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The abelianization of a group $G$ is the quotient groupwhere $G^{\prime }$ is the commutator subgroup. Its elements are cosets, and its operation is $(g{G}^{\prime })(h{G}^{\prime })=gh{G}^{\prime }$. The result is an abelian group: the two products $gh$ and $hg$ differ by a group commutator, which disappears in the quotient group.

For the symmetric group $S_3$, the sign of a permutation realizes $S_3^{\mathrm{ab}}\cong C_2$, as shown in commutator subgroup of the symmetric group on three letters. For the dihedral group of a square, abelianization of the square symmetry group gives $C_2\times C_2$: two independent yes-or-no pieces of information survive. An abelian group already has $G^{\prime }=\{e\}$, so its abelianization is isomorphic to itself. At the other extreme, a perfect group has trivial abelianization.

Calling this the largest abelian quotient means that every group homomorphism into any abelian group factors uniquely through $q$. It does not require $G$ to be finite, or compare the sizes of unrelated abelian groups. The universal property of abelianization makes this precise. It lets us solve all such mapping problems in one simpler group: any invariant that respects multiplication and takes values in an abelian group sees only $G^{\mathrm{ab}}$.

For the symmetric group $S_3$, take the transpositions $x=(12)$ and $y=(23)$. Products of permutations apply the rightmost factor first. ThenThe last permutation sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$: two swaps fail to commute by an entire three-cycle.

Every group commutator has sign of a permutation $+1$, so $S_3^{\prime }\subseteq A_3$, the alternating group of even permutations. Conversely, the displayed group commutator generates $A_3=\{e,(123),(132)\}$. Therefore $S_3^{\prime }=A_3$. The two cosets are the even permutations and the odd permutations, and their multiplication is just multiplication of signs:This is the first isomorphism theorem applied to the sign of a permutation. Once commutativity is forced, all three transpositions become indistinguishable, and both three-cycles become the identity element. The surviving information is exactly whether the permutation is even or odd. Compare the very different center of a group $Z(S_3)=\{e\}$.

Take the symmetric group $G=S_3$ and the subgroup $H=⟨(12)⟩=\{e,(12)\}$ of group order two. With the rightmost permutation applied first, its three left cosets areThe subscripts record the image of $3$: every member of $C_i$ sends $3$ to $i$. Because $H$ is the stabilizer subgroup of $3$ in the usual group action, $aH\mapsto a(3)$ is a bijection onto $\{1,2,3\}$. It is an equivariant map, since $(ga)(3)=g(a(3))$. The coset action therefore becomes the familiar group action on three letters. In particular,

The three stabilizer subgroups areTheir intersection is $\{e\}$, so the subgroup core and the group kernel are trivial. This proves that the coset action is a faithful group action even though every point has a nontrivial stabilizer subgroup. The orbit-stabilizer theorem reads $6=3\cdot 2$, while the induced group homomorphism gives six distinct permutations of the three cosets. For example, $(12)$ and $e$ agree at $C_3$, but disagree at $C_1$; inspecting one label would miss their difference.

The subgroup need not be a normal subgroup for the coset action to exist: $(123)H(123{)}^{-1}=\{e,(23)\}\ne H$. Here the group action is informative because the three conjugate stabilizer subgroups have no common nonidentity element. For the normal subgroup $A_3$ instead, nonfaithful coset action of the symmetric group on three letters shows that only the sign of a permutation survives.

For two odd fermionic operators, time ordering contributes a minus sign whenever the chronological rearrangement exchanges them. For several fields the sign is the sign of a permutation restricted to the odd fields. This convention makes the free Dirac propagator a Green function of the Dirac equation with its contact term fixed by the canonical anticommutation relations.

A group homomorphism is a map $\phi :G\to H$ preserving the group operation: $\phi (gh)=\phi (g)\phi (h)$. It also preserves the identity element and inverse elements. In an additive group, the same requirement reads $\phi (g+h)=\phi (g)+\phi (h)$. A group homomorphism preserves a way of combining inputs, even when several inputs have the same output.

Reduction modulo three from $\mathbb{Z}/6\mathbb{Z}$ to $\mathbb{Z}/3\mathbb{Z}$ is a group homomorphism: reducing a sum agrees with adding the reduced residues. For example, $2+5=1$ modulo six, and both sides reduce to one modulo three; the outputs $2$ and $2$ also add to one modulo three. This map is surjective but not injective, since zero and three both map to zero.

The sign of a permutation gives another example, from the symmetric group $S_3$ to the multiplicative group $\{1,-1\}$. A product of two odd permutations is even, matching $(-1)(-1)=1$. Its group kernel is $A_3$, the alternating group of even permutations. This shows how a group homomorphism can retain a useful feature while forgetting other details.

The kernel of a group homomorphism is the group version of a kernel. For $\phi :G\to H$, it isIt is a normal subgroup of the domain: for $n$ in the group kernel, $\phi (gn{g}^{-1})=\phi (g)e_H\phi (g{)}^{-1}=e_H$. The group kernel describes inputs whose effect disappears after mapping. For reduction modulo three from $\mathbb{Z}/6\mathbb{Z}$, it is $\{0,3\}$; for the sign of a permutation on $S_3$, it is $A_3$. A group homomorphism is injective exactly when its group kernel contains only the identity element.

For indices in $\{1,\dots ,n\}$, this Levi-Civita symbol is zero when any indices repeat, and otherwise is the sign of a permutation taking $(1,\dots ,n)$ to $(i_1,\dots ,i_n)$. Thus ${ϵ}_{1\cdots n}=1$ and exchanging any two index slots changes the sign. It expresses a determinant as $\det M={ϵ}_{i_1\cdots i_n}{M}_{i_1,1}\cdots M_{i_n,n}$ with repeated indices summed. Exactly $n!$ components are nonzero, so $\sum {ϵ}_{i_1\cdots i_n}^2=n!$.

Let $G=S_3$, the symmetric group on three letters, and take the normal subgroup $N=A_3=\{e,(123),(132)\}$, the alternating group of even permutations. Its two left cosets areProducts of permutations apply the rightmost factor first. Left multiplication by an even permutation preserves both cosets, while left multiplication by an odd permutation exchanges them. For example,Consequently all six group elements give only two distinct permutations of $\{E,O\}$. Both stabilizer subgroups equal $A_3$, so their intersection, the group kernel, is $A_3$. This is a transitive group action because $(12)$ reaches the other coset, but it is not a faithful group action because $(123)$ fixes both.

The orbit-stabilizer theorem still works perfectly: $6=2\cdot 3$. It counts reachable labels, without asserting that the induced group homomorphism is injective. The quotient group $S_3/A_3\cong C_2$ captures exactly the two visible operations, identity and exchange. This is the same group homomorphism as the sign of a permutation, with $+1$ represented by fixing both labels and $-1$ by exchanging them. Contrast the three-label example in faithful coset action of the symmetric group on three letters.

A subgroup $N\le G$ is a normal subgroup when $gN{g}^{-1}=N$ for every $g\in G$. In other words, conjugation by any group element preserves the subgroup as a set, although it may exchange its individual elements. Equivalently, $gN=Ng$ for every $g$: the left and right cosets of $N$ agree. Write $N⊴G$.

Every subgroup of an abelian group is a normal subgroup, because conjugation fixes each element. Normality can also hold in a nonabelian group: in $S_3$, the three even permutations form the alternating group $A_3=\{e,(123),(132)\}$, a normal subgroup as the group kernel of the sign of a permutation. Its two cosets separate the even and odd permutations.

In contrast, $H=\{e,(12)\}$ is a subgroup of $S_3$ but is not a normal subgroup. With function composition of permutations acting rightmost first, choose $g=(123)$. Thenwhich are different cosets. Also $g(12)g^{-1}=(23)\notin H$, directly contradicting invariance under conjugation. Thus a subgroup alone does not suffice for quotient group multiplication.

The practical value of normality is coset multiplication is well-defined exactly for normal subgroups. Each coset can act as one element of a quotient group, with results independent of representatives. Conversely every normal subgroup is the group kernel of its quotient map of groups, so normality also identifies which distinctions a group homomorphism can consistently forget.

The trivial representation and sign of a permutation giveThey are one-dimensional representations, hence irreducible representations. Their characters, in the conjugacy class order of solution, areThese are the only one-dimensional representations. All transpositions are conjugate, so they have the same scalar image $t$ in any one-dimensional representation; since a transposition squares to the identity, $t=\pm 1$. Transpositions generate the symmetric group, fixing the whole representation once $t$ is chosen.

By the sum of squares of irreducible representation dimensions, the three remaining dimensions satisfy $d_2^2+d_3^2+d_4^2=24-1-1=22$, with each $d_i\ge 2$. No $d_i$ can be at least $4$, since $16+4+4>22$. Thus the only possible dimensions are

Use the composition convention $(\sigma \tau )(i)=\sigma (\tau (i))$. Then $R_{+}(\sigma )R_{+}(\tau )e_i=e_{\sigma (\tau (i))}=R_{+}(\sigma \tau )e_i$. Since the sign of a permutation is multiplicative, the same calculation gives $R_{-}(\sigma )R_{-}(\tau )=R_{-}(\sigma \tau )$. Both send the identity to $I$ and have inverses given by ${\sigma }^{-1}$, so both are group representations.

The trace of a permutation matrix counts fixed points, and multiplying that matrix by its sign of a permutation multiplies its trace by that sign. Thus the two characters areLet $u=e_1+e_2+e_3+e_4$ and $W=\{x\in {\mathbb{C}}^4:{\sum }_i{x}_i=0\}$. Both are invariant subspaces, and ${\mathbb{C}}^4=\mathbb{C}u\oplus W$. On $\mathbb{C}u$, $R_{+}$ is the trivial representation and $R_{-}$ is the sign representation. Subtracting their characters yieldsThe character inner product verifies that it is an irreducible representation:The squares of the entries of ${\chi }_4$ are identical, so its character inner product with itself is also one. Their values at a transposition differ, so these irreducible representations are inequivalent. Therefore

Alexandre-Théophile Vandermonde presents a programme in elimination theory: remove unknowns from simultaneous polynomial equations in a way that works uniformly as the number of equations and their polynomial degrees vary. The immediate achievement is narrower than that programme: a general construction for linear equations. The academy year 1772 used in the exam is not the printing date of the memoir, which appeared in 1776. Its compact indices distinguished the equation containing a coefficient from the unknown it multiplied; they were not powers. That notational separation is crucial to interpreting the work.^[ref][ref]

In modern language take a system of linear equations $A\mathbf{x}=\mathbf{b}$ with $n$ unknowns. Its determinant is an alternating expression in the coefficients:Each product selects one coefficient from each equation and each unknown's column. The sign of a permutation determines whether it is added or subtracted. This makes the number $n$ a genuine parameter of a general rule, rather than merely supplying separately memorised formulas for two or three unknowns. The notation here is modern, not a transcription of Alexandre-Théophile Vandermonde's symbols.

For two unknowns, multiply $ax+by=u$ by $d$ and $cx+dy=v$ by $b$, then subtract. Similarly eliminate $x$. This givesThe same principle in arbitrary size is Cramer's rule. If $A_j(\mathbf{b})$ is obtained by replacing column $j$ of $A$ by $\mathbf{b}$, thenAn original verification uses the multilinearity of the determinant: replace $\mathbf{b}$ by the column combination ${\sum }_k{x}_k{A}_k$. Every resulting determinant vanishes through repeated columns except the $k=j$ term, which is $x_j\det A$. This explains how an elimination expression isolates one unknown.

The condition $D\ne 0$ is substantive. If $D=0$, multiplying by an expression which vanishes cannot justify division. For instance, $x+y=1$ and $2x+2y=3$ are inconsistent, whereas replacing $3$ by $2$ gives infinitely many solutions. Thus an elimination formula must distinguish exceptional coefficients and preserve the connection to the original system of linear equations. Nor does the displayed rule solve the much broader problem of arbitrary nonlinear equations.

The passage's importance is the move from particular elimination calculations towards a theory of alternating expressions. It would be misleading to turn this into the claim that the familiar Vandermonde determinant, formed from powers of variables, is what this extract displays. A positional upper index and an exponent have different meanings; the history of that eponym requires separate evidence. Likewise, modern matrix algebra is a useful explanatory language here, not the institutional or conceptual setting of an eighteenth-century memoir.^[ref]

In components the scalar triple product is $A_i{ϵ}_{ijk}{B}_j{C}_k$. A cyclic permutation has positive sign of a permutation, so ${ϵ}_{ijk}={ϵ}_{kij}$. Relabelling the summed indices therefore giveswhich proves $A\cdot (B\times C)=(A\times B)\cdot C$. Equivalently both expressions are the same three-column determinant.

Reversal has one inversion of a permutation for every pair of indices, hence $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=n(n-1)/2$ inversions of a permutation. The sign of a permutation givesIt is $+1$ for $n\equiv 0,1(\mathrm{mod}4)$ and $-1$ for $n\equiv 2,3(\mathrm{mod}4)$.

The determinant is unchanged by transposition: $\boxed{\det (A^T)=\det A}$. In the Leibniz formula for determinants, replacing each permutation by its inverse interchanges the row and column roles without changing its sign of a permutation.

Compose permutations from right to left. Tracking the four symbols through the two permutation cycles gives $1\mapsto 2\mapsto 1$ and $3\mapsto 4\mapsto 3$. Thus the cycle decomposition of a permutation isThe disjoint transpositions commute, and squaring their product gives the identity element, while the product itself is nonidentity. Its order of a group element is therefore two. Each transposition has negative sign of a permutation, so the product has sign $(-1{)}^2=\boxed{+1}$: it is an even permutation.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.

Use the Leibniz definition of the determinant:For the permuted-column matrix, $a_{ij}^{\sigma }=a_{i,\sigma (j)}$. Reindex the sum by $\tau =\sigma \circ \pi$. The sign of a permutation is multiplicative, so $\mathrm{sgn}(\pi )=\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau )$. It follows directly thatFor the matrix transpose, the products in its determinant are ${\prod }_i{a}_{\pi (i),i}$. Set $j=\pi (i)$ to obtain ${\prod }_j{a}_{j,{\pi }^{-1}(j)}$, and use $\mathrm{sgn}({\pi }^{-1})=\mathrm{sgn}(\pi )$. This proves $\det (A^t)=\det A$. These facts also show that equal columns, or equal rows by transposition, force the determinant to be zero: exchanging the equal pair leaves the matrix unchanged but changes the determinant's sign.

Let $M_{ij}$ be the matrix obtained by deleting row $i$ and column $j$, with the empty determinant taken as one for $n=1$. Define the cofactor $C_{ij}=(-1{)}^{i+j}\det M_{ij}$ and the adjugate matrix by $\mathrm{adj}(A{)}_{ij}=C_{ji}$. Grouping the Leibniz sum according to the column chosen in row $i$ gives the cofactor expansion $\det A={\sum }_k{a}_{ik}{C}_{ik}$: deleting that row and its chosen column leaves a permutation of $n-1$ entries, with sign factor $(-1{)}^{i+k}$. ThereforeWhen $i=j$ this is $\det A$. When $i\ne j$ it is the cofactor expansion along row $j$ of the matrix formed by replacing that row with row $i$; its remaining minors are unchanged, and its two equal rows give determinant zero. The column version proves the other product in the same way. Thus the adjugate identity is

For the real matrices $C,D$, let $p(t)=\det (C+tD)$. The determinant definition makes this a real polynomial of degree of a polynomial at most $n$. Since $C+iD$ is invertible, the allowed converse implies $p(i)\ne 0$, so $p$ is not the zero polynomial. It has at most $n$ real roots of a polynomial. Choose a real $\lambda$ outside those roots; the adjugate argument proves that $C+\lambda D$ is invertible.

Finally write the given complex similarity transformation as $P=C+iD$. The relation $AP=PB$, with $A,B$ real, yields $AC=CB$ and $AD=DB$ by comparing real and imaginary parts. Choose the preceding $\lambda$ and put $Q=C+\lambda D$. Then $AQ=QB$, soThis proves real similarity from complex similarity by constructing a real intertwiner, rather than requiring $C$ or $D$ separately to be invertible.

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.

Compose permutations from right to left. Tracking the images gives $1\mapsto 2\mapsto 1$, $3\mapsto 4\mapsto 3$, and $5\mapsto 5$, soTwo transpositions give sign of a permutation $+1$, proving that this element belongs to the alternating group $A_5$.

Its conjugacy class consists of all double transpositions. Conjugation merely relabels the letters, so no other cycle type can occur. Conversely, choose a permutation $h\in S_5$ relabelling the two pairs into any desired two pairs. If $h$ is odd, replace it by $h(12)$: the transposition $(12)$ centralizes $(12)(34)$, so this replacement is even and gives the same conjugate. Thus every double transposition is reached by conjugation inside $A_5$.

The complete list of fifteen elements, grouped by their fixed letter, isThere are five choices of fixed letter and three pairings of the other four letters, confirming the conjugacy class size $15$.