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For an odd prime number $p$, a finite p-group that is a product of $d(G)$ cyclic subgroups is a powerful p-group. Indeed, $H=G/G^p$ has exponent of a finite group $p$ and the same minimal generator number $d$. Its cyclic factorization gives $|H|\le p^d$, while its Frattini quotient has order $p^d$. Thus $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. The condition on the number of factors is essential: the nonabelian Heisenberg group over ${\mathbb{F}}_p$ has exponent $p$ and is a product of three cyclic subgroups, although its minimal generator number is two.

For coprime integers $a,n$ with $n>1$, the multiplicative order is the least positive integer $d$ such thatIt counts the distinct powers in the cyclic subgroup generated by the congruence class of $a$. Thus it is an order of a group element in the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$.^[ref]

For example, the modular exponentiation $2^k\mathrm{mod}7$, starting at $k=0$, produces $1,2,4,1,2,4,\dots$. The first return to one takes three steps, so ${\mathrm{ord}}_7(2)=3$. Starting at the congruence class three and repeatedly multiplying by two instead gives $3,6,5,3,\dots$: these are the other three units modulo seven, outside the cyclic subgroup generated by two.

Euler's theorem guarantees a return to one at exponent $\phi (n)$, where $\phi$ is Euler's totient function. Thus the multiplicative order exists for every unit.^[ref] In fact,By Lagrange's theorem, the size of the cyclic subgroup divides the size of the unit group, which is Euler's totient function. For two modulo seven, this is $3\mid 6$; the multiplicative order need not equal Euler's totient function. If $a\equiv 1(\mathrm{mod}n)$, its multiplicative order is one.

The coprime integers hypothesis is necessary: if $a^d=1+kn$, any common divisor of $a$ and $n$ divides one. For instance, powers of two modulo eight give $1,2,4,0,0,\dots$, with no positive-exponent return to one. This sequence eventually settles, but two has no multiplicative order modulo eight.

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 a positive integer $n$ and a congruence class $k$ in the additive finite cyclic group $\mathbb{Z}/n\mathbb{Z}$, the order of a group element isIndeed, write $d=\gcd (n,k)$, $n=d{n}^{\prime }$ and $k=d{k}^{\prime }$. Returning to zero after $m$ steps means $n\mid mk$, equivalently $n^{\prime }\mid m{k}^{\prime }$. Since $n^{\prime }$ and $k^{\prime }$ are coprime, this holds exactly when $n^{\prime }\mid m$; the least positive such $m$ is $n^{\prime }$. This argument also covers $k=0$, since $\gcd (n,0)=n$.

For $n=6$, the orders of $0,1,2,3,4,5$ are respectively $1,6,3,2,3,6$. The translation in a group $x\mapsto x+k$ has the same period from every starting element. Consequently its cycles each have length $n/d$ and there are $d$ of them. They are exactly the cosets of the cyclic subgroup generated by $k$, which explains both the lengths and the number of components in the picture.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$; in particular $|H|$ divides $|G|$. This follows by partitioning $G$ into cosets, each in bijection with $H$.

If $|G|=p$ is prime, take any $g\ne e$. Its cyclic subgroup has order dividing $p$, but its order is not one. Thus it has order $p$ and equals $G$. HenceThis describes a single group isomorphism type for each prime.

Now let $G=⟨x⟩$ have order $n$, and let $H$ be any subgroup. Choose the least positive integer $d$ with $x^d\in H$; this exists because $x^n=e\in H$, even when $H$ is trivial. Divide $n=qd+r$ with $0\le r<d$. Since $x^r=x^n(x^d{)}^{-q}\in H$, minimality forces $r=0$, so $d$ divides $n$. Likewise, dividing an exponent $k=qd+r$ for any $x^k\in H$ shows $r=0$. Therefore $H=⟨x^d⟩$.

Conversely, for each divisor $d$ of $n$, $⟨x^d⟩$ is a subgroup of order $n/d$. The least positive exponent property makes $d$ unique. This proves the complete subgroups and quotients of a cyclic group classification:The choices $d=n$ and $d=1$ include the trivial subgroup and the whole group.

A group $G$ is a cyclic group if some element $g\in G$ generates it:For a finite cyclic group, the order of a group element of a generator equals the number of elements of the group.

An example of a finite abelian group that is not a cyclic group is the Klein four-group, realized aswith componentwise addition modulo two. Addition commutes, so this is an abelian group of four elements. Each of its three nonzero elements satisfies $g+g=0$, so the cyclic subgroup it generates has just two elements; the identity generates only itself. No element generates all four elements.

Work in the based CW homotopy category. A homotopy-invariant functor depends on a map only through its based homotopy class. In a sequence of pointed sets, exactness at the middle means that the preceding image equals the inverse image of the distinguished point; it does not imply injectivity or surjectivity at the ends. For groups this is the familiar image-equals-kernel condition. A half-exact functor preserves such middle exactness on the appropriate elementary sequence.

For a contravariant functor $F$, the relevant elementary sequence is a cofibration $A\hookrightarrow X\to X/A$. Half-exactness isexact at $F(X)$. One should not silently turn this into a short exact sequence. In particular, arbitrary group actions and pointed sets at low degrees are not interchangeable with abelian groups.

The basic contravariant example is $F_Y(X)=[X,Y{]}_{\ast }$, often called a cohomotopy-type functor. Suppose a map $f:X\to Y$ becomes null-homotopic on $A$. A chosen null-homotopy extends it over the cone attached to $A$, giving a map from the topological mapping cone to $Y$. Conversely every such extension restricts to a null-homotopic map on $A$. Since the topological mapping cone of a subcomplex inclusion models $X/A$, this proves half-exactness, rather than merely asserting it.

Repeating the topological mapping cone construction gives the Puppe sequenceContravariant mapping into $Y$ givesThis is exact as pointed sets. An H-group structure on $Y$, for example a loop space with path concatenation, provides group structures; repeated looping gives abelian group structures in the usual ranges. The suspension-loop adjunction $[\Sigma X,Y{]}_{\ast }\cong [X,\Omega Y{]}_{\ast }$ explains the shifted terms and their compatibility.

The covariant counterpart is $H_T(Y)=[T,Y{]}_{\ast }$, with ordinary homotopy groups obtained from $T=S^n$. Its natural elementary sequence is a homotopy fiber sequence $F\to E\to B$, not an arbitrary cofibration. A map $a:T\to E$ whose composite to $B$ is null-homotopic, together with that null-homotopy, determines a map into the homotopy fibre. Conversely a fibre map projects to such a null composite. This proves exactness of $[T,F{]}_{\ast }\to [T,E{]}_{\ast }\to [T,B{]}_{\ast }$ at the middle. Adding the loop terms produces the homotopy long exact sequence. For a Serre fibration, its ordinary fibre supplies the same homotopy information.

This variance distinction matters. Include one circle, generated by $a$, in $S^1\vee S^1$, with other generator $b$. The quotient is the second circle. The kernel of ${\pi }_1(S^1\vee S^1)\to {\pi }_1(S^1)$ is the normal closure of $a$, which contains $ba{b}^{-1}$. The image of the included circle is only the cyclic subgroup generated by $a$, and does not contain that reduced word. Thus covariant ${\pi }_1$ is not half-exact on this cofibration. Fibre and cofibre statements cannot be swapped just by reversing notation.

Reduced cohomology gives group-valued contravariant examples. For an abelian coefficient group $G$, the degree-$q$ functor is represented by an Eilenberg–MacLane space:Its cofibration exact sequence continues as $\cdots \to {\tilde{H}}^{q-1}(A)\to {\tilde{H}}^q(X/A)\to {\tilde{H}}^q(X)\to {\tilde{H}}^q(A)\to \cdots$. Ordinary homotopy groups are instead covariant, with decreasing group degree across a fibre connecting map. These are the two complementary representable patterns, maps out of a test object and maps into a representing object.

Brown's representability theorem makes the contravariant pattern structural. A reduced contravariant homotopy functor on based CW complexes is represented by a CW complex if it converts arbitrary wedges to products and satisfies the Mayer-Vietoris weak-pullback gluing axiom. The latter says that compatible classes on two subcomplexes extend to their union. It contains more information than the single kernel statement of half-exactness. Consequently half-exactness alone is not a representability criterion. In a representing functor these axioms are natural: a map on a wedge is a family of based maps, while compatible maps on an intersection can be adjusted by a homotopy and glued across a homotopy pushout.

A reduced generalized cohomology theory with the usual wedge axiom supplies these axioms degree by degree. Its representing spaces $E_q$ have $E_q\simeq \Omega E_{q+1}$ by suspension compatibility, assembling an Omega-spectrum. Thus generalized cohomology can be expressed as $E^q(X)=[{\Sigma }^{\infty }X,{\Sigma }^{q}E{]}_{\mathrm{st}}$, while represented homology is $E_q(X)={\pi }_q(E\wedge {\Sigma }^{\infty }X)$. The appearance of mapping into a spectrum in one case and smashing with it in the other is another form of the same variance distinction.

There is a literal duality after stabilization and a finiteness restriction. Here a finite based CW complex is identified with its suspension spectrum. For a finite spectrum $X$, let $DX=F(X,\mathbb{S})$ be its Spanier-Whitehead dual. Then $D^{2}X\simeq X$ and $[X,Y{]}_{\mathrm{st}}\cong [DY,DX{]}_{\mathrm{st}}$ for finite spectra. Duality reverses a cofibre triangle into a fibre triangle. In particular,This turns a cohomology calculation on a finite object into a homology calculation on its dual, and vice versa. It is Spanier-Whitehead duality, not a claim that the full category of ordinary spaces is self-dual.

Half-exactness is the local image–kernel property; fibres and cofibres, representability, and finite stable duality explain how that property organizes homotopy and cohomology. The distinctions between variance, pointed sets and groups, and finite stable objects and arbitrary spaces are essential to using the duality correctly.

For a subgroup $H$ of a finite group $G$, Lagrange's theorem states $|G|=[G:H]|H|$. Thus a subgroup's order, and in particular the order of a group element, divides $|G|$.

For $|G|=10$, if every element had order one or two, the order of a finite group of exponent two would make $|G|$ a power of two, a contradiction. Hence an element has order five or ten. In the latter case $G$ is the cyclic group $C_{10}$. Otherwise choose $a$ of order five. Its cyclic subgroup $H=⟨a⟩$ has index two and is therefore a normal subgroup.

In the noncyclic case an element $b$ outside $H$ has order two: its order divides ten, it cannot have order ten, and an element of order five has trivial image in the order-two quotient group $G/H$. Every element of $G$ is then $a^j$ or $a^{j}b$. Conjugation by $b$ restricts to an automorphism of $H$, say $ba{b}^{-1}=a^r$. Since $b^2=1$, $r^2\equiv 1(\mathrm{mod}5)$, so $r=1$ or $-1$ modulo five.

For $r=1$, the generators commute and $ab$ has order ten, yielding the cyclic case. For $r=-1$, these are the relations of the dihedral group of the pentagon. The classification of groups of order ten is therefore exactly $C_{10}$ and $D_{10}$, where the subscript on $D_{10}$ denotes its order. They are not isomorphic because one is Abelian and the other is not.

For each $j\ge 0$, the elementary abelian group $P_j/P_{j+1}$ is spanned by the images of $a_i^{p^j}$. We construct exponents $e_i$ so that $g$ equals the ordered product $a_1^{e_1}\cdots a_n^{e_n}$ to successively greater precision. Suppose this product agrees with $g$ modulo $P_j$; initially this is automatic for $j=0$. Its error modulo $P_{j+1}$ can be expressed using the images of $a_i^{p^j}$. Add the corresponding digits $p^j{t}_i$ to the exponents $e_i$. Since $[P_j,G]\subseteq P_{j+1}$, moving each new factor $a_i^{p^j{t}_i}$ past the other factors changes nothing modulo $P_{j+1}$. The new ordered product therefore agrees with $g$ modulo $P_{j+1}$. Termination of the lower p-series provesNo uniqueness of this factorization is asserted for a redundant generating tuple.

The printed converse is false without an additional condition. For any odd $p$, take the Heisenberg group $G={\mathrm{UT}}_3({\mathbb{F}}_p)$ withEvery element is uniquely $x^a{y}^b{z}^c$, with $a,b,c\in {\mathbb{F}}_p$, so $G$ is a product of three cyclic subgroups. For $A$ strictly upper triangular, $A^3=0$ andin characteristic $p$, including $p=3$. Therefore $G^p=1$, whereas $[G,G]=⟨z⟩\ne 1$. This finite p-group is not powerful, disproving the literal request in the authoritative PDF.

A precise nearby correction is the minimal cyclic factorization criterion for powerful p-groups: require a product of $d(G)$ cyclic subgroups, where $d(G)$ is the minimal generator number. To prove it, put $d=d(G)$ and $H=G/G^p$. The quotient has exponent $p$ and still has minimal generator number $d$, because $G^p\subseteq \Phi (G)$. Each cyclic factor has image of order at most $p$, so $|H|\le p^d$. But $|H/\Phi (H)|=p^d$ by the Burnside basis theorem. Consequently $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. A product of exactly $d(G)$ cyclic subgroups is powerful. A direct product of cyclic groups is also powerful, since it is abelian; neither stronger hypothesis appears in the printed converse.

A cyclic group is a group generated by a single element: there is $g\in G$ such that every element is $g^n$ for some integer $n$. For a finite group of prime order $p$, choose $g\ne e$. Its cyclic subgroup $⟨g⟩$ has order dividing $p$ by Lagrange's theorem, and its order is greater than one. Hence $|⟨g⟩|=p$, and $⟨g⟩=G$. Thus every group of prime order is cyclic.

A group homomorphism from $C_{11}=⟨a⟩$ is determined by $h=\phi (a)$, since $\phi (a^j)=h^j$. Its image generator must satisfy $h^{11}=e$, so its order of a group element divides $11$. As an element of $C_{14}$ its order also divides $14$. These integers are coprime, so the only possible order is one. Thus $h=e$, andThis map is indeed a group homomorphism, so there is exactly one, the trivial homomorphism.

An equivalence relation is reflexive, symmetric and transitive: for all $a,b,c\in S$, one has $aRa$; $aRb$ implies $bRa$; and $aRb$, $bRc$ imply $aRc$. Define the equivalence class $[a]=\{b\in S:aRb\}$. Reflexivity puts $a$ in its class, so these classes cover $S$ and are nonempty. If $[a]$ and $[b]$ share $c$, symmetry and transitivity give $aRb$; the same properties show that every element of either class is in the other. Thus two classes are equal or disjoint. The distinct equivalence classes form a partition of the set.

For the relation defined by a subgroup $H\le G$, $a^{-1}a=1\in H$ proves reflexivity. If $a^{-1}b\in H$, its inverse $b^{-1}a$ belongs to $H$, proving symmetry. If also $b^{-1}c\in H$, their product is $a^{-1}c\in H$, proving transitivity. Finally,These are exactly the left cosets. The map $H\to gH$, $h\mapsto gh$, is a bijection, with inverse $x\mapsto g^{-1}x$. If $G$ is finite and has $r$ such cosets, their disjoint union gives $|G|=r|H|$. This proves the Lagrange theorem:For $g\in G$, finitely many possible powers imply $g^a=g^b$ for some $a<b$, so a positive power is the identity element. If its least positive exponent is $n$, the powers $1,g,\dots ,g^{n-1}$ are distinct: equality of two would produce a smaller positive exponent. They are closed under products and inverses by reducing exponents modulo $n$, hence form the cyclic subgroup $⟨g⟩$ of cardinality $n$. The Lagrange theorem gives $n\mid |G|$, and consequentlyNow take a positive modulus $m$. On residue classes modulo $m$, consider the classes with representatives coprime to $m$. The following integer facts are used: multiplication is associative; reducing representatives respects products; a prime dividing a product divides a factor; and the Bezout identity gives $ra+sm=1$ when $\gcd (a,m)=1$. If $a,b$ are coprime to $m$, no prime divisor of $m$ divides $ab$, so their product class is again in the set. The class of $1$ is its identity element. The Bézout equation gives $ra\equiv 1(\mathrm{mod}m)$, and $r$ is also coprime to $m$ because any common divisor of $r,m$ would divide $1$. Thus every class has an inverse, and these classes form the group of units modulo an integer.

Representing classes in $\{1,\dots ,m\}$ gives precisely the group in the question; for $m=1$ this is the one-element group. Its cardinality is the Euler totient function $\phi (m)$. Applying the finite-group exponent result yields the Fermat-Euler theorem:For $m=1$ the congruence is automatic.

First prove Simplicity of alternating groups. The $3$-cycles generate $A_n$: express an even permutation as an even number of transpositions and pair them. A pair sharing a point is a $3$-cycle, an identical pair cancels, and a disjoint pair can be written $(ab)(cd)=(acb)(acd)$. All $3$-cycles are conjugate in $A_n$ when $n\ge 5$: a conjugator in $S_n$ can have its parity corrected by a transposition on two points outside the target triple, without changing its effect on that triple.

Let $1\ne N◃A_n$ and choose $1\ne \sigma \in N$. Some $3$-cycle $\tau$ does not commute with $\sigma$. Indeed, commuting with every $3$-cycle would preserve every triple setwise; intersecting all triples containing a given point forces that point to be fixed, so would make $\sigma =1$. The nonidentity group commutator $\sigma \tau {\sigma }^{-1}{\tau }^{-1}\in N$ is supported on at most six points. Its possible nontrivial even cycle types are a $3$-cycle, a double transposition, a $5$-cycle, two disjoint $3$-cycles, or a $4$-cycle times a transposition. Each yields a $3$-cycle in $N$ by explicit group commutators:

For $(ab)(cd)$, use $\tau =(cde)$ with a fifth point $e$; conjugation reverses $\tau$, so the group commutator is ${\tau }^{-2}=\tau$. For $(abcde)$, use $(abc)$: the group commutator is $(bcd)(abc{)}^{-1}=(adb)$. The same calculation works for $(abcd)(ef)$. For $(abc)(def)$, use $(abd)$; its conjugate is $(bce)$, and the product $(bce)(abd{)}^{-1}$ is a $5$-cycle, reducing to the preceding case. Thus $N$ contains a $3$-cycle, hence the whole conjugacy class of $3$-cycles, and hence all of $A_n$. ThereforeFor the exceptional automorphism, a duad is an unordered pair of six points, a syntheme is a partition into three duads, and a pentad is a set of five synthemes covering every duad once. There are fifteen synthemes. To count pentads, fix $12|34|56$, where bars separate duads. A syntheme disjoint from it in duads is one of eight; the union of the two matchings is a six-cycle, so the stabilizer of the fixed syntheme is transitive on these eight choices. For the choice $13|25|46$, the remaining compatible synthemes are $14|26|35$, $15|24|36$, $16|23|45$ and $16|24|35$. The last shares a duad with each of the first three and cannot belong to a completion; the first three form the unique completion. Thus each of the eight choices has a unique completion through the fixed syntheme. Each completion contains four of them, so exactly two pentads contain a fixed syntheme, and there are $15\cdot 2/5=6$ pentads in total.

Here is the complete list, which also defines concrete labels for the induced action:Relabelling points gives a group homomorphism $\Phi :S_6\to S_6$ on these pentads. Direct substitution in the table givesThe restriction to $A_6$ is nontrivial, hence faithful by the simplicity just proved. Therefore the full kernel has order at most two. An order-two normal subgroup would be central, whereas $S_6$ has trivial centre: an element centralizing every transposition fixes every unordered pair and consequently every point. The kernel is trivial, and equal orders make $\Phi$ an automorphism. It preserves $A_6$, because that is the unique index-two subgroup of $S_6$: any nontrivial group homomorphism to $C_2$ sends every conjugate transposition to the nonidentity element, and is therefore the sign map. Its restriction is the pentad construction of the exceptional alternating-group automorphism. It maps a $3$-cycle to two disjoint $3$-cycles, while conjugation in $S_6$ preserves cycle type. Hence this automorphism of $A_6$ is not induced by any element of $S_6$.

For the other degrees, recover the points from $3$-cycles. An element of order three has type $3^k{1}^{n-3k}$. Its centralizer in $S_n$ has order $3^{k}k!(n-3k)!$. That centralizer contains an odd permutation: either swap two fixed points, or, when $k\ge 2$, swap two three-cycles by three transpositions. ThusFor $n=5$, only $k=1$ is possible. For $n\ge 7$, the $k=1$ value is strictly largest. For $k=2$ equality first occurs at $n=6$, and the ratio increases strictly with $n$. For $k\ge 3$, already at $n=3k$ one has $(3k-3)!>3^{k-1}k!$: it holds at $k=3$, and the inductive ratio is $(3k)(3k-1)(3k-2)>3(k+1)$. Increasing $n$ again increases the ratio. Consequently every automorphism preserves the set of $3$-cycles when $n\ne 6$.

Each cyclic subgroup generated by a $3$-cycle corresponds to its three-point support. Two distinct such subgroups commute exactly when the supports are disjoint. If they meet in one point, products of their nonidentity generators have order five. If they meet in two points, those products have order two or three. These intrinsic tests recover adjacency in the Johnson graph $J(n,3)$, so the automorphism acts on its vertices.

We can reconstruct the underlying points without an additional group-theoretic bound. The maximal cliques of a Johnson graph in $J(n,3)$ are the $n-2$ triples containing a fixed pair, and the four triples contained in a fixed four-set. To verify the classification, take adjacent triples $D\cup \{a\}$ and $D\cup \{b\}$ with $|D|=2$. Any common adjacent triple either contains $D$ or lies in $D\cup \{a,b\}$. A triple of the first kind outside that four-set is not adjacent to a triple of the second kind omitting a point of $D$, so a clique cannot mix the two alternatives. Extending a clique therefore gives exactly one of the two listed families.

For $n\ge 5$, $n\ne 6$, their sizes distinguish the pair-containing cliques. Their intersections reconstruct the pair graph $J(n,2)$: two such cliques intersect precisely when their defining pairs meet in one point. In this graph the maximal cliques are stars of $n-1$ pairs through a point and triangles of three pairs within a triple. These sizes are different for $n\ge 5$, so its automorphisms permute the stars and hence the points. Every pair is the intersection of its two point stars, and every triple is determined by its three pairs. Thus the original graph action comes from some $\pi \in S_n$.

Undo conjugation by $\pi$. The remaining group automorphism $\beta$ fixes each cyclic $3$-subgroup, so it fixes or inverts each generator. For adjacent supports, inverting exactly one generator changes the order of their product between two and three; inverting both preserves that order. Hence the two inversion choices must agree. The triple-support graph is connected, since one can replace differing points one at a time, so the choice is uniform. Inverting every $3$-cycle is impossible: take $c=(123)$, $d=(142)$, whose product is the $3$-cycle $(143)$. Then $\beta (cd)=(cd{)}^{-1}=d^{-1}{c}^{-1}$, whereas the group homomorphism property would give $c^{-1}{d}^{-1}$; these are unequal because $c,d$ do not commute. Therefore $\beta$ fixes every $3$-cycle and is the identity. Conversely, conjugation by $S_n$ acts faithfully on $A_n$, since centralizing all $3$-cycles fixes every triple and every point. The automorphisms of alternating groups from triple supports are exactly

Choose a minimal nontrivial normal subgroup $L$ of $K$ itself. Every conjugate $L^g$ for $g\in G$ is also a minimal normal subgroup of $K$, since $K$ is normal in $G$. Distinct such subgroups intersect trivially: their intersection is normal in $K$ and minimality forces either equality or trivial intersection. Moreover $[L_i,L_j]\subseteq L_i\cap L_j=1$, so distinct ones commute.

Choose a maximal family of these conjugates whose product $M=L_1\times \cdots \times L_r$ is direct. For any further conjugate $L^g$, its intersection with $M$ is normal in $K$ and is either all of $L^g$ or trivial. In the latter case it commutes with all the existing factors and could be adjoined to the direct product of groups, contrary to maximality. Thus every conjugate is contained in $M$. The product of all conjugates is consequently $M$, which is nontrivial and normal in $G$. Minimality of $K$ as a $G$-normal subgroup gives $M=K$.

Any normal subgroup of a factor $L_i$ is normal in $K$: the other direct factors commute with it, and $L_i$ normalizes it. Minimality of $L_i$ as a $K$-normal subgroup makes it simple. All factors are isomorphic because they are conjugate to $L$. This proves the direct-product structure of a finite minimal normal subgroup:An abelian simple group is cyclic of prime order, so the abelian case is elementary abelian. Now let $H$ be a maximal proper subgroup of a nonabelian simple group $X$. It is nontrivial, since $X$ has nontrivial proper cyclic subgroups. Choose a minimal normal subgroup $K$ of $H$. Then $H\le N_X(K)$. This normalizer cannot be $X$, since that would make $1<K\le H<X$ normal in $X$. Maximality givesFor each of the two groups in this question, the allowed absence of proper nonabelian simple subgroups makes the factors of such a $K$ cyclic of prime order. Hence it suffices to examine normalizers of nontrivial elementary abelian subgroups.

In $A_5$, the possible primes are two, three and five. The Sylow subgroups for the last two primes are cyclic of orders three and five. Every involution is a double transposition; for $(12)(34)$ the centralizer in $A_5$ isthe Klein four-group fixing point five. Thus an elementary abelian $2$-subgroup is cyclic of order two or a Klein four-group. The normalizer of a cyclic order-two subgroup equals its centralizer and has order four, so is contained in a point stabilizer of order twelve. A Klein four-group's common fixed point is intrinsic, and the full point stabilizer $A_4$ normalizes it; its normalizer is therefore exactly that stabilizer.

An order-three subgroup has support a triple. Its normalizer preserves the triple and the remaining pair, and is $(S_3\times S_2)\cap A_5$, of order six. An order-five subgroup acts regularly on all five points. Label the points by ${\mathbb{F}}_5$, with the subgroup acting by translations. Its normalizer in $S_5$ is the twenty affine maps $x\mapsto ax+b$. The translations are even, whereas multiplication by a generator of ${\mathbb{F}}_5^{\times }$ is a four-cycle and is odd. Exactly half these maps are even, so the normalizer in $A_5$ has order ten.

The only candidates for maximal subgroups are thus the normalizers of orders twelve, ten and six. They are all maximal: none can lie in a larger candidate, except potentially an order-six subgroup in an order-twelve $A_4$. But an index-two subgroup of $A_4$ would contain every order-three element; there are eight such elements, plus the identity, already more than six. Hence that containment is impossible. Point stabilizers are conjugate; order-three and order-five subgroup normalizers each form one class by the Sylow theorems. Therefore the maximal subgroups of A5 form preciselyTheir numbers are five, six and ten respectively, since these proper maximal subgroups are self-normalizing in the simple group.

For $X=G{L}_3(2)$, choosing independent columns gives $|X|=(8-1)(8-2)(8-4)=168$. If $t$ is an involution, write $t=I+N$. Field characteristic two gives $N^2=0$, and $\mathrm{im}N\subseteq \ker N$ in dimension three forces its nonzero rank to be one. Thus $t=I+vf$, with nonzero vector $v$, nonzero linear functional $f$ and $f(v)=0$. This transvection fixes the plane $\ker f$ and has image line $⟨v⟩$. Every flag consisting of an incident line and plane occurs, giving $7\cdot 3=21$ involutions, all conjugate. Its centralizer, and hence the normalizer of its cyclic subgroup, is the flag stabilizer of order $168/21=8$. It lies in both the line and plane stabilizers of order $24$, so cannot be maximal.

For the elementary abelian subgroups in GL3 over F2, consider commuting transvections $I+vf$, $I+wh$. Their rank-one parts satisfyIf $v,w$ are independent, equality forces both sides zero; thus $f,h$ vanish on $⟨v,w⟩$ and are the same nonzero functional. If $v=w$, they have a common image line. Consequently the two types of order-four elementary abelian subgroup areNo extra transvection can commute with all elements of $K_L$: if its image were different, its functional would have to equal each of the two independent functionals occurring in $K_L$. Dually, the same holds for $K_W$. Thus these exhaust the elementary abelian $2$-subgroups of rank at least two. The normalizer of $K_L$ is the line stabilizer and the normalizer of $K_W$ is the plane stabilizer, each of order $168/7=24$. They give different conjugacy classes because their common fixed spaces have dimensions one and two. All line stabilizers are conjugate, as are all plane stabilizers.

For an order-three element, its minimal polynomial is $(X-1)(X^2+X+1)$: it has a fixed line and an invariant complementary irreducible plane. Its normalizer preserves both and is the full $G{L}_2(2)$ on that plane, of order six, since the order-three subgroup is normal in $G{L}_2(2)$. This normalizer lies in the corresponding line and plane stabilizers, so is not maximal.

An order-seven element is irreducible. Over the finite field ${\mathbb{F}}_2$ one has $X^7-1=(X-1)(X^3+X+1)(X^3+X^2+1)$, and both cubic factors are irreducible because neither has a root in ${\mathbb{F}}_2$. Its minimal polynomial divides this square-free product and cannot be just $X-1$, since the element is not the identity. It therefore contains a cubic factor; the three-dimensional space forces that irreducible cubic to be the whole minimal polynomial. Identify the space with ${\mathbb{F}}_8$ and the element with multiplication by a generator $\alpha$ of ${\mathbb{F}}_8^{\times }$. Its centralizer consists of the seven nonzero multiplication maps. A normalizer induces an ${\mathbb{F}}_2$-automorphism of this field, so it sends $\alpha$ to $\alpha$, ${\alpha }^2$ or ${\alpha }^4$. All three possibilities occur via the field automorphisms $x\mapsto x^{2^j}$. Therefore the normalizer of the Singer cycle has order $7\cdot 3=21$, and all such normalizers are conjugate by the Sylow theorems.

The maximal-normalizer reduction leaves exactly the two order-twenty-four stabilizer types and the order-twenty-one type. Each is maximal because it cannot be contained in any larger remaining candidate; equal-order subgroups cannot properly contain one another and $21$ does not divide $24$. Thus

Let a finite group $G$ act on a set $X$. For $x\in X$, define its stabilizer subgroup $G_x=\{g:g\cdot x=x\}$ and its group orbit $Gx=\{g\cdot x:g\in G\}$. The mapis well-defined, because multiplying $g$ on the right by an element fixing $x$ does not change its image. It is onto by the definition of the orbit. If $g\cdot x=h\cdot x$, then $h^{-1}g\in G_x$, so $g{G}_x=h{G}_x$; hence it is one-to-one. Every coset has $|G_x|$ elements and the cosets partition $G$. This proves the orbit-stabilizer theorem:For a group element $x$, its cyclic subgroup $H=⟨x⟩$ has $\mathrm{ord}(x)$ elements. Apply the theorem to the action of $G$ on its left cosets $G/H$, whose stabilizer at $H$ is $H$. It follows thatThis is the needed case of Lagrange's theorem, obtained directly from the action.

To prove Cauchy's theorem for finite groups, let $p$ be a prime divisor of $|G|$ and considerThe first $p-1$ entries are arbitrary and determine the last, so $|\mathcal{X}|=|G{|}^{p-1}$ is divisible by $p$. Cyclically rotate the entries. This preserves the product constraint even if $G$ is nonabelian, sinceThus a cyclic group of order $p$ acts on $\mathcal{X}$. By the orbit-stabilizer theorem, its orbits have size one or $p$. Its fixed tuples are exactly $(g,\dots ,g)$ with $g^p=e$. If $m$ is their number, counting the remaining size-$p$ orbits gives $m\equiv |\mathcal{X}|\equiv 0(\mathrm{mod}p)$. The identity provides one fixed tuple, so $m\ge p$ and a nonidentity $g$ with $g^p=e$ exists. The order of this $g$ divides $p$: division of $p$ by its order and minimality of that order prove this directly. Since $p$ is prime and $g\ne e$, the order is exactly $p$. This is the cyclic-tuple proof of Cauchy theorem.

Now suppose every nonidentity element has order two. Then every element equals its inverse, and for any $a,b\in G$,Thus group of exponent two is abelian applies. Starting from the identity subgroup, choose an element outside the subgroup already generated. Because the group is abelian and that element squares to the identity, adjoining it doubles the subgroup: it is the disjoint union of the old subgroup and its translate by the new element. Repeating until the finite group is exhausted shows $|G|=2^r$. Conversely the direct product of groups $C_2^r$ has order $2^r$ and every nonidentity element has order two. ThereforeThe case $r=0$ is the trivial group and satisfies the condition vacuously. These finite examples are elementary abelian groups of exponent two.

An infinite example is the group of all binary sequences, with coordinatewise addition modulo two:Every sequence added to itself is zero, so every nonzero sequence has order exactly two. The sequences having a single $1$ in coordinate $j$ are all distinct, proving that the group is infinite.

First prove the torsion in an HNN extension statement directly from Britton's lemma. Let $z$ have finite order in an HNN extension of $G$. Among its conjugates, choose a reduced normal-form word with the smallest possible number $k$ of stable letters. If $k=0$, that conjugate lies in the embedded base $G$, so its finite order and the base's torsion-free group property force it to be the identity.

Suppose instead that $k>0$. There are no internal pinches in an HNN extension because the word is reduced. There is also no pinch at the join of its end with its beginning. Indeed, write it as $g_0{t}^{{\epsilon }_1}{g}_1\cdots t^{{\epsilon }_k}{g}_k$ and conjugate off $g_0$ to obtain $t^{{\epsilon }_1}{g}_1\cdots t^{{\epsilon }_k}(g_k{g}_0)$. If the joined subword $t^{{\epsilon }_k}(g_k{g}_0)t^{{\epsilon }_1}$ were a pinch, its two signs would be opposite; conjugating by $t^{{\epsilon }_1}$ and using the pinch relation would reduce the stable-letter count by two, contradicting minimality. Thus every positive power is reduced with exactly $rk$ stable letters in the $r$th power. Britton's lemma makes every such power nonidentity, contradicting finite order. We have provedThe base embeds by the same lemma, so finite order has not been lost or artificially introduced by a quotient.

For the countable torsion-free embedding with two conjugacy classes, start with $G_0=G$. Given a countable torsion-free $G_j$, list all ordered pairs $(a,b)$ of its nonidentity elements. Both elements have infinite order, so the assignment $a^r\mapsto b^r$ is an isomorphism between their infinite cyclic subgroups. Successively adjoin one stable letter for each pair with relation $t_{a,b}^{-1}a{t}_{a,b}=b$. Each HNN extension embeds the previous group and is torsion-free by the result just proved. Let $G_{j+1}$ be the union after all pairs from $G_j$ have been treated. It remains countable because only countably many generators were added, and torsion-free because any element and any purported finite-order equation occur in a finite stage.

Repeat this whole process: new stable letters and their products must also be included in later pair lists. In the final increasing unionthe original $G$ embeds and torsion-freeness and countability persist. Any two nonidentity elements lie together in some $G_j$ and are conjugate in $G_{j+1}$. Thus every countable torsion-free group embeds in a countable torsion-free group in which all nonidentity elements are conjugate. If the initial group is nontrivial there are exactly two conjugacy classes; for the trivial group the stated condition is vacuous, or one can begin with $\mathbb{Z}$ instead. The listing is set-theoretic; no algorithm for recognizing nonidentity elements is assumed.

Write $G=⟨a,b⟩$ and $c=[a,b]$. The commutator subgroup $G\prime$ is the normal closure of $c$: imposing $c=1$ makes the two generators commute, so the quotient is abelian. Part (iii), with $N=G$, makes $G\prime$ a powerfully embedded subgroup. Part (ii) therefore givesThis proves the hint, but cyclicity of the commutator subgroup alone is not yet the required cyclic normal subgroup with cyclic quotient.

If $c=1$, the group is abelian and any one of the two generator subgroups is normal with cyclic quotient. Otherwise $c\in G^p$. Part (v) permits taking a $p$th root of $c$. Whenever the chosen root is still in $G^p$, take another $p$th root. The order increases by a factor $p$ at every step, since the power being rooted is nonidentity. Finiteness forces this process to stop. Thusfor some $d$. The cyclic subgroup $A=⟨d⟩$ contains $G\prime$, and hence is normal: every conjugate of $d$ differs from $d$ by an element of $G\prime \le A$. The nonzero image of $d$ in the Frattini quotient can be completed to a basis of dimension at most two. By the Burnside basis theorem, $G=⟨d,e⟩$ for a suitable $e$; if the quotient has dimension one, $d$ already generates $G$. Therefore $G/A$ is cyclic. The powerful two-generator finite p-group is metacyclic.

The correspondence concerns uniform pro-p groups and powerful Lie lattices, with their natural morphisms. Put $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. A powerful pro-p group is a pro-p group whose closed commutator subgroup lies in its closed $p^{ϵ}$-power subgroup. It is uniform when it is topologically finitely generated and all successive indices in its lower p-series are equal. Equivalently, a finitely generated powerful pro-p group is uniform precisely when it is a torsion-free group. A powerful Lie algebra over p-adic integers means a Lie algebra over ${\mathbb{Z}}_p$ that is free of finite rank as a module and satisfiesHere a Lie algebra over a ring is a module with an alternating bilinear Lie bracket satisfying the Jacobi identity. Finite freeness is an essential part of the lattice convention. The zero-bracket module ${\mathbb{F}}_p$ satisfies the bracket inclusion but has torsion, and its additive group is not uniform. Thus allowing arbitrary torsion modules would make the categorical assertion false.

The basic structural facts behind the construction can be seen from the power layers. Write $P_n=G^{p^n}$ with $P_0=G$. Collection for a powerful pro-p group givesFor odd $p$, the power lifting is exactly the successive-correction argument of Question 3, followed by compactness; at $p=2$ the stronger fourth-power commutator condition makes the corresponding collection corrections work. The maps between successive layers are surjective linear maps. Uniformity makes their finite dimensions equal, so these maps are isomorphisms. Comparing the first layer in which two elements differ now shows that the power map $P_n\to P_{n+1}$ is injective, while the power-set formula makes it surjective. Thus all the roots used below exist uniquely in the indicated layer. If $g_1,\dots ,g_d$ lift a basis of $G/P_1$, the images of $g_i^{p^n}$ form a basis of $P_n/P_{n+1}$. Successively choosing their coefficients modulo $p$ gives the unique ordered expressionExistence follows by lifting one layer at a time and taking a limit; uniqueness follows by comparing the first nonzero coefficient layer. In particular $[G:P_n]=p^{nd}$ and these layers form a neighbourhood basis. These are the power-layer and coordinate facts needed in the proof, rather than an assumption that an arbitrary pro-p group has unique roots.

On the underlying set of $G$, set the additive zero to $1$, the additive negative of $x$ to $x^{-1}$, and defineThe p-adic integers act through limits of integer powers in the closed cyclic subgroup generated by $x$. In the second formula the product belongs to $P_n$; in the third the group commutator belongs to $P_{2n+ϵ}$. The symbols $p^{-n}$ and $p^{-2n}$ denote the unique group roots, not unrestricted division in a group. We next justify convergence and the Lie algebra laws together.

A useful concrete tool is a complete normed group algebra containing $G$. In its completed group algebra, put $b_i=g_i-1$. Ordered-coordinate uniqueness gives unique expansions ${\sum }_{\alpha \in {\mathbb{N}}^d}{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To see why, first work modulo $p$ and $P_n$: the ordered products of the $g_i$ with exponents below $p^n$ form the regular basis, and the change to the corresponding products of $g_i-1$ is triangular with diagonal entries one. Take inverse limits and then lift the coefficient digits successively in powers of $p$. Define the weightIt is a genuine separated, submultiplicative weight. Indeed reordering $b_j{b}_i$ produces a group-commutator correction in $P_{ϵ}$; the binomial expansion of a $p^{ϵ}$th power gives that correction weight at least $2ϵ$, so collection never decreases total weight. Also $v(pc)=1+v(c)$. Inverting $p$ and completing therefore gives a Banach algebra $B$ over ${\mathbb{Q}}_p$. For $g\in G$, ordered coordinates showwith value infinity at $g=1$. Thus the induced topology on $G$ is exactly its original topology. This construction supplies the ambient algebra without presupposing the uniform pro-p group and powerful Lie lattice correspondence. The weighted construction, including the case $p=2$, is described in Equivariant D-modules on rigid analytic spaces, Section 6.5.

Since $ϵ>1/(p-1)$, the p-adic logarithm and p-adic exponential converge on the relevant balls in $B$:The estimates $v_p(k!)\le k/(p-1)$ and $v_p(k)\le {\log }_{p}k$ show that term weights tend to infinity. Formal substitution is legitimate there and proves that the two maps are inverse. They preserve the leading weight. In particular $\log$ embeds the compact set $G$ homeomorphically into $B$, and $\log (g^{p^n})=p^n\log g$.

Put $a=\log x$ and $b=\log y$. Expanding the convergent Baker--Campbell--Hausdorff formula in $B$ givesThe first expansion has linear part $p^n(a+b)$ and all other terms have degree at least two in $p^{n}a,p^{n}b$. The second has leading term $p^{2n}(ab-ba)$ and all remaining terms have degree at least three. After division by the displayed powers of $p$, the extra factor of $p^n$ forces those errors to zero. The normalized expressions are logarithms of the unique roots already specified. Since $\log G$ is compact and closed in $B$, both limits belong to $\log G$. Scalar multiplication also remains in $\log G$ by continuity of powers. Consequently $\log G$ is a module over ${\mathbb{Z}}_p$, closed under the associative-algebra Lie bracket $ab-ba$. Bilinearity, alternation and the Jacobi identity follow by expanding products in the associative algebra $B$. Transporting these operations back gives exactly $L_G$ above.