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For any integer $n\ge 1$, let $R=R_{2\pi /n}$ in the stabilizer subgroup from i. Since $R_{\theta }{R}_{\varphi }=R_{\theta +\varphi }$ and $R_{\theta }=I$ exactly when $\theta \in 2\pi \mathbb{Z}$, this rotation matrix has order exactly $n$. Thereforeis a subgroup of $G$ of order $n$, including the trivial case $n=1$.

For infinitely many non-Abelian examples, let $n\ge 3$, retain $R=R_{2\pi /n}$, and take the half-turnDirect multiplication gives $J^2=I$ and $JRJ=R^{-1}$. Every word in $R,J$ therefore reduces to $R^k$ or $R^{k}J$, $0\le k<n$. The $R^k$ are distinct, as are the $R^{k}J$, and no element from the first family equals one from the second: they send $N$ to $N$ and $-N$, respectively. Hence $⟨R,J⟩$ has exactly $2n$ elements and is a dihedral group.

It is non-Abelian because $R\ne R^{-1}$ for $n\ge 3$, so $JR\ne RJ$. Different values of $n$ give different group orders, and therefore different subgroups. Thus $G$ has infinitely many non-Abelian subgroups, already among its finite subgroups.

Conjugacy classes. For $a\in G$, its conjugacy class isEvery element belongs to its own conjugacy class, by taking $g=e$. To prove uniqueness, suppose $x=ga{g}^{-1}=hb{h}^{-1}$ belongs to both $\mathrm{Cl}(a)$ and $\mathrm{Cl}(b)$. ThenFor every $t\in G$, this makes $tb{t}^{-1}$ a conjugate of $a$ as well. Hence $\mathrm{Cl}(b)\subseteq \mathrm{Cl}(a)$, and reversing the argument gives the other inclusion. Two conjugacy classes that meet are therefore equal, proving that the classes partition $G$.

Abelian case. If $G$ is an abelian group, then $ga{g}^{-1}=ag{g}^{-1}=a$ for every $g$, so $\boxed{\mathrm{Cl}(a)=\{a\}}$ for each $a$.

Centre. An element has a singleton conjugacy class precisely when $gz{g}^{-1}=z$ for every $g$, equivalently $gz=zg$ for every $g$. Thus the definition in the paper agrees with the usual centre of a group:The identity lies in $Z(G)$. If $z,w\in Z(G)$, then for any $g$, $gzw=zgw=zwg$, so $zw\in Z(G)$. Also $gz=zg$ implies $z^{-1}g=g{z}^{-1}$, so $z^{-1}\in Z(G)$. This proves it is a subgroup. Any two of its elements commute because either one commutes with every element of $G$. Consequently $\boxed{Z(G)\text{ is an abelian subgroup of }G}$.

Square symmetry. Use the paper's notation $D_4$ for the eight-element dihedral group of a square. Let $r$ be a quarter-turn and $s$ a reflection, so $r^4=s^2=e$ and $srs=r^{-1}$. A rotation $r^j$ commutes with $s$ exactly when $r^j=r^{-j}$, or $2j=0$ modulo four. This selects $e$ and $r^2$. No reflection $r^{j}s$ commutes with $r$, becauseand $r^{j-1}\ne r^{j+1}$. Therefore

Let $r$ be a quarter-turn and $s$ a reflection of the square. The dihedral group is generated by $r,s$ with $r^4=s^2=1$ and $srs=r^{-1}$, and its elements are $1,r,r^2,r^3,s,sr,s{r}^2,s{r}^3$. Two generators are minimal because this group is not cyclic. Its subgroups of order two are $⟨r^2⟩$ and $⟨s{r}^j⟩$ for $j=0,1,2,3$. Its subgroups of order four are $⟨r⟩$, $⟨r^2,s⟩$ and $⟨r^2,sr⟩$.

A group homomorphism obeys $\varphi (gh)=\varphi (g)\varphi (h)$ for all elements; a group isomorphism is a bijective group homomorphism. Take the rotation subgroup $⟨r⟩$ and defineThe quaternion element $i$ has order four, so the map is well defined modulo four, preserves products and is injective onto $\{1,i,-1,-i\}$. Surjectivity onto all eight quaternion elements is not required of a homomorphism.

A cyclic group is generated by the powers of one element. The final claim needs the usual qualification no nontrivial proper subgroup, and $|G|>1$: the trivial subgroup is proper in every nontrivial group, even one of prime order. With that interpretation, if $G=⟨g⟩$ has prime order $p$, Lagrange's theorem permits only subgroup orders one and $p$. Conversely, if its order $n>1$ is composite, choose a divisor $d$ with $1<d<n$. The subgroup $⟨g^d⟩$ has order $n/d$, strictly between one and $n$. This proves the assertion described by groups with no proper nontrivial subgroups.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

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.

Two elements are conjugate group elements if $h=xg{x}^{-1}$ for some $x\in G$. The conjugacy class of $g$ is $\{xg{x}^{-1}:x\in G\}$. Conjugacy is an equivalence relation: it is reflexive using $x=e$, symmetric by replacing $x$ with $x^{-1}$, and transitive becauseThus the classes partition the group: every element belongs to its own class, and any two classes that intersect are equal. This proves that every element belongs to exactly one class.

If $G$ is Abelian, then $xg{x}^{-1}=g$, so every class is a singleton. In general the elements whose classes are singletons are precisely those commuting with every group element. Thus the center of a group isIt contains $e$. If $a,b\in Z(G)$, then $a{b}^{-1}$ commutes with every $g$, since $b^{-1}$ does and $a$ does. The subgroup test therefore proves that $Z(G)$ is a subgroup. Any two of its elements commute with each other, so it is an abelian subgroup.

For the square centred at the origin with vertices $(\pm 1,\pm 1)$, putThe dihedral group $D_4$ consists of $R^j$ and $R^{j}S$, $j=0,1,2,3$. Its eight actions and their matrices areThese exhaust the symmetries: the image of one vertex has four choices and the image of an adjacent vertex has two choices, determining the entire square. For an explicit pair of nonconjugate elements take $R^2$ and $S$. Both have order two, but their determinants are $+1$ and $-1$. A similarity transformation preserves the determinant, so $R^2$ and $S$ are not conjugate.

To find the centre, use $SR=R^{-1}S$. A rotation $R^j$ commutes with $S$ exactly when $R^j=R^{-j}$, so $j$ is even. No reflection $R^{j}S$ commutes with $R$, since $(R^{j}S)R=R^{j-1}S$ while $R(R^{j}S)=R^{j+1}S$, and $R^2\ne I$. Therefore

The intersection $H\cap K$ contains the identity element. If $a,b\in H\cap K$, then $a{b}^{-1}$ belongs to both $H$ and $K$, so it belongs to their intersection. The subgroup test proves that $H\cap K$ is a subgroup of $G$, with the inherited operation. Explicitly, closure and inverses come from this test, and associativity is inherited from $G$.

Label the vertices of the equilateral triangle $1,2,3$ counterclockwise. For the dihedral group of order six, let $r=(123)$ be the counterclockwise rotation by $120^{\circ }$, and let $s=(23)$ be reflection in the line through vertex 1 and the opposite side's midpoint. ChooseThen $B$ rotates by $240^{\circ }$. The three Euclidean reflections $C,D,E$ fix vertices $1,3,2$ respectively. Products act on a vertex from right to left. The relations $r^3=s^2=e$ and $sr=r^{-1}s$ givewhere exponents of $r$ are reduced modulo three and those of $s$ modulo two. This constructs the entire Cayley table:For example $AC=D$ whereas $CA=E$, so the group is nonabelian.

A subgroup of order two is generated by its unique nonidentity element, which must have order two. The rotations $A,B$ have order three, and each reflection has order two. Thus there are exactlyNone is a normal subgroup. Conjugation by $A$ sends the reflection $r^{j}s$ to $r^{j+2}s$, cycling the three reflections. For example $AC{A}^{-1}=E\notin ⟨C⟩$. Each of the other two subgroups fails the same conjugation test.

Finally, take two distinct such subgroups. For instance their union is $⟨C⟩\cup ⟨D⟩=\{I,C,D\}$, but $CD=B$ lies outside it. Thus it is not closed under the inherited group operation and is not a group. The distinctness is necessary: the union of a subgroup with itself is that subgroup.

The group order $|G|$ is the number of elements of the finite group. A normal subgroup $H\le G$ satisfies $gH{g}^{-1}=H$ for every $g\in G$, equivalently its left and right cosets agree.

For the square, let $R$ be counterclockwise rotation by $\pi /2$, and $s$ reflection in the horizontal axis. The dihedral group isThe first four elements rotate by $0,90,180,270$ degrees. The other four are Euclidean reflections in the horizontal axis, the line $y=x$, the vertical axis and the line $y=-x$, respectively, for a square centred at the origin with sides parallel to the axes.

The eight nontrivial proper subgroups areTo check completeness, Lagrange's theorem allows proper orders one, two and four. The order-two subgroups are generated by the five involutions displayed. An order-four subgroup containing $R$ or $R^3$ is the rotation subgroup. Otherwise it must contain reflections; their product is a rotation, which must be $R^2$, so its reflections differ by that half-turn, giving the two listed Klein subgroups. The trivial subgroup $\{I\}$ is an additional proper subgroup. Thus the PDF's count of eight uses “proper” to exclude the trivial subgroup; with the usual inclusive definition there are nine proper subgroups.

A normal subgroup of order two is central: conjugation fixes its identity and must also fix its only other member. The half-turn $R^2$ commutes with every square symmetry, whereas the reflection subgroups are exchanged by conjugation. Hence the unique order-two normal subgroup isFor the general polygon, write $D_n=\{R^j,R^{j}s:0\le j<n\}$ with $R^n=s^2=I$ and $sRs=R^{-1}$. A rotation $R^j$ commutes with $s$ exactly when $R^j=R^{-j}$, or $2j\equiv 0(\mathrm{mod}n)$. A reflection cannot commute with $R$ when $n\ge 3$, becauseTherefore the center of a group here is $\{I\}$ for odd $n$, and $\{I,R^{n/2}\}$ for even $n$. The order-two normal subgroups correspond exactly to nonidentity central involutions, provingThis agrees with the involutions in an even dihedral group and uses the convention $|D_n|=2n$ throughout.

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.

Label the triangle vertices $1,2,3$ counterclockwise. Let $R$ rotate by $2\pi /3$ counterclockwise and let $m$ reflect across the line through vertex 1 and the midpoint of the opposite side. The dihedral group has elements $I,R,R^2,m,Rm,R^{2}m$, with $R^3=m^2=I$ and $mRm=R^{-1}$.

The action on vertices identifies them with these permutations, with the rightmost operation applied first:The minimal number of generators is two, and $\{R,m\}$ is one minimal generating set. They produce every displayed element. A one-element generating set would make the group cyclic, but every nonidentity element has order two or three, never six.

The vertex-action map preserves composition, hence is a group homomorphism from $D_3$ to the symmetric group $S_3={\Sigma }_3$. The displayed actions are all six distinct permutations of three vertices, so the map is both injective and surjective. ThereforeThis identification uses the convention $|D_3|=6$, with the subscript counting polygon vertices.

The relation $Rm=m{R}^{-1}$ implies $m{R}^j=R^{-j}m$. Move each reflection generator to the right in any word, cancel pairs using $m^2=I$, and reduce exponents modulo $N$. Every element then has a normal form $R^j$ or $R^{j}m$, $0\le j<N$.

For the actual symmetry group of a regular polygon these are $N$ distinct rotations and $N$ distinct reflections. The determinant distinguishes rotations from reflections in the planar action. Hence the source's dihedral group $D_N$ has order $2N$, with its subscript counting polygon vertices. The calculations below use that convention.

The table describes the symmetry group of a regular hexagon. Choose $R$ as a counterclockwise rotation through $\pi /3$ and $m_1$ as reflection in one symmetry axis. With $s=m_1$, the relations areThe rotation entries already are $I,R,\dots ,R^5$, and the table gives $R{m}_j=m_{j+1}$ with indices modulo six. ThereforeThe source writes $D_6$ with the subscript counting polygon vertices; this dihedral group has order 12.

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 $r^2,s,rs,r^{2}s,r^{3}s$.

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 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

No. The quaternion groupis another non-abelian group of order eight: for example $ij=k$ while $ji=-k$. Its multiplication comes from the associative quaternion algebra, so it does define a group.

It is not isomorphic to the dihedral group of the square. In $Q_8$, only $-1$ has order two; the six elements $\pm i,\pm j,\pm k$ have order four. The square's dihedral group has five elements of order two: four reflections and the half-turn. Group isomorphisms preserve element orders, so these groups are distinct. The hypothesis in part (c) excludes $Q_8$, since its only element of order two is already in each cyclic subgroup of order four.

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.

A common shift of all three angles leaves every spring length unchanged. This continuous rotational symmetry gives the zero mode: there is no restoring force against rigid rotation.

The equilateral equilibrium also has the dihedral group symmetry $D_3$. Its action on the three angular displacements preserves the uniform one-dimensional subspace and the two-dimensional sum-zero subspace. The latter is an irreducible representation over the real numbers; a reflection may reverse all angular signs as well as exchange bead labels, which does not change these invariant subspaces. The symmetric stiffness matrix must act as a scalar on the sum-zero subspace, explaining the equal vibrational angular frequencies. Directly, each sum-zero vector satisfies $(3I-11^T)\mathbf{v}=3\mathbf{v}$. This is an example of degenerate normal modes from permutation symmetry, and of normal modes of three equal masses with a symmetric gap potential.

The printed wording calls ${\Sigma }_3$ cyclic, but its meaning in question 9 is the symmetric group $S_3$. Under that interpretation, take its planar dihedral group representation, withHere $r^3=s^2=1$ and $srs=r^{-1}$. This representation is irreducible over the complex numbers: the two distinct rotation eigenspaces are exchanged by the reflection, so neither is invariant under the whole group. Its character is $2$ on the identity, $-1$ on the two 3-cycles and $0$ on the three transpositions. Squaring maps each transposition to the identity and permutes the 3-cycles. Its Frobenius-Schur indicator is thereforeIf “cyclic” were intended literally, no two-dimensional complex irreducible representation would exist, since an irreducible representation of a finite abelian group is one-dimensional.