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Interpret the functions as maps on $\mathbb{R}\setminus \{0\}$, so every reciprocal is defined. Let $e(x)=x$, $a(x)=-x$, $b(x)=1/x$ and $c(x)=-1/x$. With the left factor applied last, direct function composition gives the Cayley tableThe table proves closure and identifies $e$ as the identity. Every element is its own inverse. Associativity is inherited from composition of maps, so the set is a group. The three nonidentity elements are commuting involutions.

Its subgroups areThere are no others: a subgroup containing two distinct nonidentity elements also contains their product, the third, and hence is the whole group; a subgroup containing just one has order two.

The abstract dihedral group $D_2$ has commuting generators $R,m$ with $R^2=m^2=I$, for example a rectangle's half-turn and reflection in one axis. The assignment $I\mapsto e$, $R\mapsto a$, $m\mapsto b$, $Rm\mapsto c$ respects products and is bijective. ThusThis is the group generated by sign reversal and reciprocation, a realization of the Klein four-group. The nonzero domain is essential; treating these as functions defined at zero would not give the stated group.

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.

Put $Y_p=D(Y)$. Direct matrix multiplication givesfor every real $p$. Since $D(X)=-I_2$ commutes with all matrices and $D(Z)=-Y_p$, the remaining relations also hold:All matrices are invertible, and the relations reproduce the multiplication table of the Klein four-group. ThereforeA representation need not be an orthogonal representation; imposing orthogonality on the displayed matrices would be an extra condition.

The characters are their traces, independent of $p$:In terms of part (b), $\chi ={\chi }_{-+}+{\chi }_{--}$, so the representation is reducible. For an explicit change of basis, takeThen $S^{-1}D(Y)S=\mathrm{diag}(-1,1)$ and $S^{-1}D(Z)S=\mathrm{diag}(1,-1)$, while $D(X)$ remains $-I_2$. These are the equivalent sheared representations of the Klein four-group; the parameter changes the basis, not the two irreducible constituents.

Observe that $A=-I$, $B^2=I$ and $C=-B$. Thus $A^2=B^2=C^2=I$, while $AB=BA=C$, $AC=CA=B$ and $BC=CB=A$. The multiplication table, with the left factor labeling the row, isIt proves closure of the set. Associativity is inherited from matrix multiplication, $I$ is the identity element, and every element is its own inverse element. Therefore this set is a group. The table is symmetric, so it is an Abelian group; its three nonidentity elements have order two, identifying it as the Klein four-group $C_2\times C_2$.

The two diagonal generators are independent involutions, soIt is a Klein four-group. Write the swapping generator as $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Then $J^2=I$, so $H_2=\{I,J\}$ has order two. Moreover,Thus $J$ normalizes $H_1$. The setis a group: its products reduce usingand inverses reduce in the same way. It contains both generating subgroups, so minimality gives $G\subseteq K$; conversely every displayed product is in $G$, giving $G=K$.

There are four diagonal matrices $D$ and four off-diagonal matrices $DJ$, all distinct. ConsequentlyAlso $H_1\cap H_2=\{I\}$, since $J$ is not diagonal, and every element of $G$ has the form $D{J}^{\eta }\in H_1{H}_2$. This proves condition (I), hence the exact factorization of a group. Here $H_1$ is a normal subgroup and $G=H_1⋊H_2$ is a semidirect product. Its matrices are the two-dimensional signed permutation matrices, also realizing the dihedral group of the square.

The subgroup generated by $T,P$ is $H=\{I,T,P,-I\}$, a Klein four-group. It is not normal in $G^{\prime }$. Indeed,For $x\ne 0$ the off-diagonal entries are nonzero, whereas every member of $H$ is diagonal. Thus this conjugate of $T$ is outside $H$, givingThe normal subgroup in part (c) is the continuous boost subgroup; choosing one representative from each of its cosets does not make those representatives a normal subgroup.

Write $G=\{e,a,b,c\}$. By Lagrange's theorem, the order of a group element divides four. Since no element has order four, all three nonidentity elements have order two. Thus every element is its own inverse. For any $x,y\in G$,so the group is Abelian. Also $ab$ cannot equal $e,a$ or $b$, by cancellation and the fact that $a,b$ are distinct nonidentity elements. Hence $ab=c$. The remaining products are forced, givingThe map $e\mapsto (0,0)$, $a\mapsto (1,0)$, $b\mapsto (0,1)$, $c\mapsto (1,1)$ preserves this table under componentwise addition modulo two. Thereforethe Klein four-group, also called the Vierergruppe.

The complete subgroup list isThus there are ten subgroups: one trivial, five of order two, three of order four, and the whole group. The five order-two subgroups are mutually isomorphic. The two noncyclic order-four subgroups are mutually isomorphic Klein four-groups; neither is isomorphic to the cyclic group of order four.

For completeness, Lagrange's theorem permits only orders $1,2,4,8$. An order-four subgroup containing $U$ or $U^3$ must be $⟨U⟩$. Otherwise it can contain only $I,U^2$ and elements $U^{j}V$. The product of two distinct such elements is $U^{j-k}$. To avoid generating $U$ or $U^3$, their exponents must differ by two, giving exactly the two listed Klein four-groups. The order-two subgroups come from the five involutions $U^2$ and $U^{j}V$. This proves completeness.

LetThe product of any two distinct nonidentity elements is the third, and each squares to $e$, so $K$ is a Klein four-group. Conjugating a double transposition simply relabels its letters, keeping it in the set. Hence $K$ is a normal subgroup of $A_4$.

Choose $r=(123)$. The cosets $K,rK,r^{2}K$ are distinct and cover the twelve elements. With $\zeta =e^{2\pi i/3}$, defineThe coset decomposition makes this well-defined. Normality givesso $\Phi$ preserves products. It is onto $\{1,\zeta ,{\zeta }^2\}\cong C_3$, andThis is the tetrahedral quotient onto the cyclic group of order three. In terms of (a), it sends $C_0\cup C_2$ to $1$, $C_{+}$ to $\zeta$, and $C_{-}$ to ${\zeta }^2$. The normality check above also follows abstractly because a group homomorphism kernel is normal.

The requested two-dimensional irreducible representation does not exist for the Vierergruppe. The Klein four-group $V_4$ is an abelian group in which every nonidentity element has order two. Its complex irreducible representations are all one-dimensional. The same holds over the real numbers here: the commuting representing involutions are simultaneously diagonalizable with real eigenvalues $\pm 1$, so every two-dimensional representation has an invariant line. Each genuine irreducible character satisfies $\chi (g^2)=\chi (1)=1$, and hence has Frobenius-Schur indicator $\nu =1$.

The printed notation $C_4$ suggests a different possible interpretation, which must not be conflated with the Klein four-group. For the cyclic group of order four, a real two-dimensional irreducible representation is generated byIts character on $1,J,-I,-J$ is $(2,0,-2,0)$, giving $\nu =(2-2+2-2)/4=0$. Over the complex numbers this representation splits into two conjugate one-dimensional characters, so it is not a two-dimensional complex irreducible representation. These alternatives explain what can be computed without silently changing the printed group or the ground field.

For the symmetric group representation in part (a), the identity and three transpositions contribute $4I$ to $A={\sum }_{g}D(g^2)$. The two 3-cycles contribute $D(r)+D(r^2)=-I$. ThusMore generally, the central average of representation matrices at squares commutes with the representation, because conjugation permutes the summands. For a complex irreducible representation, Schur's lemma therefore gives $A=cI$. Taking the trace yields $c\dim D=|G|\nu$, proving the claimed formula under its correct hypotheses.

For the Klein four-group there is no two-dimensional irreducible representation, as established in part (b). Its genuine one-dimensional irreducibles have $A=4$ and $\nu =1$, so $A=(4\cdot 1/1)I$.

If the intended alternative is the real quarter-turn representation of the cyclic group of order four, its squared matrices sum to $I-I+I-I=0$. Thus $A=0=(4\cdot 0/2)I$ in that example. This equality is a direct calculation; the complex version of Schur's lemma cannot be applied to its reducible complexification.

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.

In the Klein four-group $V=\{1,a,b,ab\}$, the chainsare distinct composition series. All their composition factors are $C_2$.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same composition factors up to permutation and group isomorphism. Thus the subgroup chains may differ, while the multiset of simple quotients is intrinsic.

The cross-ratio permutation orbit is generated by the two operations $\lambda \mapsto 1/\lambda$ and $\lambda \mapsto 1-\lambda$ from part (d). The full list isThese are all the possibilities: the four permutations generated by interchanging both pairs and by exchanging the two pairs leave the cross-ratio unchanged. They form the Klein four-group inside the symmetric group on four symbols, leaving at most $24/4=6$ different values; the displayed operations attain all six rational expressions.

There are fewer than six values exactly when a nonidentity operation in this six-element action fixes $\lambda$: if two operations give the same value, composing one with the inverse of the other gives such a fixed point. For real $\lambda \ne 0,1$, the fixed-point equations areThe other two equations reduce to ${\lambda }^2-\lambda +1=0$, which has no real root. HenceEach exceptional value has the same three-element orbit $\{-1,\frac{1}{2},2\}$. Every other permitted real value gives six different values.

For the geometric interpretation, put $P_0=[1:0]=\infty$ and $P_i=[x_i:1]$ for $i>0$. ThenThe three exceptional cases becomeThus the three finite points are collinear and equally spaced, with one their midpoint, even when the $x_i$ are viewed as points of the complex plane. Equivalently they form an arithmetic progression in some order. These are the harmonic conjugate points configurations with the fourth point equal to the point at infinity of the complex projective line.

Let $A=\mathrm{diag}(1,-1)$ and $B=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. They generate the dihedral group of a square, realized by $\{\pm I,\pm A,\pm B,\pm AB\}$. Its image in the projective linear group has four elements, since opposite matrices become equal. The projective classes of $A,B$ commute because $AB=-BA$, and each squares to the identity. Sending $[A^a{B}^b]$ to $(a,b)$ gives the Klein four-group. This is an example in which a nontrivial scalar commutator disappears after passing to a projective representation.

In the quaternion group, $-1=i^2=j^2=k^2$ commutes with each generator because any element commutes with its powers. Hence $N=\{1,-1\}$ is central and is a normal subgroup. The reverse products satisfy $ji=-k$, $kj=-i$, $ik=-j$, so none of $\pm i,\pm j,\pm k$ is central. Therefore $Z(Q_8)=N$.

The four cosets are $N,iN,jN,kN$. Each nonidentity coset squares to $N$, and $(iN)(jN)=kN$. Thus the quotient group is the Klein four-group. Explicitly, mapping $\pm 1$ to its identity and mapping $\pm i,\pm j,\pm k$ to its three different nonidentity elements is a surjective group homomorphism with group kernel $N$. The first isomorphism theorem gives the title result.

The rotations in three dimensions preserving a dodecahedron form a group of order $60$. There are $24$ nonidentity rotations about the six axes through opposite face centres, $20$ about the ten axes through opposite vertices, and $15$ half-turns about the fifteen axes through opposite edge midpoints. Its nonidentity conjugacy classes have sizes $12,12,20,15$: the centralizer of a rotation of order five or three consists of rotations about its oriented axis and has order five or three, respectively, while all edge half-turns are conjugate. A normal subgroup is a union of whole conjugacy classes containing the identity. No proper nontrivial sum of these class sizes plus one divides $60$, proving that it is a simple group. The centralizers of the fifteen involutions are five distinct Klein four-groups, each containing three involutions. Conjugation permutes these five subgroups nontrivially. The kernel is a normal subgroup, so simplicity makes this nontrivial action injective. Its sign map to a group of order two must be trivial, since simplicity would otherwise force an impossible injection of sixty elements into two. Thus its image lies in the alternating group $A_5$, and equality of orders makes the image all of $A_5$.

The full symmetry group of a regular tetrahedron is the symmetric group $S_4$ acting on its vertices. Its subgroups with at most six elements have eight types up to conjugation, with the following representative subgroups and counts:A transposition is a reflection through the plane containing the two fixed vertices and the midpoint of the exchanged edge. A double transposition is a half-turn about the axis joining the midpoints of the two exchanged edges. A three-cycle is a one-third-turn about a vertex-to-opposite-face axis. A four-cycle is a quarter-turn composed with a reflection perpendicular to the opposite-edge midpoint axis; it is an improper orthogonal transformation.

The unique Klein four-group containing only double transpositions consists of the three opposite-edge half-turns and the identity element. The other three Klein four-groups consist of two perpendicular mirror reflections, their half-turn product, and the identity element. Each $S_3$ is a stabilizer subgroup of a vertex, including the three rotations and three mirror reflections preserving that vertex.

For completeness, a Klein four-group containing $(12)$ must contain the commuting involution $(34)$ or $(12)(34)$, yielding the displayed subgroup; one containing no transpositions must contain all three double transpositions. An $S_3$ subgroup has a unique cyclic subgroup of cardinality three, whose unique fixed vertex must therefore be preserved by the entire $S_3$. These observations and the disjoint cycle decomposition exhaust the types allowed by groups of order at most six. The counts sum to $25$.

Let a separable polynomial $f$ of degree four be irreducible over a field $F$. If all three pair sums $r_1={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$, $r_2={\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4$, and $r_3={\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3$ lie in $F$, its Galois group is the Klein four-group. Indeed $r_1-r_2=({\alpha }_1-{\alpha }_4)({\alpha }_2-{\alpha }_3)\ne 0$, and similarly the other pair sums differ. Every Galois automorphism therefore fixes each of the three partitions into pairs. The kernel of this action of $S_4$ on three partitions is $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Irreducibility gives a transitive group action on the four roots, so the orbit-stabilizer theorem forces $4$ to divide the group order. Hence the Galois group equals $V_4$. The argument also works in characteristic two.

With the generators $c,a$ of the direct product of cyclic groups of orders four and two, the subgroups are the trivial group, the whole group, the three order-two groups $⟨c^2⟩$, $⟨a⟩$, $⟨c^{2}a⟩$, and the three order-four groups $⟨c⟩$, $⟨ca⟩$, $⟨c^2,a⟩$. These are exhaustive by Lagrange's theorem: a proper nontrivial subgroup has order two or four; an order-four subgroup either contains an element of order four and is one of the two cyclic examples, or is precisely the four-element set killed by squaring. The latter is a Klein four-group.