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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.

Let $p$ be the smallest prime divisor of the order of a finite nonabelian simple group. A cyclic Sylow subgroup would give a normal p-complement, so if $p^3$ does not divide the order, the Sylow subgroup must be $C_p^2$. Conjugation embeds its normalizer modulo centralizer into $G{L}_2(p)$, and this quotient has order prime to $p$. For odd $p$, every other prime divisor of $|G{L}_2(p)|=p(p-1{)}^2(p+1)$ is smaller than $p$, again forcing trivial conjugation and a contradiction to the Burnside transfer theorem. For $p=2$, nontrivial conjugation must have order three, giving $12\mid |G|$. The groups $PS{L}_2(5^{2m+1})$ give infinitely many examples with exactly two factors of two in their orders.

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.

Choose $\alpha \in \Omega$. Since $G$ acts regularly, $g\mapsto g\alpha$ identifies $\Omega$ with the underlying set of $G$, and the given action becomes left multiplication $L_g(x)=gx$.

A permutation $c$ centralizing all $L_g$ satisfiesConversely every right translation commutes with every left translation. To make the multiplication convention explicit, define $R_g(x)=x{g}^{-1}$. Then $R_g{R}_h(x)=x{h}^{-1}{g}^{-1}=x(gh{)}^{-1}=R_{gh}(x)$, using rightmost-first composition. The map $g\mapsto R_g$ is injective and its image is the entire centralizer, since any possible $c(1)$ can be written $g^{-1}$. ThusThis is centralizer of a regular permutation subgroup. If one instead writes right multiplication as $x\mapsto xg$, the resulting map is an antihomomorphism; including the inverse avoids suppressing that reversal. Choosing a different base point changes the identification but not the isomorphism type.

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 $n\ge 3$, take $C_1$ to be the class of $n$-cycles, $C_2$ the class of transpositions, and $C_3$ the class of an $(n-1)$-cycle with one fixed point. With rightmost permutations applied first, a witness isMultiplying an $n$-cycle by a transposition splits it at the two exchanged letters. The resulting cycles have the lengths of the two arcs between those letters in the cyclic ordering. Thus $\alpha \beta$ has type $(n-1,1)$ exactly when the exchanged letters are cyclically adjacent.

Any product-one tuple can first be conjugated so that its first entry is $\alpha$. Its second entry is then one of the $n$ adjacent transpositions. The centralizer $⟨\alpha ⟩$ acts transitively on these choices; the third entry is forced to be the inverse product. Therefore every product-one tuple lies in a single simultaneous-conjugation orbit.

The conjugates ${\alpha }^j\beta {\alpha }^{-j}$ exchange all adjacent letters around the cycle. Their connected-edge graph generates $S_n$, so the witness, and every tuple in that orbit, generates $S_n$. HenceFor $n=3$ the last two classes coincide, which causes no difficulty: the same three adjacent swaps form one orbit and generate $S_3$.

The relevant cycle types are $(5,1)$, $(2,2,2)$ and $(3,2,1)$. Conjugate the first entry to $\alpha =(12345)$, fixing six. An entry $\beta$ of type $(2,2,2)$ pairs six with one of the five other letters. Powers of $\alpha$ move that partner to one. With the pair $(16)$ fixed, only three matchings remain:Thus there is exactly one centralizer orbit of eligible second entries, consisting of five matchings before normalization. The third entry is forced to be $(\alpha \beta {)}^{-1}$, so there is at most one inner orbit of product-one tuples of these types.

For the actual displayed $g_2$, direct multiplication giveswhich belongs to the class of the displayed $g_3$. This is a product-one witness. The three displayed permutations themselves need not multiply to one; the problem concerns their conjugacy classes.

The group $⟨\alpha ,g_2⟩$ is transitive, because $\alpha$ moves the first five points transitively and $g_2$ connects six to them. Its stabilizer of six contains the five-cycle $\alpha$ and is transitive on the other points. Hence the group is two-transitive and therefore primitive. Moreover $(\alpha g_2{)}^3=(13)$ is a transposition. The primitive transposition lemma proves that the group is $S_6$. Consequently all tuples in the sole inner orbit generate $S_6$, andAs a count check, there are $6\cdot 4!=144$ first entries and five eligible second entries for each, giving $720$ tuples. A generating tuple has simultaneous stabilizer equal to the trivial center of $S_6$, so its inner orbit also has size $720$.

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.

We prove Simplicity of the alternating group A5 by finding all its conjugacy classes. The alternating group has group order $|A_5|=5!/2=60$. Its even permutations have cycle types $1$, $(abc)$, $(ab)(cd)$ and $(abcde)$. There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles, $5\cdot 3=15$ products of disjoint transpositions, and $4!=24$ five-cycles.

For a three-cycle, a commuting permutation must rotate its three-element support and may interchange the two fixed points. Its centralizer in $S_5$ has order six, and only the three rotations are even. Thus its conjugacy class in $A_5$ has size $60/3=20$ and contains every three-cycle. The centralizer of $(12)(34)$ in $S_5$ has order eight: one may interchange the two pairs and independently interchange the elements within each pair. It contains the odd element $(12)$, so exactly half of these commuting elements are even. The conjugacy class in $A_5$ consequently has size $60/4=15$. A permutation commuting with a five-cycle is determined by the image of one point and must be a power of that cycle. All five powers are even, so each such conjugacy class has size $60/5=12$. The 24 five-cycles form exactly two classes.

A normal subgroup contains the identity and is a union of whole conjugacy classes. Its possible orders therefore lie inBy Lagrange's theorem, its order must divide 60. Only 1 and 60 in this list do so. Hence $A_5$ has no proper nontrivial normal subgroup and is a simple group.

A finite minimal normal subgroup is a characteristically simple group, and has the direct-product structure of a finite minimal normal subgroupwith isomorphic simple groups $T_i$. For completeness, choose a minimal nontrivial normal subgroup $T$ of $N$. Its distinct $G$-conjugates are minimal normal subgroups of $N$, so they centralize one another. Their product is normal in $G$, hence equals $N$. If $T$ is abelian the product is abelian and minimal primary characteristic subgroups make $N$ elementary abelian. Otherwise each $T$ has trivial center: its center is a characteristic subgroup of $T$, hence normal in $N$, and minimality would make $T$ abelian. Each new conjugate then meets the product of previous conjugates trivially, because it centralizes that product and is centerless. The product is direct. Every normal subgroup of a direct factor is normal in $N$, so minimality of that factor makes it simple.

Assume now that $N$ acts primitively and nonregularly. Any nontrivial normal subgroup of this faithful primitive action is transitive. Thus each $T_i$ is transitive. If $k\ge 2$, the transitive commuting groups $T_1$ and $T_2$ are regular: an element of one fixing $\alpha$ commutes with the other and hence fixes its entire orbit. But the centralizer of a regular group has order $n$, as its elements are determined by the image of $\alpha$. Therefore all remaining factors lie in the order-n centralizer of $T_1$, and the transitive group $T_2$ already fills it. There can be only two factors, with $|N|=n^2$. Now $|N_{\alpha }|=n$, whereas its common nontrivial subdegree $r$ divides both $n$ and $n-1$. Hence $r=1$, making $N_{\alpha }$ trivial, a contradiction. Thus $k=1$, and $N$ is simple. An abelian transitive group is regular, so $N$ is nonabelian.

Finally $C_G(N)$ is normal in the primitive group $G$. If nontrivial it would be transitive, and commuting with transitive $N$ would force $N$ regular. Thus $C_G(N)=1$, and conjugation embeds $G$ in $\mathrm{Aut}(N)$. The centerless simple group $N$ is identified with its inner automorphism group. Therefore

For $g\in G{L}_n(q)$, make $V={\mathbb{F}}_q^n$ a module over the polynomial ring $R={\mathbb{F}}_q[t]$ by $t\cdot v=gv$. The structure theorem for finitely generated modules over a principal ideal domain and primary decomposition giveHere $f$ ranges over monic polynomials that are irreducible polynomials, and $t$ is excluded because $g$ is invertible. Associate the integer partition ${\lambda }_f$ having $m_j(f)$ parts equal to $j$. It has finite support and satisfies ${\sum }_f\mathrm{deg}(f)|{\lambda }_f|=n$. Two matrices are conjugate exactly when their R-modules are isomorphic, so this gives a bijective parameterization of conjugacy classes.

The centralizer is the R-module automorphism group. Homomorphisms between distinct primary components vanish: coprimeness of their annihilating polynomials makes each such map zero. Fix one $f$ of degree $d$, put $Q=q^d$, and write $m_j=m_j(f)$. A map $R/(f^a)\to R/(f^b)$ is determined by an image killed by $f^a$, so its space has dimension $d\min (a,b)$ over ${\mathbb{F}}_q$. Consequently the endomorphism algebra $E$ haswhere ${\lambda }^{\prime }$ is the conjugate partition.

Its semisimple quotient is ${\prod }_j{M}_{m_j}({\mathbb{F}}_Q)$: reduce the maps between the equal-length summands modulo $f$ and discard the maps between unequal lengths. Composition through a different length acquires a factor of $f$ when it returns to the original length, so these reductions define an algebra homomorphism. It is onto, and its kernel is nilpotent. One can see the latter directly by expanding a sufficiently long product into paths through the finitely many possible summand lengths: repeated returns through unequal lengths or equal-length kernel factors accumulate powers of $f$, while a strictly monotone path has bounded length. Eventually every path vanishes in every summand.

An element is invertible exactly when its image in that quotient is invertible, since a nilpotent error is inverted by a finite geometric series. The fraction of units is therefore ${\prod }_j|G{L}_{m_j}(Q)|/Q^{m_j^2}={\prod }_j{\prod }_{r=1}^{m_j}(1-Q^{-r})$. Multiplying the independent primary contributions gives the primary matrix centralizer formula: