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When $[G:H]=2$, there are exactly two left cosets and two right cosets. For $g\notin H$, both $gH$ and $Hg$ are the complement $G\setminus H$; for $g\in H$, both equal $H$. This is the index-two subgroup is normal argument: $gH=Hg$ for every $g$, which is equivalent to $gH{g}^{-1}=H$. Therefore every subgroup of index two is normal.

In the dihedral group $D_{2n}=⟨r,s⟩$, the rotation cyclic subgroup $⟨r⟩$ has order $n$ and index two, so it is a normal subgroup. It is nontrivial and proper for the usual dihedral convention $n\ge 2$ (polygon symmetries have $n\ge 3$). If the degenerate convention $D_2=C_2$ were allowed, the claimed nontrivial proper subgroup would not exist; the rotation argument uses $n>1$.

For any $k\ge 3$, let $G=S_k$ and take the stabilizer subgroup $H$ of the letter $1$. Its elements freely permute the other $k-1$ letters, soConjugation by $(12)$ sends $H$ to the stabilizer of $2$, which is different: for instance $(23)$ belongs to $H$ but does not fix $2$. Thusgives a nonnormal subgroup of every prescribed index $k\ge 3$.

For a finite group $G$ and a subgroup $H$, Lagrange's theorem statesIn particular $|H|$ divides $|G|$: the left cosets partition $G$, and multiplication by a representative is a bijection from $H$ to each coset. Apply this to the cyclic subgroup $⟨g⟩$. Its size is the order of a group element $g$, so every element order divides $|G|$.

Under the square condition, every group element is its own inverse element. Consequentlyfor all $g,h$. This proves that a group of exponent two is abelian, including the trivial group.

For the fourth-power condition a counterexample to commutativity is the quaternion groupHere $i^2=j^2=k^2=-1$ and $ij=k=-ji$. Thus its six elements outside $\{\pm 1\}$ have order of a group element four, while $1$ and $-1$ have orders one and two. All fourth powers are the identity, yet $ij\ne ji$.

A permutation cycle $(a_1{a}_2\cdots a_p)$ sends $a_1$ to $a_2$, each successive listed point to the next, and $a_p$ to $a_1$, fixing every unlisted point. Its length is $p$. For any permutation $h$,Thus conjugate group elements in the symmetric group have the same cycle length. Conversely, two cycles of the same length are conjugate: send their entries, in cyclic order, to one another and extend this bijection arbitrarily to the unused points. Length-one cycles represent the identity and cause no exception to this conclusion.

An odd-length permutation cycle is an even permutation, since its sign is $(-1{)}^{p-1}$. Let $c,d$ be such cycles on $p+2$ points and choose $h\in S_{p+2}$ with $hc{h}^{-1}=d$. If $h$ is odd, let $t$ be the transposition of the two points unused by the listed cycle $c$. Then $tc=ct$, and $ht$ is even withHence all the $p$-cycles are conjugate within $A_{p+2}$. The two spare letters supply the parity correction; they are the reason the same argument need not work with only one spare letter.

In $A_4$ the answer is no. For example, $(123)$ and $(132)$ are conjugated by the odd transposition $(23)$ in $S_4$. The centralizer of $(123)$ in $S_4$ is exactly its order-three cyclic subgroup: a commuting permutation fixes the unique fixed point $4$ and acts as a power of the cycle on the other three points. All these centralizing elements are even. Every other conjugating element differs from $(23)$ by a centralizing element, so it is also odd. There is no conjugator in $A_4$. Equivalently, the eight three-cycles split into two conjugacy classes of size $12/3=4$, illustrating the alternating conjugacy class splitting criterion.

Use the standard implicit assumption that $p$ is prime, so ${\mathbb{F}}_p$ is a finite field. The matrices in question form $G{L}_2({\mathbb{F}}_p)$, the general linear group over a finite field. Products stay in the set because determinants multiply, the identity has determinant one, andhas entries in the finite field and nonzero determinant. Together with associativity this verifies the group axioms. The multiplication of matrices with column vectors is a group action because $Iv=v$ and $(gh)v=g(hv)$.

Let $C=⟨g⟩$ be the order-$p$ cyclic subgroup. By the orbit-stabilizer theorem, its group orbits on ${\mathbb{F}}_p^2$ have size either one or $p$. If $f$ is the number of fixed vectors, counting all $p^2$ vectors gives $f\equiv p^2\equiv 0(\mathrm{mod}p)$. Since the zero vector is fixed, $f\ge 1$ and hence $f\ge p$. There is a nonzero fixed vector $v$.

Extend $v$ to a basis $(v,w)$. In this basis $g$ has matrix $\left(\begin{array}{cc}1& b\\ 0& d\end{array}\right)$. From $g^p=I$ we get $d^p=1$; in the prime finite field, $d^p=d$, so $d=1$. Since $g$ has order $p>1$, it is not the identity, and $b\ne 0$. Replace $w$ by $b^{-1}w$. Then $g(b^{-1}w)=v+b^{-1}w$, and in the new basisThus all the order-p matrices in GL2 over the prime field form one conjugacy class. This proof also works at $p=2$. Primality is essential to the first assertion: for modulus four, $\mathrm{diag}(2,1)$ has nonzero determinant but no inverse, so the set defined using merely nonzero determinants would not be a group.

Suppose ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$, with $ad-bc=1$, and put $\alpha =(c{\tau }_2+d{)}^{-1}$. The identitiesshow that multiplication by $\alpha$ takes one period lattice isomorphically to the other: its basis matrix has determinant $ad-bc=1$. If $N\mid c$, the image of $1/N$ is congruent to $a/N$ modulo the target period lattice. Moreover $ad\equiv 1(\mathrm{mod}N)$, so $a$ is a unit modulo $N$. The image therefore generates the same cyclic subgroup.

Conversely, a holomorphic group isomorphism of complex tori lifts to $z\mapsto \alpha z$, with $\alpha {\Lambda }_{{\tau }_2}={\Lambda }_{{\tau }_1}$. Indeed its lift fixing zero is additive: its failure to be additive is a continuous lattice-valued function, hence zero. An additive holomorphic map is complex linear. WriteThe corresponding integral basis matrix has determinant $1$: complex multiplication preserves the real orientation, and both bases $(1,{\tau }_j)$ are positively oriented. Solving gives ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$. Preservation of the cyclic subgroup means $\alpha /N\equiv u/N(\mathrm{mod}{\Lambda }_{{\tau }_1})$ for some unit $u$ modulo $N$. Comparing the two real basis coefficients gives $N\mid c$. Thus the matrix lies in the Gamma 0 congruence subgroup, and the classification isThis is the noncompact modular curve classification; no cusp points have been added.

Let $E=\mathbb{C}/\Lambda$, and let ${\Lambda }^{\prime }$ be the inverse image of the cyclic subgroup $G$ under the quotient map. Then $\Lambda \subset {\Lambda }^{\prime }$ is an inclusion of Euclidean lattices with ${\Lambda }^{\prime }/\Lambda \cong \mathbb{Z}/N\mathbb{Z}$. The Smith normal form supplies a basis $u,v$ of ${\Lambda }^{\prime }$ for whichChoose the orientation of this basis so that $\mathrm{Im}(v/(Nu))>0$. Dividing the coordinate on $\mathbb{C}$ by $Nu$ gives an isomorphism of complex toriHere $u$ maps to $1/N$, and its class generates ${\Lambda }^{\prime }/\Lambda$. Thus the isomorphism carries $G$ to the required cyclic subgroup. Every pair is represented: $\Phi$ is surjective. This argument avoids assuming that an arbitrary integer lift of a primitive vector modulo $N$ is itself primitive.

For a nontrivial finite p-group $G$ of order $p^m$ with $m\ge 1$, partition into conjugacy classes. A noncentral class has size $[G:C_G(x)]$, a positive power of $p$ larger than one by Lagrange's theorem. The class equation therefore saysThe center of a group contains the identity, so $|Z(G)|\ge p$, proving its nontriviality. The positive-exponent qualification matters: the one-element group has no nonidentity central element.

If $|G|=p^2$, its center of a group has order $p$ or $p^2$. In the first case, the quotient group $G/Z(G)$ has prime order and is cyclic. Whenever a central quotient group is cyclic, writing all elements as $g^{i}z$ with $z$ central shows that they commute: $(g^{i}z)(g^{j}w)=g^{i+j}zw=(g^{j}w)(g^{i}z)$. Thus that case would already make $G$ Abelian and its center of a group all of $G$, a contradiction. Hence every group of order $p^2$ is Abelian.

If there is an element of order $p^2$, it is a generator of a group for $G$, giving $C_{p^2}$. Otherwise every nonidentity element has order $p$. Pick $a\ne 1$ and $b\notin ⟨a⟩$. Their cyclic subgroups have trivial intersection, and they commute; the $p^2$ distinct products $a^i{b}^j$ exhaust $G$. Hence the classification of groups of order p squared isBoth groups exist and are nonisomorphic, since only the first has an element of order $p^2$.

Every finite cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function: if $g$ has order $m$, then $g^k$ generates precisely when $\gcd (k,m)=1$. Each element generates exactly one cyclic subgroup. Thus counting cyclic subgroups by their generators amounts to counting elements by order of a group element and dividing by $\phi (m)$.

Use the cycle decomposition of a permutation in $S_5$. A permutation with $a_j$ cycles of length $j$ occurs $5!/{\prod }_j(j^{a_j}{a}_j!)$ times: order the five letters, then divide by the cyclic choices of starting point and the orderings of equal-length cycles. Its order is the least common multiple of the cycle lengths. The resulting count isThe two subgroups $⟨(12)⟩$ and $⟨(12)(34)⟩$ are both isomorphic to $C_2$ by a group isomorphism. They are not conjugate subgroups: conjugating their unique nonidentity elements would have to preserve cycle type, whereas a transposition and a product of two disjoint transpositions have different cycle types.

Lagrange's theorem states that, for a finite group $G$ and a subgroup $H$,To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left cosets $xH$, which therefore partition $G$. Left multiplication by $x$ gives a bijection $H\to xH$, so every coset has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the cyclic subgroup generated by an element shows that its order of a group element divides $|G|$.

Write the given order-two normal subgroup as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gz{g}^{-1}=z$ for all $g\in G$. Thus $z$ is in the center of a group: a normal subgroup of order two is central.

Choose $g\notin N$. By Lagrange's theorem, its order of a group element is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a subgroup of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz{)}^k=1$ implies $g^k=z^{-k}\in ⟨g⟩\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. In either case

Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

Lagrange's theorem for finite groups states that if $H\le G$ and $G$ is finite, thenso $|H|$ divides $|G|$. Indeed, the left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ to each coset.

Applying the theorem to the cyclic subgroup $⟨g⟩$ gives

In $C_3\times C_9$, an element has order dividing three exactly when its first coordinate is arbitrary and its second belongs to the unique subgroup of order three. There are $3\cdot 3=9$ such elements, and after removing the identity the answer is $\boxed{8}$.

Let $\pi :G\to G/[G,G]$ be the abelianization map. Its image $\pi (A)$ is again a $K$-approximate group. Apply the large-progression form of the Freiman-Green-Ruzsa theorem to $\pi (A)$. It gives a finite subgroup $H$, elements $x_1,\dots ,x_r$, and lengths $L_1,\dots ,L_r$, withand

The large lifted product from a coset progression applied to this progression givesBriefly, choose a section of $\pi$ on $\pi (A^6)$. Multiplication by that section is multiplicative up to $A^{12}\cap [G,G]$; lifting successively the subgroup part and each progression direction therefore places every element of $A^6\cap {\pi }^{-1}(HP)$ in the displayed product. The fiber-counting lemma for a quotient map gives $|A^6\cap {\pi }^{-1}(HP)|\ge \exp (-K^{O(1)})|A|$, which proves the estimate.

SetThe intersection of an approximate group power with a subgroup shows that each $A_i$ is a $K^{O(1)}$-approximate group contained in $A^{O(1)}$. The preimage of a cyclic subgroup of $G/[G,G]$ has step less than $s$. The same is true of the preimage of the finite subgroup $H$ because $G$ is a torsion-free group. Consequently each $⟨A_i⟩$ has step less than $s$, and the displayed estimate is the required conclusion.

Suppose $x,y$ have order three but $x\ne y,y^2$. Their cyclic subgroups have trivial intersection: any common nonidentity element would generate both subgroups and force $x=y$ or $y^2$. The nine products $x^i{y}^j$, $0\le i,j<3$, are distinct. Indeed $x^i{y}^j=x^k{y}^{\ell }$ implies $x^{i-k}=y^{\ell -j}$ in that intersection, so $i=k$ and $j=\ell$ modulo three. Nine distinct elements cannot fit in a group of order six, proving the claim without assuming commutativity.

A group has a closed binary operation, associativity, an identity, and an inverse for each element. For a finite group, Lagrange's theorem states $|G|=[G:H]|H|$ for each subgroup $H$, so $|H|$ divides $|G|$.

The order of a group element $g$ is the least positive $n$ with $g^n=e$, or infinity if none exists. In a finite group the powers form the cyclic subgroup $⟨g⟩$ of that order; Lagrange's theorem proves $\mathrm{ord}(g)\mid |G|$. If $|G|=p$ is prime and $g\ne e$, its order must be $p$. Thus $g$ generates the whole group, proving $G\cong C_p$.

If every nonidentity element has order two, every element is its own inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. This proves that a group of exponent two is abelian.

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 order of a group element $g$ is the least positive integer $q$ such that $g^q=e$, or infinity if there is no such integer. In a finite group such a $q$ exists, because two powers among $e,g,g^2,\dots$ must coincide.

The powers $e,g,\dots ,g^{q-1}$ are distinct: a repetition with exponents differing by less than $q$ would contradict its minimality. Division of any exponent by $q$ shows that these are all the elements of the cyclic subgroup $⟨g⟩$. Thus $|⟨g⟩|=q$. Applying Lagrange's theorem givesIn particular the identity has order one, and every nonidentity element has order greater than one. The divisibility assertion here concerns finite group order.

For the intended finite nontrivial case, let $G=⟨g⟩$ have composite order $N$. Choose a prime divisor $p$ of $N$, with $1<p<N$. Thenis a nontrivial proper cyclic subgroup of order $p$. Indeed $(g^{N/p}{)}^p=e$, and no smaller positive power is the identity because $g$ has order $N$. Thus $1<|H|=p<N$.

If the source also allows an infinite cyclic group, $⟨g^2⟩$ is a proper cyclic subgroup: its powers have even exponents and do not include $g$. The trivial group, of order one, is an exception to the literal formulation “not of prime order”: it has no proper subgroup at all. The nontriviality assumption is therefore necessary for that general wording. These facts are instances of subgroups and quotients of a cyclic group.

For every positive integer $m$, the group laws give $(g^{-1}{)}^m=(g^m{)}^{-1}$. ConsequentlyThe sets of positive exponents returning the two elements to the identity element are identical. If this set is nonempty, its least member is the order of a group element in both cases. If it is empty, neither element has finite order. Therefore, including the infinite-order case,For a direct divisibility argument in the finite case, if $g$ has order $n$, then $(g^{-1}{)}^n=e$, so its inverse’s order divides $n$. Reversing the roles shows $n$ divides that order, forcing equality. This is the general fact that an element and its inverse have the same order; it also reflects $⟨g⟩=⟨g^{-1}⟩$ as cyclic subgroups.

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