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A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

The centralizer of an element $x$ of a group $G$ consists of all elements that commute with $x$:It is the stabilizer subgroup of $x$ under the conjugation action, hence a subgroup of $G$. In conjugacy classes of the symmetric group on three letters, the three representative computations areFor the transposition $(12)$, a commuting permutation must preserve the exchanged pair $\{1,2\}$ and fix $3$; only $e$ and $(12)$ do so. For $r=(123)$, a commuting permutation is determined by where it sends $1$: the equation $g(r(i))=r(g(i))$ forces the images of $2$ and $3$, leaving exactly $e,r,r^2$. For $e$, every permutation commutes. These centralizer sizes $6,2,3$ give conjugacy class sizes $1,3,2$.

The centralizer is useful because it measures how much freedom remains after fixing one operation. Its index of a subgroup counts the distinct results of conjugation, and the intersection of all element centralizers is the center of a group. A centralizer need not be an abelian group: $C_G(e)=G$ even when $G$ is a nonabelian group. In an abelian group, all element centralizers equal $G$. These observations also explain why the surrounding group matters: the centralizer of $(123)$ inside the alternating group $A_3$ is the whole $A_3$, whereas its centralizer inside $S_3$ is a proper subgroup. See McKernan, the centralizer definition preceding lemma 23.6.

For an arbitrary subset $A\subseteq G$, the centralizer of a subset fixes every element of $A$ individually under the conjugation action:This is a pointwise stabilizer of the elements of $A$, not of the vertices of a polygon in a geometric realization. In particular, $r=(123)$ belongs to $C_{S_3}(r)$ although its rotation moves every vertex of a polygon of the equilateral triangle. If $A$ generates a subgroup $H$, commuting with every element of $A$ implies commuting with every product of those elements and their inverse elements; hence $C_G(A)=C_G(H)$. For the triangle rotation subgroup $H=⟨r⟩$, this gives $C_{S_3}(H)=C_{S_3}(r)=H$.

To see the commutation test directly on the equilateral triangle, compare a rotation $r$ with the reflection $s=(12)$. Their permutations satisfyThe two orders send vertex of a polygon $1$ to different places: $rs(1)=3$, but $sr(1)=1$. Thus $s\notin C_{S_3}(r)$. By contrast $r$ and $r^2$ commute, with $r{r}^2=r^{2}r=e$. The reflection reverses the chosen rotation while preserving the subgroup of all triangle rotations, so $s\in N_{S_3}(H)$. The distinction is visible before any counting: commuting preserves an operation itself; membership in the normalizer permits exchanging it with another operation in the same subgroup.

Both constructions are stabilizer subgroups for the conjugation action. The centralizer $C_G(x)$ fixes an individual element $x$; the normalizer $N_G(H)$ preserves a subgroup $H$ as a whole, while possibly moving its elements. The centralizer of a subset $C_G(H)$ fixes every member of $H$ individually, soFor a cyclic subgroup $H=⟨r⟩$, commuting with the group generator $r$ is equivalent to commuting with all its powers, giving $C_G(H)=C_G(r)$.

In the symmetric group $S_3$, let $r=(123)$ and $H=⟨r⟩=\{e,(123),(132)\}$. ThenEvery transposition sends $r$ to $r^{-1}$ under conjugation. In particular, for $t=(12)$,Thus $t$ belongs to the normalizer of $H$ but not to the centralizer of $r$. Preserving the collection of powers allows the two possible directions of its group generator to be exchanged.

This distinction lets one ask separately which operations leave a chosen operation unchanged and which operations preserve the subgroup it generates. The normalizer also supplies a group homomorphism to the automorphism group of $H$, by $g\mapsto (h\mapsto gh{g}^{-1})$; its group kernel is $C_G(H)$. In this example the group homomorphism has two possible outputs, fixing $r$ or replacing it by $r^{-1}$. The quotient group $N_{S_3}(H)/C_{S_3}(H)$ is therefore a cyclic group of group order two. Milne develops the normalizer as a stabilizer subgroup in Group Theory (v4.01) by James S. Milne.

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.

For a subgroup $H$ of a group $G$ and $g\in G$, the conjugate subgroup is the conjugate subsetIt is a subgroup because conjugation is a group automorphism. Restricting that group automorphism gives a group isomorphism from $H$ to $gH{g}^{-1}$, so conjugate subgroups have the same group order. This describes the same operations after moving their reference configuration. Conjugacy of subgroups is an equivalence relation. ^[ref]

For an equilateral triangle with vertices of a polygon labeled $1,2,3$ counterclockwise, its symmetry group is the symmetric group $S_3$, also the dihedral group of group order six. Let $r=(123)$ be the counterclockwise rotation through $120$ degrees, and let $s=(12)$ be the reflection fixing vertex of a polygon $3$. Products apply the rightmost permutation first. The cyclic subgroup $K=⟨s⟩=\{e,(12)\}$ contains the identity element and one reflection. Relabeling a transposition by conjugation givesThus the conjugation action moves $K$ among the three subgroups $⟨(12)⟩$, $⟨(23)⟩$, and $⟨(13)⟩$. Geometrically, the rotation moves the axis of the reflection to the next axis. Each reflection fixes a different vertex of a polygon, although group isomorphisms relate all three subgroups.

The normalizer tells when this change of reference leaves the subgroup unchanged. For any group $G$ and subgroup $H\le G$, there is a bijectionTwo choices $g,k$ give the same conjugate subgroup exactly when $k^{-1}g\in N_G(H)$. Consequently the index of a subgroup $[G:N_G(H)]$ counts the distinct conjugate subgroups, even for an infinite group. For a finite group the count is $|G|/|N_G(H)|$. These are cosets; they need not form a quotient group, since $N_G(H)$ need not be a normal subgroup of $G$. In the triangle example there are $6/2=3$ conjugate subgroups, while the cyclic subgroup $⟨r⟩$ has only one. ^[ref]

The group ring $\mathbb{Z}G$ becomes a $\mathbb{Z}G$-module under the conjugation action $g\cdot x=gx{g}^{-1}$. Its basis splits into conjugacy classes, and the span of the class of $x$ is the permutation module on $G/C_G(x)$.

For a subgroup $H$ of a group $G$, the normalizer isIt is the stabilizer subgroup of $H$ under the conjugation action on subgroups. Its elements can permute members of $H$; they need not commute with them. For $r=(123)$ in the symmetric group $S_3$, the cyclic subgroup $H=⟨r⟩$ has normalizer $S_3$, although the centralizer of $r$ has only three elements, as illustrated in centralizer and normalizer.

Every element of $H$ preserves $H$ under conjugation, so $H\subseteq N_G(H)$. Moreover, $H$ is a normal subgroup of $N_G(H)$, and this is the largest subgroup of $G$ with that property. Indeed, any subgroup in which $H$ is a normal subgroup consists entirely of elements preserving $H$ under conjugation. In particular, $N_G(H)=G$ exactly when $H$ is a normal subgroup of $G$.^[ref]

The orbit-stabilizer theorem counts distinct conjugate subsets of a subgroup byFor $H=⟨(123)⟩$ in $S_3$, there is just one, since $[S_3:N_{S_3}(H)]=6/6=1$. For the subgroup $K=⟨(12)⟩=\{e,(12)\}$, preserving $K$ forces its unique nonidentity element to stay fixed. Hence $N_{S_3}(K)=C_{S_3}((12))=K$, and its three distinct conjugate subsets are $\{e,(12)\}$, $\{e,(13)\}$ and $\{e,(23)\}$. This is the subgroup counterpart of counting a conjugacy class with a centralizer: the object being stabilized determines which denominator belongs in the count.^[ref]

The equality in the definition is essential. For a subgroup $H$ that is a finite group, an inclusion $gH{g}^{-1}\subseteq H$ already gives equality by equal cardinalities. For an infinite subgroup, that inference can fail: conjugation might carry it onto a proper subgroup. The normal subgroup criterion is therefore, with the explicit hypothesis $H\le G$,No finiteness or commutativity assumption is needed. A normal subgroup is preserved as a set; its elements need not lie in the center of a group.

There is a second useful count, distinct from counting conjugate subgroups. The normalizer acts on $H$ itself by group automorphisms, giving the group homomorphismIts group kernel is $C_G(H)$, so the centralizer of a subset is a normal subgroup of $N_G(H)$ and the first isomorphism theorem givesThis quotient group records the different rearrangements of $H$ induced by the surrounding group. It gives a faithful group action on $H$, but need not be the entire automorphism group of $H$. For example, take the ambient group $G=H=⟨(123)⟩$. Since $G$ is an abelian group, $N_G(H)=C_G(H)=G$, so the quotient group is trivial; nevertheless $r\mapsto r^{-1}$ is a nonidentity group automorphism of $H$. Returning to the ambient symmetric group $S_3$ makes that group automorphism available through a reflection, and $N_{S_3}(H)/C_{S_3}(H)\cong C_2$. Its two cosets are the three rotations and the three reflections.

The subgroup $H$ itself need not be contained in $C_G(H)$: their intersection is the center of a group $Z(H)$. It is contained precisely when $H$ is an abelian group. Thus $N_G(H)/H$, which exists because $H$ is a normal subgroup of its normalizer, and $N_G(H)/C_G(H)$ answer different questions and should not be substituted for one another.

For each $h\in g^G\cap H$, count the solutions of $x^{-1}gx=h$. Once a solution $x_0$ is fixed, every solution is uniquely $c{x}_0$ with $c\in C_G(g)$, the centralizer of $g$. Thus each $h$ contributes $|C_G(g)|{\chi }_W(h)$ to the sum in part (a).

The orbit-stabilizer theorem for the conjugation action gives $|C_G(g)|=|G|/|g^G|$. The intersection $g^G\cap H$ is the disjoint union of the conjugacy classes $h_1^H,\dots ,h_{\ell }^H$, and the character ${\chi }_W$ is constant on each such conjugacy class. HenceIf the intersection is empty, the empty sum gives zero. This is the conjugacy-class formula for an induced character.

The isomorphic pair is $A_8\cong PS{L}_4(2)$; $PS{L}_3(4)$ is different. All three have order $20160$, so order alone does not settle the question; explicitly $|PS{L}_3(4)|=(64-1)(64-4)(64-16)/(3\cdot 3)=20160$. We give a concrete realization and an element-order distinction.

On $W={\bigwedge }^2{\mathbb{F}}_2^4$, the Pfaffian isIt is a nonsingular plus-type quadratic form: the three displayed pairs give a hyperbolic decomposition. Change of basis multiplies the Pfaffian by the determinant, which is always one in $G{L}_4(2)$. Hence the exterior-square realization of PSL4 over F2 embeds $G{L}_4(2)$ into $O_6^{+}(2)$. The kernel of a group homomorphism is trivial: an element acting trivially on the exterior square fixes every decomposable two-space, then every one-space as an intersection of two-spaces, and hence is a scalar; over ${\mathbb{F}}_2$ this scalar is one.

For a second model let $E$ be the even-weight vector subspace of ${\mathbb{F}}_2^8$ and quotient by the all-one vector. The dot product descends to a nondegenerate alternating bilinear form on the six-dimensional quotient, andis well defined, since complementing an even subset changes half its weight from $w/2$ to $4-w/2$. It has $(1+70+1)/2=36$ zero vectors, so is plus type. Coordinate permutations give a faithful $S_8$ action: fixing the quotient fixes each weight-two subset, because its complement has weight six, and fixing all two-subsets forces the identity permutation.

From the independently derived symplectic order and quadratic orbit count below,Thus $O_6^{+}(2)\cong S_8$ and the image of $G{L}_4(2)=S{L}_4(2)=PS{L}_4(2)$ is its index-two subgroup $A_8$. The index-two subgroup is unique: a nontrivial group homomorphism $S_8\to C_2$ sends every transposition to the same nonidentity element and is the sign map.

To separate $PS{L}_3(4)$, observe that $A_8$ contains $(12345)(678)$ of order 15. An odd-order projective element of $PS{L}_3(4)$ lifts to a semisimple linear operator, since its order upstairs divides three times an odd order in characteristic two. The possible eigenspace degree patterns are $1+1+1$, $2+1$ and $3$. In the split case all eigenvalues have order dividing three, so the projective element has no order 15. In the $2+1$ case the eigenvalues are $\lambda ,{\lambda }^4,{\lambda }^{-5}$ with $\lambda \in {\mathbb{F}}_{16}^{\times }$; its fifth power is scalar because ${\lambda }^{15}=1$, so its projective order divides five. In the irreducible case the determinant-one torus has order $4^2+4+1=21$ and quotienting its scalar subgroup of order three gives projective order dividing seven. There is therefore no element of order 15 in $PS{L}_3(4)$.

Several small-parameter coincidences connect the finite classical groups to permutation groups. The projective-line action immediately gives $PS{L}_2(2)\cong S_3$, $PS{L}_2(3)\cong A_4$, $PG{L}_2(3)\cong S_4$ and $PS{L}_2(4)\cong A_5$ by faithful actions of degrees three, four and five and their orders. Also $PS{L}_2(5)\cong A_5$: its involutions have Klein-four centralizers, giving five Sylow subgroups; their conjugation action is faithful and embeds its order-60 simple group into $S_5$. Simplicity here can be checked from class sizes $1,15,20,12,12$: no proper nontrivial normal class union has order dividing 60. The same five-subgroup action extends to $PG{L}_2(5)$, with trivial kernel of a group homomorphism and order 120, giving $S_5$. Indeed its kernel intersects $PS{L}_2(5)$ trivially, so centralizes that normal subgroup. Commuting projectively with both upper and lower elementary unipotents forces a representing matrix to be scalar, so this kernel is trivial.

The less immediate coincidence $PS{L}_2(9)\cong A_6$ can be exhibited on the ten unordered triple partitions used in Question 4. Let $a=(123)$, $b=(456)$, $w=(12)(36)$, assign $\infty$ to $123\mid 456$ and $0$ to $126\mid 345$, and label the other partitions by applying $a^x{b}^y$ to the zero partition, with $t=x+iy\in {\mathbb{F}}_9$ and $i^2=-1$. Inspection of these nine partitions givesTranslations and inversion generate $PS{L}_2(9)$: conjugating the upper unitriangular matrices by inversion gives the lower ones, and elementary elimination generates $S{L}_2(9)$. Their projective action has order $9(9^2-1)/2=360$. These partition permutations come from even permutations of six letters, whose partition action is faithful as proved below; hence they fill $A_6$.

Finally, on the even-weight module of ${\mathbb{F}}_2^6$ modulo its all-one vector, coordinate permutations preserve a nondegenerate alternating bilinear form in dimension four and act faithfully, again by the weight-two-subset argument. Both $S_6$ and $S{p}_4(2)$ have order 720, proving $S{p}_4(2)\cong S_6$, with commutator subgroup $A_6$. These examples illustrate both the need to quotient group centers and the low-rank exceptions to general simplicity. They do not make equal orders sufficient for group isomorphism, as the two order-20160 examples already demonstrate.

A group action is a map $(g,x)\mapsto g\cdot x$ from $G\times X$ to $X$ satisfying $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. Define the orbit of a group action $Gx=\{g\cdot x:g\in G\}$ and the stabilizer subgroup $G_x=\{g:g\cdot x=x\}$. The latter is a subgroup: the identity fixes $x$, products of elements fixing $x$ fix $x$, and inverses do too.

For finite $G$, the orbit-stabilizer theorem statesTo prove it, send a left coset $g{G}_x$ to $g\cdot x$. If $g{G}_x=h{G}_x$, then $h^{-1}g\in G_x$ and $g\cdot x=h\cdot x$. Conversely, equality of the two images implies $h^{-1}g\in G_x$, hence equality of the cosets. Every point of the orbit of a group action has such a preimage. This is therefore a bijection, and each coset has $|G_x|$ elements, proving the formula.

For the action on subgroups, conjugation maps a subgroup $K$ to a subgroup $gK{g}^{-1}$ because it preserves products and inverses. Furthermore $eK{e}^{-1}=K$ andThus this is the conjugation action. The stabilizer subgroup of $H$ is its normalizer $N_G(H)$, and $H\subseteq N_G(H)$ since each $h\in H$ conjugates $H$ onto itself. Hence the number $k$ of distinct conjugate subgroups isNow suppose $H$ is proper. Put $m=|G|/|H|\ge 2$ and list the $k$ distinct conjugate subgroups as $H_1,\dots ,H_k$. Each has $|H|$ elements and all contain the same identity. Counting the identity just once, even if other elements overlap, givesThus some element of $G$ lies in no conjugate of $H$. This proves conjugates of a proper subgroup do not cover a finite group by an explicit count; it does not require the conjugates to intersect only at the identity.

Work with finite groups, as required for these Sylow subgroup counts. Let $P$ act by conjugation on the set of Sylow $p$-subgroups. The stabilizer of $Q$ is $P\cap N_G(Q)$. Since $Q$ is the normal Sylow subgroup of its own normalizer, every $p$-subgroup of that normalizer lies in $Q$. Consequently the stabilizer is exactly $P\cap Q$, and the orbit-stabilizer theorem gives orbit size $[P:P\cap Q]$. The orbit of $P$ has size one. Every other orbit size is a power of $p$ at least $p^a$, hence is divisible by $p^a$. Adding orbit sizes proves the strengthened Sylow congruence from intersections:Now suppose $G$ is simple of order $2^e\cdot 15$. If $e=0$, its Sylow $5$-subgroup is normal because its number divides three and is one modulo five. If $e=1$, an involution in the regular permutation action of $G$ swaps fifteen pairs, so its sign is negative. The sign of a permutation would give a nontrivial group homomorphism $G\to C_2$ with a proper nontrivial normal kernel. Both cases contradict simplicity. Thus $e\ge 2$.

The Sylow theorems make $n_2$ a divisor of fifteen. It is not one, since a nontrivial Sylow subgroup would then be normal. It is not three, since the conjugation action on three Sylow subgroups would give a faithful group homomorphism $G\hookrightarrow S_3$, impossible by order. If $n_2=5$, take a Sylow subgroup $P$ itself: its normalizer has subgroup index five. If $n_2=15$, not all distinct Sylow subgroups can intersect $P$ in subgroup index at least four, since that would give $15\equiv 1(\mathrm{mod}4)$. Choose $Q\ne P$ with $[P:P\cap Q]=2$. Put $R=P\cap Q$, of order $2^{e-1}$. It has subgroup index two in both $P,Q$, so is normal in each. Hence $N_G(R)$ contains both distinct Sylow subgroups and has order greater than $2^e$. It is proper: otherwise $R$ would be a nontrivial proper normal subgroup of $G$. Its order is therefore $2^e\cdot 3$ or $2^e\cdot 5$. The latter would give an index-three subgroup and again an impossible faithful action on three cosets. Thus $[G:N_G(R)]=5$.

In either case we have the requested subgroup of order $2^e$ or $2^{e-1}$ with index-five normalizer. The action on the five cosets of that normalizer is nontrivial and faithful by simplicity. Its image lies in $A_5$, since composing with sign cannot give a nontrivial map from this simple group of order greater than two to $C_2$. Thus $2^e\cdot 15$ divides $60$, forcing $e=2$ and equality of orders. This proves the simple groups with order a power of two times fifteen conclusion:For $S{L}_2(5)$, choose the first column: there are $5^2-1=24$ nonzero choices. For each, determinant one is a nonzero linear equation in the second column and has five solutions. Therefore $|S{L}_2(5)|=120$. A central matrix commuting with $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ has zero lower-left entry and equal diagonal entries; commuting also with $\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ makes the upper-right entry zero. It is scalar. Determinant one then givesTo establish the projective special linear group over the field with five elements isomorphism without assuming its simplicity, compute its classes. For determinant-one matrices, the characteristic polynomial is $X^2-tX+1$ with trace $t$. The possibilities over ${\mathbb{F}}_5$ exhaust the group:For trace zero the eigenvalues are $2,3$, so the determinant-one centralizer is the split diagonal torus of order four. For traces $\pm 1$, the discriminant is the nonsquare $2$: the centralizer is the norm-one subgroup of ${\mathbb{F}}_{25}^{\times }$, of order six. In both semisimple cases the determinant map on the full linear centralizer is surjective onto ${\mathbb{F}}_5^{\times }$; consequently each trace type is a single special-linear class. For the noncentral traces $\pm 2$, a Jordan block has full linear centralizer $\left\{\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right):a\ne 0\right\}$ with determinant $a^2$. The determinant image consists of the two squares, so each general-linear class splits into two special-linear classes; the determinant-one centralizer has order $2\cdot 5=10$. These sizes sum to $120$.

Quotienting by $Z$ identifies a matrix with its negative. The trace-zero class gives one projective class of size fifteen; the two traces $\pm 1$ give one class of size twenty; the four noncentral unipotent classes give two classes of size twelve. Their respective projective orders are $2,3,5,5$, by the characteristic polynomials. A normal subgroup is a union of classes containing the identity. The nontrivial proper possible union sizes are $13,16,21,25,28,33,36,40,45,48$, none of which divides $60$. Thus the quotient is simple. Applying the order-$60$ result above givesFinally, if $A^2=I$ over an odd-characteristic field, its minimal polynomial divides the square-free polynomial $(X-1)(X+1)$, so it is diagonalizable with eigenvalues $\pm 1$. In dimension two, determinant one requires both eigenvalues equal. Hence $A=I$ or $A=-I$. This proves the unique involution in SL2 over an odd field assertion: the only element of order two in $S{L}_2(5)$ is $-I$. If an index-two subgroup $H$ existed, it would be normal and have order sixty. By Cauchy theorem it would contain an involution, hence contain $Z$. Then $H/Z$ would be a normal subgroup of order thirty in the simple quotient $PS{L}_2(5)$, a contradiction. Thus $S{L}_2(5)$ has no index-two subgroup.

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.

For each $g\in G$, conjugation sends a subgroup $H$ to $gH{g}^{-1}$, a subgroup of the same order and hence still proper. Identity conjugation fixes every subgroup, andThese verify the group action laws.

The stabilizer subgroup of $B$ for this conjugation action is its normalizer $N_G(B)$. Since $B\le N_G(B)$, the orbit-stabilizer theorem gives the number of distinct conjugates asAll these conjugate subgroups contain the identity, so counting that element only once givesThe last inequality uses the properness of $B$, so $[G:B]>1$. There is therefore an element outside this union, which is not conjugate to any element of $B$. This proves that conjugates of a proper subgroup do not cover a finite group, including the case $B=\{1\}$.

For a group action, the orbit-stabilizer theorem identifies the orbit of a group action of $x$ with the left cosets of its stabilizer subgroup:This is well defined and bijective because $gx=hx$ exactly when $h^{-1}g\in G_x$. In particular, for finite $G$,Under the conjugation action $g\cdot x=gx{g}^{-1}$, a singleton orbit of a group action means that $gx{g}^{-1}=x$ for every $g$. This is equivalent to $gx=xg$ for every $g$, so precisely these elements constitute the centre of a groupThe identity is central, and if $x,y$ commute with every element then so do $xy$ and $x^{-1}$. Thus $Z(G)$ is a subgroup. Each central element is fixed under every conjugation, giving $gZ(G)g^{-1}=Z(G)$: the centre is a normal subgroup.

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

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.

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.

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

The normalizer of $P$ in $G$ isIt is the stabilizer of the subgroup $P$ under the conjugation action of $G$ on its subgroups.