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Use the dihedral group presentation $G=⟨r,s:r^{2n}=s^2=1,srs=r^{-1}⟩$. Every element is $r^j$ or $s{r}^j$. A nonidentity rotation has order two only when its exponent is $n$, while $(s{r}^j{)}^2=1$ for every $j$. Thus the complete list isThe half-turn $r^n$ is central, so its conjugacy class is the singleton $\{r^n\}$ and its normal closure is $\{1,r^n\}$.

For reflections, $r^k(s{r}^j)r^{-k}=s{r}^{j-2k}$ and $s(s{r}^j)s^{-1}=s{r}^{-j}$. Hence the conjugacy class is exactly the $n$ reflections whose exponents have the same parity as $j$:The subgroup generated by this class contains $r^2$, as the product of two consecutive listed reflections, and contains $s{r}^j$. Conversely those two generators contain every listed reflection. Their subgroup is invariant under conjugation by both $r$ and $s$, so the smallest normal subgroup containing $s{r}^j$ isof order $2n$ and index two. This involutions in an even dihedral group classification distinguishes the two reflection classes from the central half-turn.

Here a reduced root system is crystallographic, as appropriate to a semisimple Lie algebra. It is a finite spanning set $\Phi \subset E\setminus \{0\}$ in a real inner-product space of dimension $r$, invariant under each Weyl reflectionwith $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$ for every pair of roots and $\Phi \cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$. A base of a root system $\Delta =({\alpha }_1,\dots ,{\alpha }_r)$ is a basis such that the coefficients of every root are integers that are either all nonnegative or all nonpositive. These define the positive system of a root system. We use the existing Cartan matrix conventionUsing the alternative denominator $({\alpha }_i,{\alpha }_i)$ transposes all the matrices below.

The classification of rank-two root systems gives $A_1\times A_1$, $A_2$, $B_2=C_2$, and $G_2$. In orthonormal coordinates we may choose the following simple roots and Cartan matrices:

Each complete root system consists of these positive roots and their negatives. To see why the list is exhaustive, the off-diagonal Cartan matrix entries of two distinct simple roots are nonpositive integers, and their product is $4{\cos }^2\theta <4$. Thus the product is $0,1,2$ or $3$, corresponding to simple-root angles $\pi /2,2\pi /3,3\pi /4$ or $5\pi /6$. These determine the four systems and their length ratios.

The Weyl group of a rank-two root system is generated by the two simple Weyl reflections. Their product is a rotation of order $m=2,3,4,6$, respectively, so the groups are the dihedral groups of order $2m=4,6,8,12$. The first is also the Klein four-group and the second the symmetric group $S_3$. The root-orthogonal lines cut the plane into $2m$ Weyl chambers, each of angle $\pi /m$. A chosen fundamental chamber of a root system is $(x,{\alpha }_1)>0$, $(x,{\alpha }_2)>0$; its walls are the two root-orthogonal lines. The Weyl group acts simply transitively on these open chambers.

The crystallographic hypothesis matters: if the integer-pairing condition is dropped, reduced noncrystallographic systems of type $I_2(m)$ also occur. They are not additional Lie-algebra root systems.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.

No prime and no presentation of $G_3$ have p-deficiency at least one. Set $r=xy$. The relations giveEvery element has form $r^j$ or $r^{j}x$ with $0\le j<1024$. Conversely, the usual rotations and reflections of a regular $1024$-gon satisfy the presentation and give $2048$ distinct elements. Hence $G_3$ is the finite dihedral group of order $2048$. The criterion p-deficiency at least one implies infinitude excludes every alternative presentation and every prime. As a check, the given presentation has

Yes, for $p=2$. The displayed group presentation is that of the infinite dihedral group. Put $u=at$ and $v=t$. Then $v^2=1$ andConversely, from $u^2=v^2=1$, set $a=uv$ and $t=v$; then $ta{t}^{-1}=vu=a^{-1}$. These inverse substitutions giveEach relator has free-group $2$-root exponent one, soThe change of presentation matters: the original presentation's mixed conjugation relator has weight one and would give a smaller p-deficiency.

False. Realize the symmetries of a square as a dihedral group of order eight, generated by $r=(1234)$ and $s=(24)$, with $r^4=s^2=e$ and $srs=r^{-1}$. Take $H=K=⟨s⟩$. Its double cosets are preciselyThese are disjoint and exhaust the eight symmetries. There are three double cosets, and $3\nmid 8$. This proves that double-coset counts need not divide group order. Specifying the order-eight realization avoids competing conventions for the notation of a dihedral group.

A set $S$ of prime numbers has Dirichlet density $\delta$ if the following limit exists:Equivalently, the denominator can be ${\sum }_p{p}^{-s}$, since that sum is $\log (1/(s-1))+O(1)$. Adding or removing finitely many prime numbers leaves the Dirichlet density unchanged.

Put $\alpha =\sqrt[4]{2}>0$ and $L=\mathbb{Q}(\alpha ,i)$. This is the splitting field of $X^4-2$, whose four roots of a polynomial are $\alpha ,i\alpha ,-\alpha ,-i\alpha$. The polynomial is an Eisenstein polynomial at $2$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=4$. Since $\mathbb{Q}(\alpha )$ is contained in the real numbers, it does not contain $i$, giving $[L:\mathbb{Q}]=8$. Thus $L/\mathbb{Q}$ is a Galois extension of degree $8$. Its Galois group is the dihedral group of order $8$: the field automorphisms $r(\alpha )=i\alpha$, $r(i)=i$ and $s(\alpha )=\alpha$, $s(i)=-i$ satisfy $r^4=s^2=1$ and $srs=r^{-1}$.

For an odd prime number $p$, the condition $p\equiv 1(\mathrm{mod}4)$ means that the finite field ${\mathbb{F}}_p$ contains all fourth roots of unity. Under this condition, $2$ is a quartic residue exactly when $X^4-2$ has a root in ${\mathbb{F}}_p$; multiplying that root by $1,i,-1,-i$ gives all four distinct roots. Conversely, complete splitting of $X^4-2$ over ${\mathbb{F}}_p$ gives both a fourth root of $2$ and a primitive fourth root of unity, so it also forces $p\equiv 1(\mathrm{mod}4)$.

The polynomial discriminant of $X^4-2$ is $-2^{11}$. Hence every odd prime number is unramified in $L$, and the root-splitting criterion is equivalent to its Frobenius automorphism acting trivially on all the roots. Since the roots generate $L$, this is equivalent to the Frobenius automorphism being the identity, or to $p$ being a completely split prime of $L/\mathbb{Q}$. The Chebotarev density theorem says that the unramified prime numbers with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|\mathrm{Gal}(L/\mathbb{Q})|$. Here $C=\{1\}$, so

Conversely, assume $H$ is a normal subgroup and the quotient group $G/H$ is an abelian group. Commuting the two cosets $aH$ and $bH$ givesso every group commutator $a^{-1}{b}^{-1}ab$ belongs to $H$. This proves the converse and completes the equivalence.

For the dihedral group of order ten, choose a rotation $r$ and a reflection $s$ with $r^5=s^2=1$ and $srs=r^{-1}$. Its rotation subgroup $⟨r⟩$ is a normal subgroup, and the quotient by it has order two, hence is an abelian group. Consequently the commutator subgroup is contained in $⟨r⟩$. On the other hand,Since $r^{-2}$ generates the order-five cyclic group $⟨r⟩$, the commutator subgroup is exactly $⟨r⟩$. Every abelian quotient must kill this subgroup. There are only two subgroups containing it, since the remaining quotient has prime order. The complete list, including the trivial quotient, isThis is the order-ten instance of the abelianization of an odd dihedral group.

Use the coefficients in the authoritative PDF, including their factor $i$. Write $a=2\sqrt{3}i$ andThe numerator and denominator are coprime: a common zero would, by subtraction, require $z=0$, where both equal one. Therefore the degree of a rational map of the Riemann sphere is four. Direct algebra givesThe domain transformations $z\mapsto iz$ and $z\mapsto (iz+1)/(1-iz)$ are sphere rotations as special-unitary Möbius transformations: respectively a quarter-turn about the third axis and a one-third turn cycling the three coordinate axes. In particular the latter sends $0\mapsto 1\mapsto i\mapsto 0$, the north, first-axis and second-axis directions. They generate the order-24 rotational symmetry group of a cube, isomorphic to the symmetric group $S_4$. Their target transformations are rotations: $w\mapsto 1/w$ is a half-turn about the first target axis and $w\mapsto \omega w$ is a one-third turn about the third target axis. This proves whole-map combined equivariance, not just symmetry of selected roots.

The target rotation image is the order-six dihedral group $D_3$. The spatial half-turns about all three coordinate axes act trivially on the map: $R(-z)=R(z)$ and $R(1/z)=R(z)$ generate a Klein four-group kernel. Thus the full Skyrmion symmetry combines the spatial cubic rotations with compensating isospin rotations, while its angular energy and Skyrme baryon density have pure spatial cubic symmetry.

The critical directions make the geometry explicit. Its Wronskian of a rational map isThe five finite ramification points are $0,\pm 1,\pm i$; infinity supplies the sixth, since in the local coordinate $w=1/z$ the map starts with $R=1+2a{w}^2+O(w^4)$. These are the six coordinate-axis directions, or face centres of a cube. The angular Jacobian of a rational map vanishes there, consistent with a cube-shaped shell whose density is concentrated away from its face centres. Any further rotational equivariance would have to preserve this set, so the spatial proper rotation group is exactly the cubic group already generated above.

There is also a reflection relation $R(\overline{z})=1/\overline{R(z)}$. Together with the proper rotations, it makes the angular density invariant under the full order-48 symmetry group of a cube, usually denoted $O_h$. The target operation in this reflection relation is orientation reversing; it should not be mistaken for a proper isospin rotation. In the full Skyrme model, reflections are expressed using the field parity operation $U(\mathbf{x})\mapsto U^{†}(-\mathbf{x})$ together with a compensating isospin rotation.

This is the cubic charge-four rational-map ansatz:The cubic rational-map ansatz for four Skyrmions is a useful approximation and starting point for the cubic four-Skyrmion, whose lowest spin-zero, isospin-zero quantized state models an alpha particle. A radial minimization and, for precision, unrestricted field relaxation are still required. The TeX's missing $i$ changes this map; the six critical directions alone would not detect the error, because the same Wronskian zero set persists when $a$ is real. The actual rotational equivariance identities are the stronger check.

Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

For order $48=2^4\cdot 3$, the Sylow theorems give $n_2\mid 3$ and $n_2\equiv 1(\mathrm{mod}2)$, hence $n_2=1$ or $3$. If $n_2=1$, the Sylow subgroup of order $16$ is a nontrivial proper normal subgroup. If $G$ were simple in the other case, it would be nonabelian (a finite abelian simple group has prime order), and the preceding result would force $48\mid 3!/2=3$, impossible. Thus every group of order 48 is not simple.

An example with no normal Sylow subgroup for $p=2$ isIn $S_4$, $P=⟨(1234),(13)⟩$ is a dihedral group of order $8$: conjugation by $(13)$ inverts $(1234)$. Thus $P\times C_2$ has order $16$ and is a Sylow subgroup of $G$. The only transpositions in $P$ are $(13)$ and $(24)$; conjugating $(13)$ by $(23)$ gives $(12)\notin P$. Therefore $P$ and $P\times C_2$ are not normal. By conjugacy, no other Sylow subgroup for $p=2$ is normal either.

For a normal subgroup $H⊴G$, the quotient group $G/H$ is the set of cosets $gH$ with multiplication $(gH)(kH)=gkH$. If $gH=g^{\prime }H$ and $kH=k^{\prime }H$, normality moves the intervening elements of $H$ past $k$, proving that the product is independent of representatives.

If $g$ has finite order $m$, then $(gH{)}^m=H$, so $\mathrm{ord}(gH)$ divides $m$. For finite $G$, the greatest element order in $G/H$ is therefore no greater than that in $G$.

Every kernel is normal because $\phi (gh{g}^{-1})=\phi (g)\phi (h)\phi (g{)}^{-1}=1$ whenever $h\in \ker \phi$. Conversely, if $H$ is normal, the quotient map $G\to G/H$ has kernel $H$. Thus $H⊴G$ exactly when it is the kernel of a group homomorphism.

A group is metacyclic when it has a cyclic normal subgroup with cyclic quotient. A dihedral group has its cyclic rotation subgroup normal and quotient $C_2$, so every dihedral group is metacyclic.

Up to isomorphism the groups of order eight areAll except $C_2^3$ are metacyclic: use a cyclic subgroup of order four in the three noncyclic examples that have one. In $C_2^3$, every cyclic subgroup has order at most two and its quotient is not cyclic. The normal subgroups of $A_4$ are $1,V_4,A_4$; none gives the required cyclic subgroup and quotient, so $A_4$ is not metacyclic. Finally, $S_1,S_2,S_3$ are metacyclic, while $S_4$ and $S_5$ are not, by their normal-subgroup structures. Hence