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For a prime divisor $p$ of the order of a finite group, rotate the tuples $(g_1,\dots ,g_p)$ whose product is the identity. Rotation preserves that constraint even for a nonabelian group, by conjugating the product. Every nonfixed orbit has $p$ elements, while the fixed tuples correspond to $g^p=1$. There are $|G{|}^{p-1}$ tuples, so the number of fixed tuples is divisible by $p$. Since the identity is one, another solution exists and has order exactly $p$. This supplies an elementary proof of Cauchy theorem for groups.

If every nonidentity element of a finite group has order equal to a prime $p$, its order is a power of $p$. For any prime divisor $q$ of the group order, Cauchy theorem for groups gives an element of order $q$, forcing $q=p$. This statement does not imply that the group is abelian; odd-characteristic upper unitriangular groups provide counterexamples.

The map $g\mapsto g^{-1}$ partitions $S$ into disjoint two-element sets $\{g,g^{-1}\}$, so $|S|$ is even. Since $|G|$ is even, $|G\setminus S|$ is even as well. This complement contains the identity and consists of elements with $g^2=e$. It therefore contains at least one further element $h\ne e$, and that element has order two. Thus every finite group of even order contains an involution, without using Cauchy theorem for groups.

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

By Cauchy theorem for groups, choose $a$ of order $p$ and $b$ of order two. The subgroup $P=⟨a⟩$ has index two, hence is normal, and $b\notin P$. Thus $a,b$ generate $G$. Conjugation by $b$ has $ba{b}^{-1}=a^r$ with $r\not\equiv 0(\mathrm{mod}p)$ and $r^2\equiv 1(\mathrm{mod}p)$. Since $p$ is prime, $r\equiv 1$ or $-1$.

If $r=1$, the two generators commute and $ab$ has order $2p$, so $G$ is cyclic. If $r=-1$, then $(ab{)}^2=e$ and $ab\ne b$. The two involutions $b,ab$ generate $G$, because $(ab)b=a$. Hence $G$ is dihedral in the stated convention. Thuswhere $D_{2p}$ denotes the dihedral group of order $2p$. In both cases $P$ is a proper nontrivial normal subgroup, so $G$ is not simple.

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

For a finite p-group, the class equation iswhere the sum is over noncentral conjugacy classes. Each index is a positive power of $p$, hence divisible by $p$. Therefore $p$ divides $|Z(G)|$, and the center of a group is nontrivial.

If $|G|=p^n$ with $n>1$ and $Z(G)\ne G$, this is already a proper nontrivial normal subgroup. If $Z(G)=G$, the group is Abelian; the established case of Cauchy theorem for groups gives a subgroup of order $p$, which is normal and proper. Thus

Augustin-Louis Cauchy's importance lies in the interaction of foundations, methods and applications. He made the justification of limits, series and integrals a programme for teaching and research; developed complex analysis into a source of powerful general methods; and contributed to permutation groups, geometry and continuum mechanics. His work is best assessed through particular achievements and their limitations, rather than by treating every theorem now bearing his name as a modern statement he already possessed. His engineering training and subsequent mathematical career also make a rigid separation of pure and applied work historically unhelpful.^[ref]

The Cours d'analyse by Augustin-Louis Cauchy (1821) makes convergence a condition requiring argument. A series is studied through its partial sums; a limit describes a variable approaching a fixed value; an infinitesimal is a variable tending to zero, rather than simply a mysterious fixed quantity smaller than every positive number. The introduction explicitly challenges unrestricted transfers of rules from finite expressions to infinite ones. This is not an abandonment of calculation: it seeks to identify when calculations are valid.^[ref]

For example, the Cauchy criterion for series asks whether tails can all be made small:This is a modern formulation of the convergence requirement, with the completeness of the real numbers understood. Merely requiring $a_n\to 0$ is weaker. In the harmonic series, grouping the terms from $2^j+1$ to $2^{j+1}$ gives a block of at least $1/2$, so its tails cannot satisfy the criterion. In contrast, the root test compares a tail with a geometric series: if $\mathrm{limsup}|a_n{|}^{1/n}<q<1$, then eventually $|a_n|\le q^n$, which proves absolute convergence. Such arguments explain the usefulness of convergence tests instead of reducing Augustin-Louis Cauchy's contribution to a vocabulary of definitions.

His Résumé des leçons sur le calcul infinitésimal by Augustin-Louis Cauchy (1823) extends the programme to differentiation and integration. For a continuous function on a compact interval, an integral can be approached through sums of values times subinterval lengths. A modern justification uses uniform continuity: for a partition of mesh at most $\delta$, two choices of sample points give sums differing by at most $(b-a){\omega }_f(\delta )$, where the modulus of continuity ${\omega }_f(\delta )$ tends to zero. Refinement then establishes the common limit. This reconstruction makes clear why control of error, rather than a geometrical picture of area alone, is decisive. It does not attribute the later full Riemann integral theory for discontinuous functions to the 1823 text.^[ref]

The achievement has limits. The 1821 Sum theorem of Cauchy asserts continuity of sums of convergent series of continuous functions in terms that, under a pointwise reading, are insufficient. The original proof treats the tail as negligible without establishing the uniform control needed near the point concerned. A modern diagnostic is $f_n(x)=x^n$ on $[0,1]$: every $f_n$ is continuous, but its pointwise limit is zero for $x<1$ and one at $1$. Thus a proof must distinguish pointwise convergence from uniform convergence. Augustin-Louis Cauchy's use of variable and infinitesimal language deserves historical interpretation, but it does not license silently importing modern hypotheses into his printed theorem.^[ref]

In complex analysis, the Mémoire sur les intégrales définies, prises entre des limites imaginaires by Augustin-Louis Cauchy (1825) exemplifies a new productive setting for integration. A contour integral depends on a path in the complex plane; deforming that path and accounting for isolated singularities can evaluate otherwise difficult definite integrals. In modern form, Cauchy's integral theorem applies to a holomorphic function on a suitable domain and Cauchy integral formula states, for a circle enclosing $z$ and lying with its interior in that domain,The importance is structural: local values and derivatives are determined by data on a surrounding path. Historical proofs and hypotheses should be distinguished from later general forms of the theorem.^[ref]

A short modern consequence measures the power of this method. For an entire function bounded by $M$, the derivative version of Cauchy integral formula on a circle of radius $R$ gives $|f^{\prime }(z)|\le M/R$. Since $R$ is arbitrary, $f^{\prime }=0$, proving Liouville's theorem. If a nonconstant polynomial $P$ had no root, then $1/P$ would be an entire function, bounded outside a large disc by the leading term and inside the disc by compactness. Liouville's theorem would force it to be constant, a contradiction. This proves the fundamental theorem of algebra. The chain is an assessment of the reach of the methods, not a claim that Augustin-Louis Cauchy alone originated every named result.

Augustin-Louis Cauchy's algebra concerns concrete permutations as well as analytical formulas. His 1845 investigations developed the theory of substitutions; Cauchy theorem for groups says that a prime number dividing a finite group order occurs as an element order. One can appreciate its generality from a modern argument: rotate the coordinates of the tuples $(g_1,\dots ,g_p)$ satisfying $g_1\cdots g_p=e$. Rotation preserves this condition, since $g_2\cdots g_p{g}_1=g_1^{-1}(g_1\cdots g_p)g_1=e$. There are $|G{|}^{p-1}$ such tuples; nonfixed rotation orbits have $p$ elements, and fixed tuples correspond to $g^p=e$. When $p\mid |G|$, the number of fixed tuples is divisible by $p$, so besides $e$ there is an element of element order $p$. This is explanatory mathematics, not Augustin-Louis Cauchy's original proof. The dating itself needs care: Peter Michael Neumann's archival argument places the major burst of work in 1845 despite an 1844 date traditionally attached to a volume.^[ref]

Finally, Augustin-Louis Cauchy's work on elasticity shows the strength of making a local mathematical object out of physical interactions. Instead of describing only a force perpendicular to a surface, one studies the traction on a plane through a point, with its direction dependent on the plane's unit normal. In modern notation the relation is $\mathbf{t}(\mathbf{n})=\mathbf{\sigma }\mathbf{n}$, where the Cauchy stress tensor contains the necessary directional information. Together with balance of linear momentum, this yields equations for the internal motion of a body. The modern tensor notation is later explanatory language; the conceptual contribution is the reduction of the forces on arbitrary internal planes to a finite collection of local components.^[ref]

Augustin-Louis Cauchy's lasting contribution was a powerful research programme, not a finished foundation. His insistence on justified passage to a limit reshaped analysis, while complex integration, substitutions and local stresses opened methods whose value extended beyond their first applications. The inadequacy of a pointwise convergence hypothesis, the need to state domains in complex analysis, and the later construction of the real numbers show that nineteenth-century rigour remained a collective and revisable achievement. These qualifications strengthen an assessment of his influence by identifying what later mathematicians could extend and repair.

Let $L$ be the splitting field of the separable polynomial. Its Galois group is $G={\mathrm{Aut}}_K(L)$, acting faithfully on its $n$ distinct roots, hence as a subgroup of $S_n$. An automorphism preserves each root's minimal polynomial. Conversely an isomorphism $K(\alpha )\to K(\beta )$ for roots of the same irreducible factor extends to an embedding of $L$ in an algebraic closure, and normality of the splitting field makes the image $L$ itself. Thus the root orbits are precisely the irreducible factors' root sets. The action is transitive exactly when the polynomial is irreducible.

If $n=p$ is prime, a transitive action has $p\mid |G|$ by orbit-stabilizer, so Cauchy theorem for groups supplies an element of order $p$. Its permutation must be a $p$-cycle. Conversely a $p$-cycle already acts transitively on all $p$ roots. This proves the stated equivalence for prime degree.

The factorization theorem modulo a good prime says that if an integer polynomial retains its degree and has square-free reduction modulo $p$, and that reduction's irreducible factor degrees are $d_1,\dots ,d_r$, its Galois group over $\mathbb{Q}$ contains a permutation with disjoint cycle lengths $d_1,\dots ,d_r$. Equivalently the finite-field Galois group, generated by Frobenius on the roots, occurs as an unramified decomposition subgroup. The degree-retention qualification matters for nonmonic polynomials.

For the given quintic, Eisenstein criterion at three proves irreducibility. Modulo two it factors asNeither factor has a root in ${\mathbb{F}}_2$, so both are irreducible, and they are distinct; the reduction is separable. Thus $G$ contains an element of cycle type $(2)(3)$, whose cube is a transposition. It is also transitive of degree five.

To prove that this forces the full symmetric group here, form the graph on the five roots with an edge for each conjugate in $G$ of that transposition. The graph is $G$-invariant, and transitivity makes its connected components all the same size. That size divides five; there is an edge, so the size cannot be one. Hence the graph is connected. Transpositions along the edges of a connected graph generate all transpositions (move along a path and conjugate successive swaps), and hence generate $S_5$. They all belong to $G$, giving

Here is an involution form of the double-coset criterion for a one-point extension. Embed $G$ in $\mathrm{Sym}(\Omega \cup \{\infty \})$ by fixing $\infty$, choose $a\in \Omega$, and put $H=G_a$. The necessary and sufficient condition is the existence of an involution $t$ such thatProducts in this part are compositions of maps, with the rightmost map applied first. The criterion is a finite test on permutations of the enlarged set; it does not presuppose that a larger group has already been found.

For sufficiency, put $K=G\cup GtG$. Multiplying a double coset by an element of $G$ on either side leaves it unchanged. A product of two elements of $GtG$ reduces to $g_{1}tgt{g}_2$. If $g\in H$, then $tgt\in H\le G$; otherwise the last condition puts it in $GtG$. Hence $K$ is closed under multiplication. It contains the identity and is closed under inverses, since $t^{-1}=t$, so it is a group.

Every element $g_{1}t{g}_2$ sends $\infty$ to $g_1(a)\in \Omega$, whereas elements of $G$ fix $\infty$. Thus $K_{\infty }=G$. Moreover the orbit of $\infty$ contains $a$ and then all of $\Omega$, so $K$ is transitive and is the desired one-point extension of a permutation group. For the order check, $G\cap tGt=H$: an element of the intersection fixes both $\infty$ and $a$, and the reverse containment follows from $tHt=H$. Consequently $|GtG|=|G{|}^2/|H|=|\Omega ||G|$, as required.

For necessity, let $K$ be an extension. It is doubly transitive group action, of order $|\Omega |(|\Omega |+1)|G_a|$, which is even. By Cauchy theorem for groups it contains a nontrivial involution, and a pair interchanged by this involution can be carried to $(\infty ,a)$ by double transitivity. Conjugating gives an involution $t$ interchanging these two points. Their common point stabilizer is $H$, so $tHt=H$. Finally $G$ has just two orbits on the enlarged set, namely $\{\infty \}$ and $\Omega$, giving $K=G\cup GtG$. More explicitly, for $k\in K\setminus G$, choose $g\in G$ with $g(a)=k(\infty )$; then $t{g}^{-1}k$ fixes $\infty$ and belongs to $G$. If $g\notin H$, the element $tgt$ moves $\infty$, so it belongs to $GtG$, completing necessity.

A composition series is a subnormal chain whose nontrivial successive quotients are simple. The Jordan–Hölder theorem asserts that any two composition series of a finite group have the same length and the same multiset of simple factors, up to isomorphism. Existence follows by successively choosing maximal proper normal subgroups until the identity is reached.

For uniqueness, induct on the group order. Let $A,B$ be the penultimate terms of two series. If $A=B$, apply induction to that subgroup and add the common top factor. If $A\ne B$, their normality and maximality give $AB=G$. Put $C=A\cap B$. The isomorphism theorems giveBoth are simple. Take any composition series of $C$. Appending $A$ to it gives a series for $A$, so induction identifies the factors of the original series inside $A$ with the factors of $C$ together with $G/B$. Similarly the factors inside $B$ are those of $C$ together with $G/A$. Adding the respective top factors produces the same multiset in both cases. This proves uniqueness, including multiplicities and length.

A normal subgroup $K$ of order two is central because its automorphism group is trivial. By Jordan–Hölder, a composition series through $K$ leaves the simple quotient $G/K\cong A_5$. The image of $Z(G)$ in this quotient is central and hence trivial, since $A_5$ is nonabelian simple. ThusThe symmetric group $S_5$ has trivial group center: conjugation of each transposition by a central permutation must preserve its two-element support, forcing the permutation to fix every point. Since its Jordan–Hölder factors are $A_5$ and $C_2$, a normal subgroup of order two would equal that trivial group center, a contradiction.

Identify $C_2=\{1,-1\}$ and put $D=1\times A_5$ in $C_2\times S_5$. Its three maximal normal subgroups areThey are the three kernels of nonzero maps from the abelianization $C_2^2$ to $C_2$. There is no quotient $A_5$: a central $C_2$ must map trivially, while $S_5$ has no quotient $A_5$. The latter follows from its normal subgroups $1,A_5,S_5$, which can themselves be obtained by intersecting with the simple subgroup $A_5$ and using its trivial centralizer. The complete list of composition series isWithin $M_1$ the two direct factors are the only maximal normal subgroups; within each of $M_2,M_3\cong S_5$ the only such subgroup is its copy of $A_5$. Hence there are exactly four series.

For $S_5\times S_5$, put $D=A_5\times A_5$, $X=A_5\times 1$, $Y=1\times A_5$ andAgain these are exactly the maximal normal subgroups: they are the three cyclic quotient kernels, and there is no simple quotient $A_5$ of the whole product. All series aretogether withTo verify exhaustiveness, $M_L$ has maximal normal subgroups $D$ and $1\times S_5$, and $M_R$ has $D$ and $S_5\times 1$. A quotient $A_5$ of a direct product can receive a nontrivial image from only one simple factor, since the images commute and $A_5$ has trivial group center. The remaining maximal-normal possibility is $M_D$. Its derived subgroup is $D$, because both alternating factors are perfect. A hypothetical quotient $A_5$ would project one alternating factor isomorphically and kill the other, but conjugation by an odd diagonal element induces an outer automorphism on that surviving factor, impossible inside $A_5$. Odd conjugation is outer because otherwise an odd permutation times an even one would centralize $A_5$; a permutation commuting with every three-cycle fixes every three-element support and is the identity. Thus $D$ is the only maximal normal subgroup of $M_D$. Finally the only maximal normal subgroups of $D$ are $X,Y$: normal subgroups of a product of two centerless simple groups are products of the factors, as follows by commutating with each factor. This proves exactly eight series and the classification of composition chains in products with S5.

For $S{L}_2(5)$, commuting with both elementary upper and lower transvections forces a central matrix to be scalar. A scalar $\lambda I$ has determinant ${\lambda }^2$, soIf $g^2=I$, characteristic five makes $g$ diagonalizable with eigenvalues in $\{1,-1\}$. Determinant one rules out one of each, so the unique nonidentity involution is $-I$. An index-two subgroup would have order $60$ and, by the Cauchy theorem for groups, contain that involution and hence the group center. Its image in $PS{L}_2(5)$ would have index two. The simplicity and identification $PS{L}_2(5)\cong A_5$ are established by the independent matrix argument in Solution 4; a simple group has no proper index-two subgroup. Therefore $S{L}_2(5)$ has no subgroup of index two.

For a profinite group, its profinite Frattini subgroup isUse $G$ for an empty intersection. This is a closed normal subgroup, invariant under continuous group automorphisms, since these automorphisms permute the maximal open subgroups. Let $H=\overline{⟨X⟩}$. If $H=G$, its image certainly topologically generates $G/\Phi (G)$. Conversely, suppose $H\ne G$. For $g\notin H$, the open normal subgroup basis of a profinite group gives $N{⊴}_{o}G$ with $g\notin HN$: choose a sufficiently small neighbourhood $gN$ disjoint from the closed subgroup $H$. The proper subgroup $HN/N$ of the finite group $G/N$ lies in a maximal proper subgroup. Its preimage $M$ is a maximal proper open subgroup of $G$, contains $H$, and contains $\Phi (G)$. Since $H\Phi (G)$ is compact and therefore closed, density of the image of $X$ in $G/\Phi (G)$ would force $H\Phi (G)=G$, contradicting $H\Phi (G)\le M<G$. Generation can therefore be tested modulo the profinite Frattini subgroup.

For a finite p-group, every maximal proper subgroup is normal and has index $p$. One proof uses the normalizer condition for finite p-groups: maximality forces its normalizer to be the whole group, and its simple $p$-group quotient has order $p$. Therefore every such maximal subgroup contains powers and group commutators. Conversely, the quotient by the algebraic $G^p[G,G]$ is an elementary abelian p-group; its nonzero elements are separated by linear maps to ${\mathbb{F}}_p$. Their kernels give maximal subgroups. This proves the finite Frattini subgroup formula.

Now take a pro-p group and put $D=\overline{G^p[G,G]}$, retaining the closure of the algebraically generated subgroups. A maximal open subgroup is the preimage of a maximal subgroup of a finite $p$-group quotient, so it contains $D$. If $g\notin D$, some finite quotient of the profinite group $G/D$ detects $gD$. That quotient is elementary abelian, and a linear functional nonzero at $gD$ gives a continuous group homomorphism $G\to {\mathbb{F}}_p$ whose kernel is a maximal open subgroup missing $g$. HenceThe closure bar is present in the original PDF; the converted TeX fraction is an extraction error.

For the finite-index assertion, first pass from the arbitrary abstract subgroup $K$ to its normal core of a subgroup $C$. The action on the finitely many cosets of $K$ shows $[G:C]<\infty$, without assuming that $K$ or $C$ is closed. Write $m=[G:C]=p^{r}q$ with $(p,q)=1$, and let $X_m=\{h^m:h\in G\}$. By Lagrange theorem, $X_m\subseteq C$. For $N{⊴}_{o}G$, the quotient $G/N$ is a finite p-group of order, say, $p^s$. Choose an integer $a$ with $aq\equiv 1(\mathrm{mod}{p}^s)$. Thenso $g^{p^r}\in X_{m}N$ for every such $N$. The set $X_m$ is the continuous image of the compact space $G$, hence is closed. Consequently $g^{p^r}\in {\bigcap }_N{X}_{m}N=X_m\subseteq C$. Every element of the abstract finite quotient $G/C$ thus has order dividing $p^r$. By Cauchy theorem for groups, no prime other than $p$ divides its order. Therefore $[G:C]$, and its divisor $[G:K]$, are powers of $p$. This proof does not assume continuity of the finite quotient map.

To prove openness under finite generation, first establish commutator collection with fixed generators in nilpotent groups. If a nilpotent group $P$ is generated by $a_1,\dots ,a_d$, every element of $P\prime$ isInduct on the nilpotency class $c$. The assertion is trivial for $c=1$. Apply the induction hypothesis in $P/{\gamma }_c(P)$; the remaining error lies in the central subgroup ${\gamma }_c(P)=[{\gamma }_{c-1}(P),P]$. Since commutators in this last layer are central, collecting against words in the $a_i$ expresses the error as ${\prod }_i[y_i,a_i]$ with $y_i\in {\gamma }_{c-1}(P)$. Also $[P\prime ,{\gamma }_{c-1}(P)]\le {\gamma }_{c+1}(P)=1$. Thus replacing each $x_i$ by $x_i{y}_i$ multiplies the desired product by precisely that central error, proving the formula.

Let $a_1,\dots ,a_d$ now topologically generate $G$. Every finite continuous quotient is a nilpotent group, so the formula holds there with these fixed generators. For $z\in \overline{[G,G]}$, the sets of tuples $(x_1,\dots ,x_d)\in G^d$ solving $z\equiv {\prod }_i[x_i,a_i](\mathrm{mod}N)$ are nonempty closed subsets of a compact space. They have the finite intersection property, by replacing finitely many $N$ by their intersection. A common tuple solves the equality in $G$. It follows that the algebraic commutator subgroup is the compact image of $G^d$ under this product map, and is closed.

Its quotient $A=G/[G,G]$ is a topologically finitely generated abelian pro-p group. The continuous map ${\mathbb{Z}}_p^d\to A$ taking $({\lambda }_i)$ to ${\prod }_i{\bar{a}}_i^{{\lambda }_i}$ is onto, because its compact image contains a dense generating subgroup. Therefore $A/pA$ is generated as an elementary abelian group by at most $d$ elements, and $pA$ is compact. The preimage $G^p[G,G]$ of $pA$ is closed of finite index, hence open: its complement is a finite union of closed cosets. The finite-index result and the maximal-subgroup property of finite p-groups show that any abstract index-$p$ subgroup of $G$ is normal. It contains this open subgroup, so is itself open.

Finally, a finite-index normal subgroup $C$ has finite $p$-group quotient. If $[G:C]>1$, choose a normal index-$p$ subgroup in that quotient and call its preimage $H$. The preceding argument makes $H$ open. An open subgroup of a finitely generated pro-p group is finitely generated: intersect a finitely generated dense abstract subgroup with $H$, use the Schreier lemma, and observe that the intersection is dense in the open subgroup $H$. Induction on $[G:C]$ now makes $C$ open in $H$, hence in $G$. Returning to the normal core of a subgroup, $K$ is a union of open cosets of $C$. Every finite-index subgroup of a finitely generated pro-p group is open.

We first prove the needed instance of Cauchy theorem for groups, rather than invoke it. Let a prime number $q$ divide $|G|$ and consider the setThere are $|G{|}^{q-1}$ such tuples: choose the first $q-1$ entries freely, and the last is forced. Cyclic rotation preserves this set, because moving the first entry to the end conjugates the product, which remains $1$. Each rotation orbit of a group action has either one or $q$ elements. Indeed a nontrivial shift fixing a tuple generates all shifts, since its step has an inverse modulo the prime $q$. The fixed tuples are precisely $(g,\dots ,g)$ with $g^q=1$.

Counting nonfixed orbits in multiples of $q$, and using $q\mid |G{|}^{q-1}$, shows that the number of solutions of $g^q=1$ is divisible by $q$. The identity supplies one solution, so there is a nonidentity solution. If its order of a group element is $d$, divide $q$ by $d$: the remainder would give a smaller positive exponent producing the identity, so $d\mid q$. Hence $d=q$. This proves the required prime-order existence result by the cyclic-tuple proof of Cauchy theorem.

Our hypothesis says that this nonidentity element has order $p$, so $q=p$. Every prime divisor of $|G|$ is therefore $p$, which givesThus a finite group of prime exponent has prime-power order. The case $n=0$ includes the trivial group. No group-theoretic counting theorem was assumed in the tuple argument.

Multiplicativity of the determinant gives $\delta (gh)=\delta (g)\delta (h)$ and $\delta (1)=1$, so $\delta$ is a linear character. Its finite image is a cyclic group of roots of unity. If it contains $-1$, its order is $2m$ and the character of a representation ${\delta }^m$ maps onto $\{\pm 1\}$. The kernel of a group homomorphism is the preimage of its identity, so $\ker ({\delta }^m)$ is a normal subgroup of index $2$. The subgroup $\ker \delta$ itself need not have index $2$.

By Cauchy theorem for groups, a group $E$ of order $2k$ has an involution $g$. In the regular representation, left multiplication by $g$ is a permutation consisting of $k$ disjoint transpositions, so its determinant is $(-1{)}^k=-1$ when $k$ is odd. The preceding argument gives a normal subgroup of index two.

For a nonabelian simple group of order below $80$, Burnside's theorem requires at least three distinct prime factors. An odd order with three factors is at least $3\cdot 5\cdot 7=105$. An even order with exactly one factor of $2$ has a normal subgroup of index $2$. Thus its order must be divisible by $4$ and by at least two odd primes. Below $80$ the only possibility is $4\cdot 3\cdot 5=60$, since the next is $4\cdot 3\cdot 7=84$. Hence $\boxed{|H|=60}$.

Every finite quotient of the abelian group $A_{\neg p}$ is abelian. Part iii excludes elements of prime order $q\ne p$ by the Cauchy theorem for groups, so any finite quotient is a finite abelian $p$-group. If its exponent divides $p^n$, the quotient map kills $p^n{A}_{\neg p}$ and therefore factors throughEvery quotient of a cyclic group is cyclic, so the finite quotient is isomorphic to $\mathbb{Z}/p^k\mathbb{Z}$ for some $k\le n$. Conversely, reduction modulo $p^k$ gives a surjection $A_{\neg p}\to \mathbb{Z}/p^k\mathbb{Z}$. Thus these are exactly the nontrivial finite quotients, together with the trivial case $k=0$.

If $\gcd (m,|H|)=1$, choose $r$ with $mr\equiv 1(\mathrm{mod}|H|)$ by the Bezout identity. The Lagrange theorem gives $x^{|H|}=1$ for every $x\in H$, so $(x^m{)}^r=x$ and $(x^r{)}^m=x$. Thus $x\mapsto x^m$ is bijective, even though it need not be a homomorphism.

Conversely, if a prime $p$ divides both $m$ and $|H|$, the Cauchy theorem for groups gives $x\ne 1$ with $x^p=1$. Then $x^m=1$, so the power map sends both $x$ and the identity to the identity and is not injective. This proves the power-map criterion for a finite group.

A group action of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The orbit and stabilizer of $x$ areFor a finite group, the mapis a well-defined bijection from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the orbit-stabilizer theorem is

The Cauchy theorem for groups says that if a prime number $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, letThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition becauseEvery orbit has size one or $p$. The fixed points are exactly the tuples $(g,\dots ,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.

Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the conjugation action on the set $\mathcal{X}$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal{X}$ therefore has size $11$, soDistinct members of $\mathcal{X}$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal{X}|\le 33$, forcing $|\mathcal{X}|=1$. Thus $H$ is a normal subgroup.

Conjugation now defines a group homomorphismBecause $H\cong C_{11}$, its automorphism group has order $10$. By the Lagrange theorem, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the center of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\mathrm{lcm}(11,3)=33$. Therefore

The Lagrange theorem states that $|H|$ divides $|G|$ for every subgroup $H$ of a finite group $G$, with $|G|=|G:H||H|$. The left cosets partition $G$, and multiplication by a representative bijects $H$ with each coset, proving the formula.

The intersection $H\cap K$ contains the identity and is closed under $x{y}^{-1}$, so it is a subgroup. Its order divides both $|H|$ and $|K|$; if these are coprime, $H\cap K=\{e\}$.

The order $o(x)$ is the least positive $m$ with $x^m=e$. Division $k=qm+r$, $0\le r<m$, shows that $x^k=e$ exactly when $r=0$, so exactly when $o(x)\mid k$.

If $x,y$ commute and have coprime orders $m,n$, then $(xy{)}^{mn}=e$. Conversely $(xy{)}^r=e$ implies $x^r=y^{-r}$ lies in $⟨x⟩\cap ⟨y⟩=\{e\}$, so $m,n\mid r$ and $mn\mid r$. Hence $o(xy)=mn$.

Cauchy theorem for groups says that every prime divisor $p$ of $|G|$ occurs as the order of an element of $G$. For $|G|=26$, the Sylow $13$-subgroup $P\cong C_{13}$ is unique and normal, and Cauchy's theorem supplies a complement $C_2$. Thus $G=C_{13}⋊C_2$. The homomorphism $C_2\to \mathrm{Aut}(C_{13})\cong C_{12}$ is either trivial or has the unique image of order two, inversion. These give exactly $C_{26}$ and the dihedral group of order $26$.

The Cauchy theorem for groups says that if a prime $p$ divides $|G|$, then $G$ contains an element of order $p$.

ConsiderThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. Cyclic rotation acts on $X$: if the product is one, thenEvery orbit has size one or $p$. The fixed points are exactly the constant tuples $(x,\dots ,x)$ satisfying $x^p=1$. Since $|X|\equiv 0(\mathrm{mod}p)$, the number of fixed points is divisible by $p$. The identity supplies one, so there is a nonidentity $x$ with $x^p=1$. Its order divides the prime $p$ and is not one, hence is $p$.

For a finite group $G$, the map $x\mapsto x^m$ is bijective exactly when $m$ is coprime to $|G|$. If the two numbers are coprime, an inverse exponent modulo $|G|$ gives an inverse map. If a prime $p$ divides both, the Cauchy theorem for groups supplies a nonidentity element sent to the identity.