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A mixed spreadsheet reference fixes exactly one coordinate. $A2 fixes column A and lets the row follow the copy; B$1 fixes row 1 and lets the column follow the copy. Put $2,3,4$ in A2:A4 and $5,6,7$ in B1:D1. Store =$A2*B$1 in B2, then fill the rectangle B2:D4. The colon denotes the inclusive rectangular range between the named corner spreadsheet cells. Each spreadsheet formula reads its row's factor from column A and its column's factor from row 1, producing a multiplication table.

Copying from B2 to D4 moves two columns right and two rows down. The mixed spreadsheet references become $A4 and D$1, so D4 contains =$A4*D$1 and has value $4\cdot 7=28$. For comparison, under that same displacement A2 becomes C4, $A$2 remains $A$2, $A2 becomes $A4, and A$2 becomes C$2. These four cases express the independent choice for each coordinate. Microsoft's official table demonstrates the same four possibilities for a two-row, two-column copy. ^[ref]

The filled rows have values $(10,12,14)$, $(15,18,21)$, and $(20,24,28)$. The multiplication table works because the left factor moves only vertically and the top factor moves only horizontally. Without the anchors, copying =A2*B1 from B2 to D4 would give =C4*D3, reading interior results rather than the intended boundary inputs.

// Target: computer-science.bigb

A multiplication table lists the products of the factors labeling its rows and columns. Each cell is determined by one multiplication; the same arrangement can serve as a spreadsheet tabular model.

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.

Definitions and labels. A complex group representation on a vector space $V$ is a group homomorphism $\rho :G\to \mathrm{GL}(V)$, so $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. It is an irreducible representation if $V\ne 0$ and its only invariant subspaces are $0$ and $V$.

Read permutation matrices by their action on column basis vectors. The labels in the original PDF areIn particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.

Multiplication and conjugacy. Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete multiplication table, with row factor multiplied by column factor, isConjugating a permutation relabels its cycles. Thus the conjugacy classes areof sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the symmetric group on three letters.

Dimensions. The sum of squares of irreducible representation dimensions equals the group order, which here is six. A three-dimensional irreducible representation would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex irreducible representations, one for each conjugacy class, including the trivial representation of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. HenceOne can also see reducibility directly: every permutation matrix fixes $(1,1,1{)}^T$.

Characters. The character of a representation is the trace of its representing matrix, constant on each conjugacy class. The given permutation representation has character equal to the number of fixed basis vectors:in class order $(I;A,B;C,D,E)$. A trivial representation has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm 1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. Character orthogonality with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the sign representation without constructing its matrices.

Write the two-dimensional character as $(2,x,y)$. Character orthogonality with the first two rows givesso $x=-1$, $y=0$. Its squared norm is $(4+2)/6=1$, as required. The complete character table is thereforeThe last row belongs to the standard representation of the symmetric group.

To decompose the given representation, use the character inner productIts multiplicities areThusConcretely the invariant line is spanned by $(1,1,1{)}^T$ and its invariant complement is $\{(x,y,z):x+y+z=0\}$. Their characters are exactly the first and third rows, agreeing with the decomposition obtained entirely from character orthogonality.

Write $r=g_1$. The matrix $D(s)$ is a reflection by part (a). Reflection reverses a rotation angle, giving $D(s)D(r)D(s)=D(r{)}^{-1}$. Faithfulness lifts this matrix identity back to the group:Moving each $s$ past a rotation reduces every word to $r^i$ or $r^{i}s$, $0\le i<p$. These $2p$ elements are distinct: the rotations are distinct by the order of $r$, and reflections have determinant $-1$, so cannot equal a rotation. ThereforeHere the subscript denotes group order. This is the reflection case of finite subgroups of the plane orthogonal group.

Every $r^{i}s$ has square $r^{i}s{r}^{i}s=r^i{r}^{-i}=I$, so there are $p$ reflection subgroups of order two. Within the rotations, the even-$p$ assumption from part (b) supplies the additional half-turn $r^{p/2}$. Distinct involutions generate distinct two-element subgroups. ThusFor comparison, without the evenness requirement the corresponding odd-$p$ count would be $p$. The even case agrees with involutions in an even dihedral group.

For $p=4$, takeDirect multiplication gives $SRS=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=R^3$, so $srs=r^3=g_3$. The product rules arewith exponents modulo four. They yield the complete multiplication table, where the row element multiplies the column element on the left:This is the symmetry group of a square, the dihedral group of order eight, often written $D_4$ when its subscript counts vertices instead. Its five order-two subgroups are generated by $r^2,s,rs,r^{2}s,r^{3}s$.

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 cyclic subgroup $A=⟨a⟩$ has order four and index two. Every index-two subgroup is normal: its complement is both the only nontrivial left coset and the only nontrivial right coset. Thus conjugation by $b$ is an automorphism of $A$. It must send its generator $a$ to another element of order four, namely $a$ or $a^3$. Since $b^{-1}=b$,equivalently $ab=ba$ or $ab=b{a}^3$, as required. Also $b\notin A$, since otherwise $a,b$ could generate at most four elements.

In the commuting case, every word reduces uniquely to $a^j{b}^{\epsilon }$ with $0\le j<4$ and $\epsilon \in \{0,1\}$. The eight expressions are distinct because $A$ and $Ab$ are disjoint. Multiplication is addition of the two exponents modulo four and two, respectively. ThereforeThis gives the multiplication table independently of any other properties of $G$, proving uniqueness up to group isomorphism.

In the inversion case, the relation $ba=a^{-1}b$ gives the same eight unique normal forms, now with multiplicationwith exponents reduced modulo four and two. Thus this case also has a uniquely determined group structure. A faithful geometric realization is the dihedral group of a square, with $a$ a quarter-turn and $b$ a reflection in a line through opposite edge midpoints. For example, on the plane,These preserve the square with vertices $(\pm 1,\pm 1)$ and satisfy $a^4=b^2=I$, $bab=a^{-1}$. They generate eight distinct transformations, so

The products in (a), together with $(-A)B=A(-B)=-(AB)$ and $(-A)(-B)=AB$, give closure of the eight-element set. Associativity is inherited from matrix multiplication, $W$ is the identity element, and the inverses from (a), with $(-A{)}^{-1}=-A^{-1}$, remain in the set. Thus the group axioms hold. This is the quaternion group $Q_8$.

In the element order specified in the question, the multiplication table isInterpreting “character of the group” as the character of a representation specified by these matrices, take the trace of each matrix:A character belongs to a representation; the underlying abstract group can have other characters.

Label the triangle's vertices counterclockwise as $1,2,3$. Let $r$ rotate it by $2\pi /3$ and let $s$ reflect it in the axis through vertex 1 and the midpoint of the opposite side. ThenThe six dihedral group elements areThe first three are the identity and the rotations by $120^{\circ },240^{\circ }$; the last three are reflections fixing vertices $1,3,2$, respectively. These are all possible symmetries: the image of one vertex has three choices, and the order of the other two has two choices.

Use rightmost-first composition. From the generator relation,with exponents reduced modulo three and two. The row-times-column multiplication table isThus every symmetry is expressed in the two group generators $r,s$.

The relations in (a) reduce every word in the generators to $U^a{V}^b$, with $0\le a<4$ and $b\in \{0,1\}$. These eight matrices are distinct: the $b=0$ elements have third diagonal entry $1$ and the $b=1$ elements have entry $-1$, while the four planar blocks are distinct in each family. Their multiplication iswith exponents reduced modulo four and two. This proves closure. The identity element is $U^0{V}^0$, the inverse of $U^a$ is $U^{-a}$, and each $U^{a}V$ is its own inverse. Associativity is inherited from matrix multiplication, proving the group axioms. This is the eight-element dihedral group $D_8$.

Set $g_j=U^j$ and $g_{j+4}=U^{j}V$ for $j=0,1,2,3$, consistently with the two labels specified in the question. The row-times-column multiplication table isThe rotation powers commute with one another, and $U^a$ commutes with $U^{c}V$ exactly when $a$ is even. Two elements $U^{a}V,U^{c}V$ commute exactly when $a-c$ is even. These statements also check the symmetric portions of the table.

Here $a^2=-I$, $a^3=-a$ and $a^4=I$, so $K$ is a cyclic group of order four. Its multiplication table isWrite the specified image of $a$ as $A$. Direct matrix multiplication givesA group representation is a group homomorphism to a general linear group, so the images of all powers of $a$ are forced:All four images are invertible. Defining $\Phi (a^j)=A^j$ is well-defined modulo four because $A^4=I_2$, and $A^{j+\ell }=A^j{A}^{\ell }$ verifies every product in the table. These observations prove both existence and uniqueness of this representation with the prescribed image.

By (d), every group of group order three is a cyclic group. Writing its elements as $e,a,a^2$, every product is determined by $a^3=e$; the multiplication table isThe map $a^j\mapsto j(\mathrm{mod}3)$ is a bijective group homomorphism to addition modulo three. Thereforeso there is exactly one such group up to isomorphism.