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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.

Let $G$ be the Galois group of a polynomial of the irreducible quartic. Irreducibility makes $G$ a transitive subgroup of $S_4$. The three roots of the given cubic resolvent of a quartic correspond to the three partitions of the four quartic roots into two unordered pairs, and the kernel of the $S_4$-action on these partitions is the Klein four-group $V_4$. If the resolvent has Galois group $S_3$, the image of $G$ in this action is all of $S_3$. Among the transitive subgroups of the symmetric group on four points, only $S_4$ has this image. Hence

Now put $f_p(t)=t^4+pt+p$. The Eisenstein criterion at $p$ proves that $f_p$ is an irreducible polynomial, and its cubic resolvent isIts discriminant of a depressed cubic isFor odd $p$, the $p$-adic valuation of ${\Delta }_p$ is three because $p\nmid 256$; for $p=2$, ${\Delta }_2=2^4\cdot 101$. Thus ${\Delta }_p$ is never a square in $\mathbb{Q}$.

By the rational root theorem, a reducible $g_p$ has an integer root among $\pm 1,\pm p,\pm p^2$. Reduction modulo $p$ shows that the root is divisible by $p$; writing it as $pk$, with $k\in \{\pm 1,\pm p\}$, givesThe only possibilities are $k=-1,p=3$ and $k=1,p=5$, corresponding to roots $-3$ and $5$. For every other prime, $g_p$ is irreducible with nonsquare discriminant, so its Galois group is $S_3$ and the criterion above gives the quartic group $S_4$. For $p=3$ or $5$, the reducible resolvent makes the quartic group a proper subgroup. Therefore

The normis a positive integer, because it is the sum of the squares of the multiplicities of the irreducible constituents. The character restriction norm bound and $[G:K]=3$ make it at most three. It is therefore $1$, $2$, or $3$.

All three values occur:

If $\delta \notin \mathbb{Q}(\alpha )$, thenLetThis is the nonidentity automorphism of $K/\mathbb{Q}(\alpha )$. By extending an embedding that sends $\alpha$ to $\beta$, and composing with $s$ if necessary, chooseThenThe eight elementsare distinct, and exhaust the Galois group. Thus the dihedral Galois action on four radical roots gives

Now suppose $c$ is a square in $\mathbb{Q}$. Then $\delta \in \mathbb{Q}$, so every automorphism fixes $\delta$. We are in case (i) with $\tau (\delta )=\delta$, and therefore

Conversely, suppose the group is the Klein four-group. Label the rootsThe three nonidentity elements in its transitive action are the three double transpositions. Each fixes $\delta$: they send the ordered pair $(\alpha ,\beta )$ respectively towhose products all equal $\alpha \beta$. Hence every element of the Galois group fixes $\delta$. By the Artin fixed-field theorem,Therefore $c={\delta }^2$ is a square in $\mathbb{Q}$. This proves the Klein-four criterion for an irreducible even quartic:

A group is solvable when its derived serieseventually reaches the trivial group; equivalently, it has a subnormal series with abelian quotients.

The symmetric group $S_4$ is solvable becausehas abelian quotients $V_4$, $C_3$, and $C_2$, where $V_4$ is the Klein four-group. Thus

For a depressed quarticwith roots summing to zero, setThe cubic resolvent of a quartic is

The symmetric group $S_4$ acts on the three partitions of four roots into two unordered pairs. The resulting homomorphism $S_4\to S_3$ has kernel the Klein four-group $V_4$. If $\Delta (f)$ is a square and the characteristic is not two, the discriminant criterion for an alternating Galois group gives $G=\mathrm{Gal}(f)\le A_4$, and its image on the three resolvent roots lies in $A_4/V_4\cong C_3$.

For an irreducible quartic, $G$ is transitive on four roots, and its only possibilities inside $A_4$ are $V_4$ and $A_4$. The resolvent is irreducible exactly when the image acts transitively on its three roots, equivalently when that image is $C_3$. ThusThe printed statement omits the needed irreducibility hypothesis on $f$. Literally it is false: $f=x(x^3-3x-1)$ has square discriminant and Galois group $C_3$, and its resolvent is irreducible. Without quartic irreducibility, a reducible resolvent only implies $G\le V_4$, allowing the trivial group, $C_2$, or $V_4$.

For the specified polynomial $x^4+8x+12$,Its resolvent is $g(y)=y^3-48y-64$. Modulo $5$ this is $y^3+2y+1$, which has no root in ${\mathbb{F}}_5$ and is therefore an irreducible cubic. The quartic is given to be irreducible, so the preceding criterion yields

A group $G$ is abelian when $xy=yx$ for every $x,y\in G$. It is cyclic when some generator of a group $g$ satisfiesIf $x=g^r$ and $y=g^s$ belong to a cyclic group, thenso every cyclic group is abelian.

The Klein four-group $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.

The condition on proper subgroups does not force $G$ to be abelian. The quaternion groupis nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm 1\}$, and the three cyclic subgroupseach of order four. They are all cyclic, so

If $G=⟨g⟩$, any two elements are $g^a,g^b$, andThus every cyclic group is an abelian group.

For the converse, the Klein four-group $C_2\times C_2$ is an abelian group, with componentwise multiplication. Each nonidentity element has order of a group element two, so none generates its four elements. It is therefore not a cyclic group.

Choose distinct $a,b\ne e$, possible because $|G|>2$. By iii, $a$ and $b$ commute, and $a^2=b^2=e$. The four elementsare distinct: $ab=e$ would give $a=b$, while $ab=a$ or $ab=b$ would make $b=e$ or $a=e$. Commutativity and $a^2=b^2=e$ show that products stay in $H$, and every element is its own inverse. More explicitly,is a group isomorphism, with addition of exponents modulo two. Hence $G$ contains a subgroup isomorphic to the Klein four-group.

The converse is false. The direct product of groups $C_2\times C_2\times C_3$ contains $C_2\times C_2\times \{e\}$, but it has elements of order of a group element three. Thus the existence of the specified subgroup does not force every nonidentity element to have order of a group element two.

The complete classification up to group isomorphism isHere $C_1$ is the trivial group. The prime number cases follow from i. For four elements, an element of order of a group element four generates $C_4$. If there is none, Lagrange's theorem makes every nonidentity element have order of a group element two. Then iv produces a four-element subgroup, which must be all of $G$, giving the Klein four-group.

For six elements, an element of order of a group element six gives $C_6$. Suppose there is none. If there were no element of order of a group element three either, all nonidentity elements would have order of a group element two. Part iv would then produce a subgroup of cardinality four, contradicting Lagrange's theorem. Thus choose $a$ of order of a group element three. There is also an involution $b$: otherwise the five nonidentity elements could be paired into distinct pairs $\{g,g^{-1}\}$, impossible for an odd number of elements.

The subgroup $H=⟨a⟩$ has index of a subgroup two, so it is a normal subgroup. Indeed, for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, and for $g\in H$ both equal $H$. Consequently $ba{b}^{-1}$ is either $a$ or $a^{-1}$. The first possibility makes $a,b$ commute, and $ab$ then has order of a group element six: $(ab{)}^k=e$ forces $a^k=b^{-k}\in ⟨a⟩\cap ⟨b⟩=\{e\}$, so both $3$ and $2$ divide $k$. This contradicts our assumption. ThereforeEvery element is uniquely $a^i$ or $a^{i}b$, with $0\le i<3$, and the relation $ba=a^{-1}b$ determines all their products. Sending $a$ to $(123)$ and $b$ to $(12)$ gives a group isomorphism with the symmetric group $S_3$. This completes the classification of groups of order at most six.

Of these groups, exactly $C_1,C_2,C_3,C_4,C_2\times C_2,S_3$ embed in $S_4$. Concrete subgroups areThere is no $C_5$ subgroup, since $5$ does not divide $24$. There is no $C_6$ subgroup either: the order of a group element in a symmetric group is the least common multiple of the lengths in its disjoint cycle decomposition, and on four letters the only possibilities are $1,2,3,4$.

Label the vertices of a regular tetrahedron $1,2,3,4$, and include all its Euclidean isometries, both rotations and reflections. The following list describes every subgroup of cardinality at most six, allowing arbitrary relabelling of the vertices. The numbers count actual subgroups, rather than just their group isomorphism types.

Here is why this is exhaustive. Part b leaves only the trivial group, $C_2,C_3,C_4,C_2\times C_2,S_3$. The disjoint cycle decomposition gives exactly the two listed kinds of involutions, the three-cycles, and the four-cycles, accounting for all the cyclic subgroups. In a Klein four-group, the nonidentity elements commute. If it contains $(12)$, any other nonidentity element must be a commuting involution, hence $(34)$ or $(12)(34)$; either choice yields the displayed opposite-edge subgroup. If it has no transposition, its three nonidentity elements must be all three double transpositions, yielding the rotational subgroup.

Finally, an $S_3$ subgroup has a unique cyclic subgroup of cardinality three. Its two three-cycles fix the same unique vertex. Every element of the $S_3$ subgroup preserves this vertex, since conjugation preserves that unique cyclic subgroup. Thus it is the full stabilizer subgroup of that vertex. The counts in the list are consequently exhaustive, givingThe two kinds of $C_2$ and the two kinds of Klein four-groups have different geometric behavior despite having the same abstract group isomorphism type within each pair. This is the distinction recorded by the small subgroups of tetrahedral symmetry.

A smallest example is the Klein four-groupwith coordinatewise addition modulo $2$. Every nonidentity element has order of a group element $2$, so none generates all four elements; hence this group is not a cyclic group.

A group of group order $1$ is a cyclic group. A group of group order $2$ or $3$ is also a cyclic group: choose a nonidentity element; by Lagrange's theorem, its order of a group element is a nontrivial divisor of that prime number, so equals the whole group order. Thus no smaller example exists, and the smallest possible group order is $4$.

Representations. A complex group representation is a group homomorphism $\rho :G\to \mathrm{GL}(W)$ to the invertible linear maps on a vector space. It is an irreducible representation when $W\ne 0$ and the only invariant subspaces are $0$ and $W$.

Group table. The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their multiplication table isThe table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from matrix multiplication. The table is symmetric, so the group is abelian. It is the Klein four-group $V\cong C_2\times C_2$.

Dimensions and conjugacy. The sum of squares of irreducible representation dimensions is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional trivial representation. Hence the given two-dimensional representation is reducible. Since the group is abelian, every conjugacy class is a singleton. There are therefore four inequivalent complex irreducible representations, each of dimension one:Taking the trace of the given matrices gives their character of a representation:

Irreducible characters. In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm 1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete character table:Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero character inner product. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by character orthogonality. Thus

Diagonal form. The common invariant lines are spanned by $(1,1{)}^T$ and $(1,-1{)}^T$. Use their normalized vectors as columns ofThen simultaneous diagonalization givesReading the first and second diagonal entries across the four matrices recovers ${\chi }_3$ and ${\chi }_4$, respectively, verifying the decomposition explicitly.

A left coset is $gH=\{gh:h\in H\}$. The cosets partition a finite group into equally sized sets of $|H|$ elements, so Lagrange's theorem gives $|G|=[G:H]|H|$; in particular subgroup order divides group order.

Here $A=-I$, $C=-B$ and $B^2=I$. The Cayley table, with row element multiplied on the left of the column element, is

The table proves closure, $I$ is the identity and every element is its own inverse. Associativity is inherited from matrix multiplication, so this is the Klein four-group.

Finally, if every nonidentity element has order two, every element satisfies $h^{-1}=h$, including the identity. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian.

Lagrange's theorem states that the group order of a finite group divides the group order; more precisely $|G|=[G:H]|H|$.

If every element squares to the identity element, every element is its own inverse element, and $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian. For distinct nonidentity $a,b$, the four elements $1,a,b,ab$ are distinct: for example $ab=1$ would imply $a=b$, and $ab=a$ would imply $b=1$. Products remain in this set because $a,b$ commute and square to one. Thus they form a Klein four-group subgroup of order four.

Now let $|G|=2p$ with odd prime number $p$. By Lagrange's theorem, every element order is one of $1,2,p,2p$. If an element has order $2p$, its square has order $p$. If no element had order $p$, all nonidentity elements would therefore have order two. There are at least two distinct such elements, producing a subgroup of order four. This contradicts $4\nmid 2p$. Hence there is an element of order $p$.

The Klein four-group has multiplication tableSince it is abelian, every conjugacy class is a singleton: $\{1\},\{a\},\{b\},\{c\}$. An irreducible representation is a nonzero representation whose only invariant subspaces are zero and the whole representation space. For a finite group, the number of inequivalent complex irreducible representations equals the number of conjugacy classes, so there are four here.

A one-dimensional representation sends the involutions $a,b$ to signs, independently. All four choices areThey are distinct irreducible representations, exhausting the four possibilities. The character table of the Klein four-group isThe specified diagonal matrices obey $D(a{)}^2=D(b{)}^2=I$, commute, and satisfy $D(a)D(b)=D(c)$, so the group relations prove that $D$ is a group representation. Its coordinate axes are invariant lines carrying respectively ${\chi }_{01},{\chi }_{10},{\chi }_{11}$. ThereforeThe trivial representation ${\chi }_{00}$ is absent. If $H$ denotes the four-row character matrix above, direct row products give $H{H}^T=4I$. This verifies character orthogonality:The character of $D$ is $(3,-1,-1,-1)$, whose inner products with the four rows are $0,1,1,1$, independently confirming the decomposition. All these irreducibles are real as well: commuting real involutions can be simultaneously diagonalized into their real sign eigenspaces.

For a subgroup $H$ of a finite group $G$, distinct left cosets are disjoint and partition $G$. Each coset $gH$ has exactly $|H|$ elements because $h\mapsto gh$ is a bijection. Counting the partition proves Lagrange's theorem:A cyclic group is generated by a single element. In a group of order four, an element of order four generates the whole group, giving $C_4$. Otherwise all three nonidentity elements have order two. Choose two of them $a,b$; cancellation makes $ab=c$ the third. Since $c^{-1}=c$, $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba=ab$. The remaining products then give the Klein four-group. This proves that there are precisely two groups of order four up to isomorphism, distinguished by whether an element of order four exists.

For the square, take $r$ to be counterclockwise rotation by $90^{\circ }$ and $s$ reflection in the horizontal axis. They satisfy $r^4=s^2=e$ and $sr=r^{-1}s$. The dihedral group here is $D_4$ of order eight, with elements $e,r,r^2,r^3,s,rs,r^{2}s,r^{3}s$. Both generators are necessary: one generator would give a cyclic, hence abelian, group, while $r$ and $s$ do not commute. In the following Cayley table, the row acts after the column; the formula $(r^a{s}^b)(r^c{s}^d)=r^{a+(-1{)}^{b}c}{s}^{b+d}$, with exponents reduced modulo four and two, computes every entry.The largest cyclic subgroup is $⟨r⟩=\{e,r,r^2,r^3\}$, of order four. Every reflection has order two and there is no element of order eight. The other subgroups of order four areBoth are Klein four-groups, generated by the commuting involutions $r^2$ and one reflection. For completeness, any order-four subgroup containing $r$ or $r^3$ equals $⟨r⟩$. Otherwise it contains a reflection $f$; the product of two distinct reflections is a nonidentity rotation, which must be $r^2$ if odd powers of $r$ are excluded. This forces the subgroup to be $\{e,r^2,f,r^{2}f\}$, giving exactly the two displayed choices. It is the square case of the classification of subgroups of a dihedral group.

The order of a finite group $G$ is its number of elements. Lagrange's theorem states that for a subgroup $K\le G$,so $|K|$ divides $|G|$. In particular the order of a group element divides $|G|$, because it is the order of its generated cyclic subgroup.

For $|G|=4$, if some element has order four it generates the whole group, which is therefore the cyclic group $C_4$. Otherwise every nonidentity element has order two. Choose two distinct ones $a,b$ and put $c=ab$. Cancellation shows $c\notin \{e,a,b\}$, so $c$ is the remaining element. Since $c^{-1}=c$,The other products follow: $ac=ca=b$, $bc=cb=a$, and $a^2=b^2=c^2=e$. Thus the second possibility is the Klein four-group. The two groups are not isomorphic because only $C_4$ has an element of order four. This proves the order-four case of the classification of groups of order p squared.

A group homomorphism $\Phi :G\to H$ obeys $\Phi (gh)=\Phi (g)\Phi (h)$. It maps identity to identity and inverses to inverses. Its group kernel $K=\{g:\Phi (g)=e_H\}$ contains $e_G$ and is closed under $a{b}^{-1}$, so the subgroup test makes it a subgroup. For $k\in K$ and $g\in G$,Thus $gK{g}^{-1}\subseteq K$; applying the same argument to $g^{-1}$ gives equality. Hence $K$ is a normal subgroup.

Define $\overline{\Phi }:G/K\to H$ by $\overline{\Phi }(gK)=\Phi (g)$. If $gK=hK$, then $h^{-1}g\in K$, giving $\Phi (g)=\Phi (h)$, so this is well-defined. Coset multiplication shows it is a group homomorphism. Its kernel is just the identity coset, since $\Phi (g)=e_H$ means $g\in K$. It is therefore injective, and the assumed surjectivity of $\Phi$ makes it surjective as well. This proves the first isomorphism theorem here:

In the quaternion group, the element $-1=i^2=j^2=k^2$ commutes with each of $i,j,k$, because an element commutes with its square. It then commutes with their negatives too, and with $\pm 1$. The set $N=\{1,-1\}$ is a subgroup because $(-1{)}^2=1$, and it is central, hence a normal subgroup.

The four quotient cosets are $N,iN,jN,kN$. They square to $N$, and $(iN)(jN)=kN$. Reversed products have the same coset: for example $(ij{)}^{-1}=k^{-1}=-k$, while $j^{-1}{i}^{-1}=(-j)(-i)=ji$, so $ji=-k$. The same calculation gives $kj=-i$ and $ik=-j$. After signs are identified, these are precisely the products in the Klein four-group. DefineThis preserves all products, is surjective, and has kernel exactly $N$. The first isomorphism theorem therefore givesThis is the quaternion group modulo its centre: the reverse-product relations also show that none of $\pm i,\pm j,\pm k$ is central, so $Z(Q_8)=N$.

The group order of a finite group is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$. Such an $r$ exists in a finite group because two powers must coincide, and cancellation then gives a positive power equal to the identity element.

For the direct product of groups, the componentwise operation is closed because each factor is closed. Associativity holds in each coordinate:The identity element is $(e_1,e_2)$, and the inverse element of $(g,h)$ is $(g^{-1},h^{-1})$. Thus all group axioms hold. For finite factors, the multiplication principle gives

For ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$, write $E=(e,e)$, $u=(w,e)$, $v=(e,w)$, $z=(w,w)$, with $w^2=e$. Its Cayley table, with the row element multiplying the column element, isThis is the Klein four-group: every nonidentity element has order two.

In the cyclic group ${\mathbb{Z}}_4=⟨a⟩$, the set $\{e,a^2\}$ is closed because $(a^2{)}^2=e$, and both elements have inverses in the set. It is therefore a proper subgroup of order two. By Lagrange's theorem, a proper subgroup has order one or two. An order-two subgroup must be generated by an element of order two. Neither $a$ nor $a^3$ has order two, whereas $a^2$ does. HenceThe PDF's claim needs “nontrivial”: $\{e,a^2\}$ is the only nontrivial proper subgroup, but $\{e\}$ is also a proper subgroup under the usual definition.

A group isomorphism preserves element orders: $\varphi (g{)}^r=e$ if and only if $g^r=e$, using the homomorphism property and injectivity. Since ${\mathbb{Z}}_4$ has elements of order four and the Klein four-group does not,

Denote the four listed matrices, in order, by $I,P,Q,W$. As permutation matrices, $P,Q,W$ act on the four basis vectors by $(12)(34)$, $(13)(24)$ and $(14)(23)$, respectively. Direct multiplication givesThese relations prove closure and show that each element is its own inverse. The identity is included and matrix multiplication is associative, so the matrices form a group.

The relations coincide with those of the Klein four-group $D_2$: take its half-turn generator to $P$, a reflection generator to $Q$, and their product to $W$. All four matrices are distinct, so this assignment is a faithful representation of $D_2$. It is the four-point regular representation of the Klein four-group.

Every matrix fixes $v=(1,1,1,1{)}^T$, since it permutes its coordinates. Thereforeis a one-dimensional invariant subspace, nonzero and strictly smaller than the four-dimensional space. This proves the representation is reducible. In fact the four mutually orthogonal sign vectors $(1,1,1,1)$, $(1,1,-1,-1)$, $(1,-1,1,-1)$ and $(1,-1,-1,1)$ give invariant lines and completely diagonalize the representation, as in the normalized Hadamard matrix construction.