Codex @codex  0

Write the finite cyclic group as $C_n=⟨g:g^n=e⟩$, the source's $Z_n$. It is Abelian, so $hg{h}^{-1}=g$ and every conjugacy class is a singleton: there are $n$ such classes.

Over the complex numbers, all its irreducible representations are one-dimensional. One direct proof is that the matrix representing $g$ has an eigenvector over $\mathbb{C}$; its line is preserved by every power of that matrix and hence by the whole group. Irreducibility forces the line to be the whole representation space. Its eigenvalue must be an $n$th root of unity.

The complex irreducible representations of a finite cyclic group are thereforeTheir generator values are distinct, so no two are equivalent. There are $n$ of them, and $n$ squared dimensions equal one, exhausting the sum of squares of irreducible representation dimensions: ${\sum }_j{d}_j^2=n=|G|$. Thus the list is complete.

For $n=3$, let $\omega =e^{2\pi i/3}$, so ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$. The character table isThe complex-field convention is essential. The preceding part used real matrices, but that does not make the one-dimensional classification valid over $\mathbb{R}$: conjugate nonreal characters combine into the two-dimensional irreducible real representations of a finite cyclic group. The character table and dimension-sum result used here have their usual complex meaning.

The conjugacy class of $g$ is $\{hg{h}^{-1}:h\in G\}$. Rotations commute with one another, while conjugation by a reflection changes $R^j$ to $R^{-j}$. Thus the rotation classes are $\{I\}$, $\{R,R^4\}$ and $\{R^2,R^3\}$.

For reflections,Because multiplication by two permutes the residues modulo five, these conjugates exhaust all five reflections. Consequently the complete class partition isTheir sizes $1+2+2+5=10$ account for every element, as in the conjugacy classes of a finite dihedral group.

A normal subgroup is a union of complete conjugacy classes. The trivial subgroup is normal, and the rotation subgroup is normal because conjugation preserves it; equivalently it has index two. None of the five order-two subgroups is normal, because it contains just one member of the five-element reflection class. Thus the proper normal subgroups are exactly

The conjugacy class of $g$ isIf $N$ is a normal subgroup and $g\in N$, then normality gives $hg{h}^{-1}\in N$ for every $h$. Thus the whole conjugacy class of each member of $N$ is contained in $N$, andConjugacy classes partition the group, so this is a union of complete disjoint classes, never selected fractions of classes.

Use the identity relation as printed in the PDF, $XYZ=I$. Since $Z^2=I$, multiplying on the right by $Z$ gives $XY=Z$. Alsobecause $X,Y,Z$ are involutions. Thus $X$ and $Y$ commute, and so does their product $Z$ with either of them. Therefore the group is Abelian, and is the Klein four-group.

Each conjugacy class is a singleton. Over the complex numbers every irreducible representation of a finite abelian group is one-dimensional. Choose independent signs $\epsilon =\chi (X)$ and $\eta =\chi (Y)$; the relations force ${\epsilon }^2={\eta }^2=1$ and $\chi (Z)=\epsilon \eta$. Hence the complete character table isThese four distinct one-dimensional representations exhaust the irreducible possibilities: there are four conjugacy classes, and their squared dimensions already sum to $|G|=4$. The rows also satisfy character orthogonality. The local transcription damages the final symbol in the defining relation; the PDF makes the identity $I$ unambiguous.

In a symmetric group, two permutations are in the same conjugacy class exactly when they have the same cycle type. For $S_4$ the five possibilities and their sizes areThere are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ transpositions, three pairings into two unordered pairs, $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles, and $(4-1)!=6$ four-cycles. Together with the identity these counts sum to $24=|S_4|$.

For a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations is its number of conjugacy classes, and the sum of squares of irreducible representation dimensions equals the group order. There are therefore five irreducibles for $S_4$, with squared dimensions summing to $24$.

First establish that part (c) found every one-dimensional representation. In such a representation, conjugate elements have the same scalar image. All transpositions are conjugate and square to the identity, so they all have one common image $s$ with $s^2=1$. Transpositions generate $S_4$, hence $s=1$ gives exactly the trivial representation, while $s=-1$ gives exactly the sign representation. There are precisely two degree-one irreducibles.

The other three dimensions $d_1,d_2,d_3$ are integers at least two and satisfyNo one of them can be at least four, since $4^2+2^2+2^2=24>22$. They are therefore all two or three. Starting from three twos gives sum of squares $12$; each replacement of a two by a three adds $5$. To reach $22$ requires exactly two replacements. Thus the complete list isThe count of conjugacy classes guarantees that these are all the inequivalent complex irreducible representations, not just dimensions of a proposed incomplete list.

For a finite group over the complex numbers, the sum of the squared dimensions of its inequivalent irreducible representations equals the group order. There is always a one-dimensional trivial representation, and part (b) supplies an irreducible representation of dimension three. Thus the remaining squared dimensions sum toThey must be two further dimensions equal to one. HenceThe number of inequivalent irreducible representations equals the number of conjugacy classes, so $A_4$ has four classes. There are only three cycle types in part (a); consequently cycle type cannot classify conjugacy within $A_4$.

More explicitly, the identity is a singleton class, and conjugation by three-cycles permutes the three double transpositions transitively. For a three-cycle, the centralizer in $A_4$ consists of its three powers, so its conjugacy class has $12/3=4$ elements. The eight three-cycles therefore split into two classes of four:For example, $(123)$ and $(132)$ have the same cycle type but are not conjugate in $A_4$. Any permutation conjugating the first to its inverse must fix its unique fixed vertex and reverse the orientation of the remaining three labels, so it is odd. They are conjugate in $S_4$, illustrating exactly why restricting the ambient group changes the answer.

The elements $1,-1$ commute with the whole quaternion group, so they form singleton classes. Each of the three remaining pairs is a conjugacy class: for example conjugation by $j$ sends $i$ to $-i$, and likewise cyclically. No such pair can mix with a different pair, because the quotient by $\{\pm 1\}$ is abelian and its cosets $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$ are preserved by conjugation. HenceFor a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations equals the number of conjugacy classes, so there are five. The sum of squares of irreducible representation dimensions is the group order, eight. With five positive integer dimensions, the only possibility isIndeed five one-dimensional contributions give five, and replacing one by a two-dimensional contribution adds exactly three; any further enlargement would exceed eight. These are the dimensions used in the character table of the quaternion group.

Rotations commute with one another, while reflection conjugation sends $R^j$ to $R^{-j}$. Thus the rotational conjugacy classes are pairs $\{R^j,R^{-j}\}$, except when the two entries coincide. For reflections, direct use of the relations givesIf $n$ is odd, multiplication by two permutes residues modulo $n$, so all reflections are conjugate. The complete class list isTheir sizes are $1$, then $(n-1)/2$ classes of size two, then one class of size $n$, giving $(n+3)/2$ classes and $2n$ elements in total.

If $n$ is even, the half-turn $R^{n/2}$ is central and forms its own singleton. Conjugation changes a reflection exponent by an even amount or negates it, so it preserves parity and reaches every exponent of that parity. The list isThere are $n/2+3$ classes; their sizes are $1,1$, then $n/2-1$ pairs, then two classes of size $n/2$. Geometrically the two reflection classes correspond to vertex-axis and edge-axis reflections, in either order depending on the choice of $m$. These are the conjugacy classes of a finite dihedral group.

Let $n=p\ge 3$ be prime. Any subgroup contained in $⟨R⟩\cong C_p$ is either $\{I\}$ or the whole rotation subgroup. Both are normal: the trivial subgroup always is, and the rotation subgroup has index two, or equivalently is preserved by $mR{m}^{-1}=R^{-1}$.

If a normal subgroup contains any reflection, it contains its entire conjugacy class. Part (b) shows that this is every reflection for odd $p$. Products of reflections include a generator of the rotations, since $(Rm)m=R$. Such a normal subgroup must therefore be the whole dihedral group.

Thus the complete list of proper normal subgroups, including the trivial subgroup, isThis also agrees with classification of subgroups of a dihedral group: the reflection subgroups of order two are proper but are not normal when $p$ is odd.

Each reflection has order two, giving exactly the three order-two subgroupsConjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$, cycling the three reflections. None of these subgroups is therefore normal.

The two nonidentity rotations generate the same order-three subgroup:It is invariant under conjugation by $r$, and $srs=r^{-1}$ shows invariance under $s$ as well; since $r,s$ generate the group, this proves normality. The complete partition into conjugacy classes isThe rotations are mutually conjugate by $s$, and the reflections by $r$; no rotation is conjugate to a reflection because conjugation preserves the order of a group element.

The character of a representation is the trace of its representing matrix. For the rotation powers,Each $U^{j}V$ has a planar reflection block of trace zero and third diagonal entry $-1$, giving character $-1$. In the chosen element order,Equivalently its values on the conjugacy classes $\{I\}$, $\{U^2\}$, $\{U,U^3\}$, $\{V,U^{2}V\}$, $\{UV,U^{3}V\}$ are $3,-1,1,-1,-1$.

Act on the four labelled vertices. A proper rotation gives an even permutation, identifying the rotational tetrahedral symmetry group with $A_4$, an alternating group. Geometrically there are one identity, three half-turns about opposite-edge midpoint axes, and two nontrivial rotations about each of four vertex-face axes. ThereforeThe tetrahedral rotation conjugacy classes areTo justify the splitting, the centralizer of a 3-cycle within $A_4$ is its three-element cyclic subgroup, so its conjugacy class has size $12/3=4$. Its inverse is not conjugate to it in $A_4$: the permutations that reverse that cycle while preserving its fixed vertex are odd. Conjugating $(123)$ by the even permutations produces the displayed $C_{+}$, and taking inverses gives $C_{-}$. The three half-turns are conjugate, since rotations move their axes into one another. Their centralizer has order four and their class therefore has size three.

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

Write the dihedral group as $D_6=⟨r,s\mid r^6=s^2=1,srs=r^{-1}⟩$, where $r$ rotates through $\pi /3$. There are six rotations and six reflections, so the order is 12. Conjugation by $s$ reverses a rotation, and conjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$. Consequently the conjugacy classes and their sizes areThe last two classes distinguish the reflection axes through opposite vertices and through opposite edge midpoints. Their sizes sum to 12.

A solvable group has a finite subnormal series from the identity to the group with abelian group factors. An equivalent definition uses the derived series: set $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$; the series must reach the identity. For completeness, an abelian-factor series forces successive derived subgroups into its successive terms, while a terminating derived series itself supplies such a series.

If $H\le G$, induction gives $H^{(j)}\le G^{(j)}$: every commutator of elements of $H^{(j)}$ is also a commutator in $G^{(j)}$. Thus termination for $G$ implies termination for $H$, proving that every subgroup of a solvable group is solvable.

In $S_4$, useThe Klein four-group is invariant under conjugation in $S_4$; the other normality assertions follow from the alternating group being the kernel of the sign map. The quotient group factors are $V_4,C_3,C_2$, all abelian. Hence $S_4$ is solvable.

For the last example take $A_5\times C_2$. Its subgroup $\{1\}\times C_2$ is nontrivial, proper and normal, so the product is not a simple group. Projection onto $A_5$ is surjective, and every quotient of a solvable group is solvable, because a homomorphism carries each derived subgroup onto the derived subgroup of its image. But the nonabelian simple group $A_5$ has $A_5^{\prime }=A_5$: its derived subgroup is nontrivial and normal. Thus the product is not solvable either. One standard justification of the simplicity used here is that the nonidentity conjugacy class sizes of $A_5$ are $15,20,12,12$; no proper nontrivial union with the identity has size dividing $60$, as a normal subgroup would require by Lagrange's theorem for finite groups.

Choose an inscribed regular tetrahedron from one of the regular compounds of five tetrahedra in the dodecahedron, and let $H$ be its rotational stabilizer. Its twelve tetrahedral rotations preserve the dodecahedron. Concretely, with $\varphi =(1+\sqrt{5})/2$, use dodecahedron vertices $(\pm 1,\pm 1,\pm 1)$ together with cyclic coordinate placements of $(0,\pm {\varphi }^{-1},\pm \varphi )$. The tetrahedron has the four cube vertices whose coordinate product is $+1$. Cyclic coordinate permutations and even numbers of coordinate sign changes preserve both vertex sets, have determinant $+1$, and give twelve distinct rotations. No other rotation can preserve this tetrahedron, because a regular tetrahedron has only twelve orientation-preserving symmetries.

Their action on its four vertices is faithful and consists exactly of the even permutations: the eight rotations through $\pm 120^{\circ }$ induce three-cycles, the three half-turns induce double transpositions, and there is the identity. Thus $H\cong A_4$, independently of part (ii).

Finally a nonidentity element of $H$ has order two or three. In the dodecahedron's full rotation group, all fifteen half-turns about opposite-edge axes form one conjugacy class, and all twenty third-turns about opposite-vertex axes form one conjugacy class. These geometric conjugacies follow by carrying the corresponding oriented axes to one another by rotations. A normal subgroup containing such an element must contain its whole class, which already has more than twelve elements. Hence no nonidentity normal subgroup of $G$ is contained in $H$, verifying the required extra property.

Let $G$ act by left multiplication on its five left cosets of $H$. This gives a group homomorphism $\phi :G\to S_5$. Its kernel is the core of a subgroup, ${\bigcap }_{g\in G}gH{g}^{-1}$: an element fixes every coset precisely when all its conjugates lie in $H$. It is a normal subgroup contained in $H$, so the hypothesis forces the kernel to be trivial. Thus $G$ embeds in $S_5$.

The image has order $60$ and index two, hence is the kernel of a surjection $S_5\to C_2$. All transpositions are conjugate and generate $S_5$, so under any such surjection they must all have the same nonidentity image. This determines the map uniquely as the sign homomorphism. Its kernel is $A_5$, proving $G\cong A_5$.

To prove simplicity directly, the nonidentity conjugacy classes in $A_5$ have sizes $15,20,12,12$. A double transposition has centralizer of order four in $A_5$; a three-cycle has centralizer of order three; a five-cycle has centralizer of order five, yielding the two classes of size twelve. The class sizes together with the identity account for all sixty elements. A normal subgroup must be a union of these classes containing the identity, and its order must divide sixty. None of the proper nontrivial sums of these sizes with one is a divisor of sixty. Therefore $A_5$, and hence $G$, is simple.

An element $x$ is central exactly when $gx=xg$ for every $g\in G$. By part (b), this is exactly the assertion that every $g$ stabilizes $x$. That means every conjugate of $x$ is $x$; conversely, a singleton conjugacy class has this property for every $g$.In particular, the singleton classes are in one-to-one correspondence with the elements of the centre.

Act by left multiplication on the five cosets of $H$. The kernel is the core ${\bigcap }_{g\in G}gH{g}^{-1}$, a normal subgroup contained in $H$, so the hypothesis makes it trivial. Thus $G$ embeds in $S_5$, and its image has order $60$ and index two. An index-two subgroup is normal, so the quotient is of order two. Every three-cycle maps trivially to that quotient, since its order divides three. Three-cycles generate $A_5$, so the image contains $A_5$; their equal order gives $\boxed{G\cong A_5}$.

For completeness, prove simplicity rather than merely invoking it. The nonidentity even cycle types in $S_5$ are three-cycles, products of two disjoint transpositions and five-cycles. In $A_5$, a three-cycle has centralizer of order three, so its conjugacy class has size $20$. A double transposition has centralizer of order four in $A_5$, giving class size $15$. A five-cycle has centralizer its cyclic subgroup of order five, giving class size $12$; the $24$ such elements therefore split into two classes of size $12$. These account for all $60$ elements with the identity.

A normal subgroup is a union of whole conjugacy classes, containing the identity, and its order must divide $60$. No nonempty proper sum of the class sizes $15,20,12,12$, plus one, is a divisor of $60$ greater than one: the possible proper nontrivial sums are $13,16,21,25,28,33,36,40,45,48$. Only the full union has order $60$. Hence $\boxed{A_5\text{ is simple}}$, and so is $G$.