapsto t_g$ represents right multiplication on the cosets. DefineThe product has no order ambiguity in the abelianization $H/H^{\prime }$. Replace each representative $t$ by $a_{t}t$, with $a_t\in H$. The new factor is $a_{t}h(t,g)a_{t_g}^{-1}$. In the Abelian quotient, the extra products cancel because $t\mapsto t_g$ permutes $T$. Thus the transfer is independent of the transversal. Moreover,Multiplying and using the same permutation of representatives gives $V(g_1{g}_2)=V(g_1)V(g_2)$. This proves that group transfer is a well-defined group homomorphism.

For the Burnside transfer theorem, let $P\subseteq Z(N_G(P))$ be a Sylow $p$-subgroup. In particular $P$ is Abelian. We first prove the fusion fact that two elements of $P$ which are conjugate in $G$ must be equal. Suppose $u\in P$ and $v=g^{-1}ug\in P$. Both $P$ and $g^{-1}Pg$ centralize $v$, and they are Sylow subgroups of $C_G(v)$: they have the maximal $p$-order possible even in $G$. By Sylow theorems inside this centralizer, choose $h\in C_G(v)$ with $h^{-1}{g}^{-1}Pgh=P$. Thus $gh\in N_G(P)$ andBut $gh$ centralizes $P$ by hypothesis, so $v=u$.

Evaluate transfer to $P$ on $u\in P$. Partition the right cosets of $P$ into cycles under multiplication by $u$. For a cycle of length $r$ beginning at $Pt$, the product of its transfer factors telescopes to $t{u}^r{t}^{-1}\in P$. It is conjugate to $u^r\in P$, so the fusion fact makes it equal to $u^r$. The cycle lengths sum to $m=[G:P]$. Since $P$ is Abelian, all factors commute andThe integer $m$ is prime to $p$, and this power map is an automorphism of the finite Abelian $p$-group $P$: choose an inverse exponent modulo the exponent of $P$. Therefore $V_{G,P}:G\to P$ is surjective. Its kernel $K$ is normal and the first isomorphism theorem givesThus $K$ is the required normal p-complement; it also has trivial intersection with $P$.

Now suppose $P$ is cyclic of order $p^a$, where $p$ is the smallest prime divisor of $|G|$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, whose order is $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$ and $P$ is Sylow in $N_G(P)$, the quotient has order prime to $p$. Any remaining prime factor would divide $p-1$ and hence be smaller than $p$, but would also divide $|G|$, an impossibility. Thus $N_G(P)=C_G(P)$ and $P\subseteq Z(N_G(P))$. The Burnside transfer theorem proves that a cyclic Sylow subgroup at the least prime divisor gives a normal complement.

Apply this to a finite nonabelian simple group. If $p^3\nmid |G|$, its Sylow $p$-subgroup has order $p$ or $p^2$. A cyclic Sylow subgroup would give a normal complement; simplicity would then force either a trivial complement and a $p$-group, or an improper complement, neither possible for a nonabelian simple group. Groups of order $p^2$ are Abelian: their nontrivial group center, supplied by the class equation, is either the whole group or has cyclic quotient, which also forces the group Abelian. Thus the only remaining Sylow possibility is $P\cong C_p^2$.

The conjugation quotient $N_G(P)/C_G(P)$ now embeds in $G{L}_2(p)$ and is still prime to $p$. Its possible prime divisors come fromIf $p$ is odd, every prime divisor of $p-1$ is smaller than $p$, and every prime divisor of the even number $p+1$ is at most $(p+1)/2<p$. Since $p$ was the least prime divisor of $|G|$, the quotient must be trivial. Then the Burnside transfer theorem again contradicts simplicity. If $p=2$, the only nontrivial odd divisor of $|G{L}_2(2)|=6$ is three. Trivial conjugation would again give a normal complement, so three must divide $|G|$; the Sylow subgroup has order four. We have proved the least-prime divisibility constraint for a finite simple group:

For infinitely many instances of the latter alternative only, takeThese are simple, as permitted in the question (and also covered by the next solution). For $q=5^{2m+1}$, the group order is $q(q^2-1)/2$. Since $q\equiv 5(\mathrm{mod}8)$, $q-1$ is divisible by four but not eight and $q+1$ by two but not four. Hence the order is divisible by four but not eight. Also $q\not\equiv 0(\mathrm{mod}3)$, so $3\mid q^2-1$. Their smallest prime divisor is two andTheir orders strictly increase with $m$, so they give infinitely many distinct examples.

For the centralizer of a subset, the identity commutes with every $h\in A$. If $g,k$ commute with each member of $A$, then for every such $h$,Also $gh=hg$ implies $g^{-1}h=h{g}^{-1}$ after multiplication by $g^{-1}$ on both sides. Thus identity, products and inverses lie in $C(A)$, proving that it is a subgroup.

For the conjugation action, identity acts trivially, andThese are the defining laws of a left group action. Its orbits of a group action are the conjugacy classes. The stabilizer of $h$ is exactly its centralizer, since $gh{g}^{-1}=h$ if and only if $gh=hg$. The orbit-stabilizer theorem therefore givesA class has one element exactly when that element commutes with all of $G$, so the singleton classes are precisely the elements of $C(G)$, the centre of a group. Partitioning $G$ into all its conjugacy classes now proves the class equationFor $|G|=p^r$, Lagrange's theorem for finite groups makes every class size a power of $p$. A nonsingleton class has size at least $p$ and is therefore divisible by $p$. Reducing the class equation modulo $p$ yields $|C(G)|\equiv 0(\mathrm{mod}p)$, because $r>0$. The identity lies in $C(G)$, so its order is positive and divisible by $p$. ConsequentlyThis proves the nontrivial center of a finite p-group directly from the conjugation classes.

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. For $|G|=p^r$ with $r\ge 1$, let $G$ act on itself by conjugation. The orbit-stabilizer theorem makes the size of the conjugacy class of $g$ equal to $[G:C_G(g)]$, a power of $p$. A class has size one exactly when $g\in Z(G)$; every other class size is divisible by $p$. The class equation therefore givesso $p$ divides $|Z(G)|$. Since the identity belongs to the centre of a group, its size is a positive multiple of $p$. This proves the nontrivial center of a finite p-group.

If $G/Z(G)$ is cyclic, choose a representative $a$ of a generator. Every element is $a^{m}z$ for an integer $m$ and $z\in Z(G)$. Any two such elements commute, since their central factors commute with everything and their remaining factors are powers of the same element. Thus $G$ is an abelian group, $Z(G)=G$, and the cyclic quotient by the centre of a group is trivial.

For a non-abelian group of order $p^3$, the centre of a group has order $p$, $p^2$ or $p^3$ by the proved result and Lagrange's theorem. The last would make the group abelian. The middle would make $G/Z(G)$ have prime order and hence cyclic, also making $G$ abelian by the preceding argument. Therefore its centre has order $p$.

Finally write the given upper-unitriangular matrix as $M(a,b,c)$. Direct matrix multiplication givesIt commutes with every $M(a^{\prime },b^{\prime },c^{\prime })$ exactly when $a{c}^{\prime }=a^{\prime }c$ for all $a^{\prime },c^{\prime }\in F$. Taking $(a^{\prime },c^{\prime })=(0,1)$ forces $a=0$, and taking $(1,0)$ forces $c=0$; those conditions also suffice. Thus the centre of a group in this Heisenberg group over a prime field is

A nilpotent group has a finite central seriesEquivalently the upper central series, defined by $Z_{j+1}/Z_j=Z(G/Z_j)$, reaches $G$.

We first prove the normalizer condition for nilpotent groups. If $H<G$, choose the least $j$ with $Z_j⊈H$. Then $Z_{j-1}\le H$, and an element $z\in Z_j\setminus H$ satisfies $[z,H]\le Z_{j-1}\le H$. Thus $z$ normalizes $H$, giving $H<N_G(H)$.

Let $P$ be a Sylow subgroup of a finite nilpotent $G$, and write $N=N_G(P)$. If $N<G$, the normalizer condition gives some $g\in N_G(N)\setminus N$. The subgroups $P$ and $P^g$ are both Sylow subgroups of $N$, so Sylow theorem supplies $n\in N$ with $P^{gn}=P$. But then $gn\in N_G(P)=N$, forcing $g\in N$, a contradiction. Therefore every Sylow subgroup is a normal subgroup of $G$.

For normal Sylow subgroups $P,Q$ at different primes, $[P,Q]\le P\cap Q=1$. They commute, have coprime orders, and their product has order $|G|$. Multiplication consequently gives an isomorphism from their external direct product of groups onto $G$.

Conversely, a nontrivial finite p-group has nontrivial center of a group: its class equation has every noncentral conjugacy-class size divisible by $p$, so $|Z(G)|$ is a positive multiple of $p$. Induction on order, applied to $G/Z(G)$, gives a central series for $G$. Finite direct products of these central series give a central series for the product, padding shorter series by the full factor. Thus the finite nilpotent group decomposition is

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

For a nontrivial finite p-group $G$ of order $p^m$ with $m\ge 1$, partition into conjugacy classes. A noncentral class has size $[G:C_G(x)]$, a positive power of $p$ larger than one by Lagrange's theorem. The class equation therefore saysThe center of a group contains the identity, so $|Z(G)|\ge p$, proving its nontriviality. The positive-exponent qualification matters: the one-element group has no nonidentity central element.

If $|G|=p^2$, its center of a group has order $p$ or $p^2$. In the first case, the quotient group $G/Z(G)$ has prime order and is cyclic. Whenever a central quotient group is cyclic, writing all elements as $g^{i}z$ with $z$ central shows that they commute: $(g^{i}z)(g^{j}w)=g^{i+j}zw=(g^{j}w)(g^{i}z)$. Thus that case would already make $G$ Abelian and its center of a group all of $G$, a contradiction. Hence every group of order $p^2$ is Abelian.

If there is an element of order $p^2$, it is a generator of a group for $G$, giving $C_{p^2}$. Otherwise every nonidentity element has order $p$. Pick $a\ne 1$ and $b\notin ⟨a⟩$. Their cyclic subgroups have trivial intersection, and they commute; the $p^2$ distinct products $a^i{b}^j$ exhaust $G$. Hence the classification of groups of order p squared isBoth groups exist and are nonisomorphic, since only the first has an element of order $p^2$.

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.

For a finite group, the conjugacy classes partition the group, and a class is a singleton exactly when its element belongs to the centre of a group. Since the centre is trivial, the class equation becomeswhere the ${\mathcal{C}}_j$ are all the nonidentity conjugacy classes. If the prime number $p$ divided every $|{\mathcal{C}}_j|$, reduction modulo $p$ would give $0=1$, because $p\mid |G|$. Therefore at least one of these class sizes is not divisible by $p$. Primality now givesIts size is greater than one because the centre is trivial. This proves the prime-to-p conjugacy class lemma.

The conclusion requires $p$ to be prime. The printed question does not explicitly impose this hypothesis. If arbitrary composite divisors are allowed, take the symmetric group $S_3$ and $p=6$: its centre is trivial, its two nonidentity conjugacy classes have sizes $3$ and $2$, and neither is coprime to $6$. Thus the argument proves the intended prime case and also identifies why the unrestricted literal reading is false.

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.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

Use the class equation for the action of $G$ on itself by conjugation:where the sum runs over noncentral conjugacy classes. By Lagrange's theorem, each noncentral class has size a positive power of $p$ greater than one. Thus $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. Since the identity element belongs to the center of a group,For $|G|=p^3$, the possible orders of the center of a group are $p,p^2,p^3$. Order $p^3$ would make $G$ an abelian group. Order $p^2$ would make the quotient group $G/Z(G)$ have prime order and hence be a cyclic group. But a cyclic quotient by the center forces $G$ to be Abelian: if $gZ(G)$ generates the quotient, any two elements are $g^{a}z$ and $g^{b}w$ with $z,w\in Z(G)$, and their products commute. Both larger orders contradict the hypothesis. Consequently

Proceed by induction on $|G|$. If $p\mid |Z(G)|$, Cauchy's theorem for finite groups gives a central subgroup $C$ of order $p$. By induction $G/C$ has a subgroup of order $p^{n-1}$; its inverse image in $G$ has order $p^n$.

If $p\nmid |Z(G)|$, the class equation shows that some noncentral conjugacy class has size not divisible by $p$. For a representative $x$, its class size is $[G:C_G(x)]$, so $p^n\mid |C_G(x)|$. This centralizer is proper because $x$ is noncentral, and induction gives it, hence $G$, a subgroup of order $p^n$. This proves Sylow's first 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 $6$.